ProofAtlas

The ProofAtlas ranking

Top 500 open problems in mathematics
ranked by LLM-assessed importance

Explore questions in mathematics, computation, and mathematical physics, with exact statements, status notes, and sources.

This is a model-based assessment of mathematical importance, not an expert consensus or a forecast of which problems will be solved next.

Resolved problems from earlier editions remain available with their original ranks and sources.

2 entries have conditional ranks. Each entry explains why.

Edition 17 subjectsHow the ranking worksDownload ranking (JSON) Changes from previous edition (JSON)
How the ranking works

What importance means here. Each entry is a specific mathematical question. Importance considers the significance of a resolution, centrality within its field, reach across fields, scholarly and historical fame, broader recognition, and scientific or practical impact.

How problems are ranked. LLMs compare pairs of problems. A reliability-weighted model combines those judgments into the ranking, with calibration across model families. The model-family weights are OpenAI 1.00, Claude 1.00, GLM 0.95, and DeepSeek 0.90. These are modeling choices, not measured probabilities of correctness.

What a rank means. Nearby ranks can be uncertain. The expanded entries include available ranking ranges and comparison counts. Filtering or alphabetizing the directory preserves every problem’s rank in this edition.

Diagnostics from the preceding source fit. After the reported resolutions, edition ranks were compacted without a new numerical fit. The displayed uncertainty ranges and comparison counts come from the preceding fit of 1,280 targets; the current eligible pool contains 1,132. These diagnostics were not recomputed for the smaller pool.

Related and nested problems. A substantial subproblem or generalization may receive its own rank when it has independent mathematical standing. Families identify narrow groups of related questions; an implication alone does not merge their families. The ranking process checks family crowding, including a diagnostic retaining the highest-ranked member of each family. Family membership does not itself change scores, impose a quota, or create an automatic penalty.

Dates and sources. The edition date identifies this list. A status review date, when recorded, appears separately inside each entry; it does not promise that the literature has been checked since then. Source links identify the exact formulation and available status evidence.

The ranking preserves a fixed edition. Separately dated status notes cover 168 of 500 entries; this is not a fresh literature review of the whole list. Recent resolution claims and formulation concerns are marked below with their historical ranks retained.

Find an open problem

Select a problem to read its exact statement, why it matters, status, sources, and related research.

Edition ranks stay fixed
  1. Rank001P versus NPWhen a yes-or-no problem has short evidence for each yes answer that can be checked in polynomial time, must a deterministic computer also be able to solve the problem in polynomial time? Here polynomial time measures how the work grows with the input size.Theoretical computer scienceClassical complexity separations · 6 problems

    Exact statement

    Determine whether P = NP, that is, whether every decision problem whose solutions can be verified in polynomial time by a deterministic Turing machine can also be solved in polynomial time by a deterministic Turing machine.

    Why it matters

    Equality would give polynomial-time algorithms for every NP decision problem; separation would establish that some efficiently checkable problems resist all such algorithms. This addresses computational efficiency as input size grows.

    Status

    Open

    P versus NP remains open on the maintained Clay statement and the bounded claim sweep; no accepted exact-target resolution was established.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    • #34 · Deterministic Polynomial-Time Solvability of PPAD (P versus PPAD)Related problemDeterministic Polynomial-Time Solvability of PPAD (P versus PPAD)

      Scope qualification: Deterministic total-search PPAD solvability has independent standing through End-of-Line and bimatrix equilibrium. P=NP implies polynomial selectors; failure of a PPAD selector implies P!=NP. The opposite outcomes do not provide converse implications. No unconditional both-outcome resolution implication is adopted.

    • #29 · Polynomial Hierarchy Noncollapse ConjectureRelated problemPolynomial Hierarchy Noncollapse Conjecture

      Scope qualification: Full PH noncollapse has independent complexity-theory standing. Its affirmative branch implies P!=NP; a higher-level collapse does not itself answer P versus NP. This is a branch-dependent relation, not an unconditional implication between complete resolution events. No unconditional both-outcome resolution implication is adopted.

    • #33 · P versus NP ∩ coNPRelated problemP versus NP ∩ coNP

      Related questions with different scopes.

      Scope qualification: These are related complexity questions. No implication between these exact statements is established here; their shared family does not assert one.

    Problem family

    Classical complexity separations · 6 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 1–1.

    Source-fit opponent count: 39.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  2. Rank002Riemann HypothesisDo all zeros of the Riemann zeta function inside the strip with real part between zero and one lie exactly on its middle line?Number theory & arithmetic geometryRiemann hypotheses and exceptional zeros · 3 problems

    Exact statement

    For the meromorphic continuation of the Riemann zeta function zeta(s)=sum_{n>=1} n^(-s), initially defined for Re(s)>1, every zero s in the open critical strip 0<Re(s)<1 has Re(s)=1/2. This is the Riemann Hypothesis for zeta alone; it does not assert simplicity of its zeros.

    Why it matters

    A proof would sharply control fluctuations in the distribution of primes: the prime-counting function would differ from its logarithmic-integral approximation by at most O(√x log x). A disproof would invalidate this equivalent error bound.

    Status

    Open

    Earlier note (2026-09-14): Riemann remains open. September 2026 progress on the proportion of simple critical-line zeros is partial, not proof that every nontrivial zero lies on the line. Uncorroborated or inaccessible proof claims were not accepted. October 7, 2026: OpenAI’s October 6 release reports partial progress. A zero-free half-plane Re(s)>7/8 for zeta excludes only the right-hand part of the critical strip. The target requires every nontrivial zero to have real part exactly 1/2; no conclusion excluding zeros between 1/2 and 7/8 is supplied. This is partial zero-location progress, not the Riemann Hypothesis. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    Riemann hypotheses and exceptional zeros · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 2–2.

    Source-fit opponent count: 53.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  3. Rank003Yang-Mills Existence and Mass GapCan a nontrivial Yang–Mills quantum field theory in four-dimensional spacetime be constructed rigorously for every compact simple gauge group—the choice of internal symmetry—with a positive, finite gap between vacuum energy and the remaining energy spectrum?Mathematical physics

    Exact statement

    For every compact simple gauge group G, prove that a nontrivial four-dimensional quantum Yang-Mills theory on R^4 exists with all existence requirements of Jaffe-Witten, Quantum Yang-Mills Theory, sections 3-4, including axiomatic properties at least as strong as their cited references [45,35], the required gauge-invariant local fields and short-distance compatibility specified there. Its vacuum Hamiltonian H has no spectrum in (0,Delta) for some Delta > 0, and the supremum of such Delta is finite. The official reconstruction and existence requirements are retained by reference, not replaced by an unrestricted equivalence of arbitrary Euclidean and Minkowski theories. An isolated first excited eigenstate, confinement and asymptotic completeness are not additional requirements.

    Why it matters

    The construction would turn the specified four-dimensional quantum gauge-field model into a theory satisfying the required mathematical axioms. Its spectrum would have a positive, finite gap above vacuum energy, without requiring an isolated first excited eigenstate.

    Status

    Open

    The universal four-dimensional Yang–Mills existence/mass-gap target remains open. Three-dimensional outlines, individual gauge-group claims and withdrawn postings do not establish the full target.

    Status reviewed

    Sources

    Problem family

    Yang-Mills Existence and Mass Gap · 1 problem in this edition. Family membership alone does not assert an implication.

    • #3 · Yang-Mills Existence and Mass Gap (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 3–4.

    Source-fit opponent count: 47.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  4. Rank004Hodge ConjectureHodge theory singles out certain rational cohomology classes, which encode topological features of a smooth complex projective variety. Must every such class be a rational combination of classes represented by algebraic subvarieties?Geometry & topologyHodge conjectures · 3 problems

    Exact statement

    Determine whether every rational Hodge class on a smooth complex projective algebraic variety is a rational linear combination of classes of algebraic subvarieties (algebraic cycles).

    Why it matters

    Cohomology records features of a variety, while algebraic cycles are built from its algebraic subvarieties. A proof would show that every rational class of the specified Hodge type can be accounted for by those algebraic pieces.

    Status

    Open

    Earlier note (2026-09-14): The rational Hodge conjecture remains open. Conditional witness axioms, selected Fermat varieties and inaccessible full-proof claims do not establish the all-varieties target. October 7, 2026: OpenAI’s October 6 release reports partial progress. The target quantifies over every smooth complex projective variety and every rational Hodge class. The family proves that assertion for CM abelian varieties, arbitrary products of projective K3 surfaces, and the specified additional abelian/cover families, in the stated codimensions. These are substantial subclasses, not all smooth projective varieties; none of the abstracts gives a reduction of the universal target to them. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    Hodge conjectures · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 4–7.

    Source-fit opponent count: 32.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  5. Rank005Navier-Stokes Existence and Smoothness: Fefferman AFor an incompressible fluid filling three-dimensional space, with any positive viscosity and no external force, does every smooth, rapidly decaying initial velocity give a flow that stays smooth forever, with uniformly bounded total kinetic energy?Analysis & PDEIncompressible fluid regularity · 2 problems

    Exact statement

    Prove Fefferman statement (A): for every viscosity nu > 0 and every real smooth divergence-free Schwartz vector field u0 on R^3, with force f identically zero, there are real velocity u and pressure p in C-infinity(R^3 x [0,infinity)) satisfying partial_t u + (u dot grad)u = nu Delta u - grad p, div u = 0 and u(x,0) = u0(x). The velocity energy integral over R^3 of |u(x,t)|^2 is bounded above by one finite constant for all t >= 0; that constant may depend on u0 and nu. Schwartz means every spatial derivative decays faster than every inverse polynomial, as in official condition (4). Smoothness includes time zero. The selected domain is whole space, not the torus.

    Why it matters

    This asks whether the specified three-dimensional viscous-flow equations can evolve every allowed starting flow for unlimited time without developing a singularity, while keeping total kinetic energy bounded.

    Status

    Open

    The entry concerns Fefferman A: unforced whole-space Navier–Stokes regularity for smooth rapidly decaying divergence-free data. OpenAI's September claim addresses forced alternatives C/D, not A. This exact target remains open; the broader prize announcement does not establish an accepted proof.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    Incompressible fluid regularity · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 4–8.

    Source-fit opponent count: 45.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  6. Rank006Birch and Swinnerton-Dyer ConjectureFor an elliptic curve over the rational numbers, its rank counts the independent infinite-order generators of its group of rational points. Does this rank equal the order of vanishing of its L-function at s = 1—the multiplicity of the zero there, taken as zero when the function is nonzero?Number theory & arithmetic geometryElliptic curve ranks and congruent numbers · 3 problems

    Exact statement

    Prove or disprove that, for every elliptic curve E over the rational numbers, the rank of the finitely generated abelian group E(Q) of rational points equals the order of vanishing at s = 1 of the associated Hasse–Weil L-function L(E, s).

    Why it matters

    A proof would identify an algebraic measure of the curve’s rational points with an analytic measurement of its L-function. A counterexample would locate a curve where those two measurements differ.

    Status

    Open

    Earlier note (2026-09-14): BSD rank equality remains open. The located all-rank working-draft claim lacks independent acceptance; recent uses and special cases do not supply a universal proof. October 7, 2026: OpenAI’s October 6 release reports partial progress. The target is equality of analytic and Mordell-Weil ranks for every E/Q, rather than the full leading-term formula. This family gives that equality and the stronger leading-term conclusion when the full q-power Selmer corank is zero or one for some prime q. It leaves curves outside that hypothesis untreated, so it is a low-corank subcase only. The companion proves rank equality for E/Q with nonzero rational two-torsion and full 2-power Selmer corank zero or one, allowing arbitrary reduction at 2. The BSD rank target includes every E/Q without either restriction, so this is a limited subcase and does not follow as a full closure from the H10 result. The low-corank 2-converse explicitly gives equality of analytic and Mordell-Weil ranks for E/Q whose full 2-power Selmer corank is zero or one. The family also supplies analytic rank densities and their mean over twists, but an average or density-one assertion does not prove rank equality for every curve or every twist. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    • #145 · Congruent Number ProblemSubproblemCongruent Number Problem

      A proof of this assertion would also establish #145.

    • #178 · Parity conjecture for elliptic curvesRelated problemParity conjecture for elliptic curves

      Related questions with different scopes.

      Scope qualification: The BSD entry concerns elliptic curves over the rationals, whereas the parity entry covers every number field. An implication between these exact scopes is not established here; this does not assert logical independence.

    Problem family

    Elliptic curve ranks and congruent numbers · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 4–8.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  7. Rank007Global Langlands Functoriality ConjectureLanglands L-groups encode symmetries used to organize automorphic representations. For quasisplit reductive groups over number fields, does every admissible map between their L-groups transfer these representations while matching their local arithmetic data at all but finitely many places?Number theory & arithmetic geometryLanglands functoriality and modularity · 2 problems

    Exact statement

    For every number field F, quasisplit connected reductive F-groups G-prime and G, and admissible L-homomorphism rho:L(G-prime)->L(G) (over Gamma_F, commuting with its projections, in Arthur Principle1.1's convention), every automorphic representation pi-prime of G-prime(A_F), in the inclusive Langlands sense cited there, has an automorphic representation pi of G(A_F) whose unramified Satake classes equal the images under rho_v of those of pi-prime outside a finite set of places. This is existence of a weak transfer, not uniqueness or a prescribed transfer function.

    Why it matters

    A proof would guarantee the stated transfer across every group map in scope, linking automorphic representations through their local parameters. The existence claim permits more than one transfer and leaves its exceptional places finite.

    Status

    Open · subcases solved

    Global functoriality remains open, with solved subcases. Existing restricted transfers and recent conjectural use do not establish the full specified transfer theorem; evidence of its current general status remains limited.

    Status reviewed

    Sources

    Problem family

    Langlands functoriality and modularity · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 4–8.

    Source-fit opponent count: 39.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  8. Rank008Existence of One-Way FunctionsDo one-way functions exist: functions that a deterministic computer can evaluate in polynomial time, but that every randomized polynomial-time algorithm reverses on a random input’s output with success eventually smaller than every inverse power of the input length? Reversing means finding any input with that output.Cryptography, coding, information & optimizationOne-way functions and permutations · 2 problems

    Exact statement

    Prove or disprove that there exists a deterministic polynomial-time computable function family f_n:{0,1}^n->{0,1}^{poly(n)} such that every probabilistic polynomial-time algorithm inverts f_n on a uniformly random input with only negligible probability.

    Why it matters

    A construction would unconditionally separate efficient forward computation from hard inversion on random inputs, even when the inverter uses randomness. Every randomized polynomial-time inverter would have only negligible success probability as input length grows.

    Status

    Open

    Unconditional existence of one-way functions remains open. Conditional constructions and hardness characterizations do not constitute an unconditional solved subcase of this existential target.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    One-way functions and permutations · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 5–8.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  9. Rank009The abc Conjecture of Masser and OesterleSuppose coprime positive integers satisfy a + b = c, and R is the product of the distinct primes dividing abc. For every positive ε, is c less than C(ε) times R^(1 + ε), with one constant C(ε) working for every such triple?Number theory & arithmetic geometryabc and Szpiro conjectures · 2 problemsOpen · disputed claim

    Exact statement

    Prove that for every epsilon > 0, there exists a constant C(epsilon) such that for all coprime positive integers a + b = c, c < C(epsilon) rad(abc)^{1+epsilon}.

    Why it matters

    A proof would uniformly restrict how large an additive triple can be relative to its distinct prime factors. A disproof would find a positive epsilon for which no constant controls all such triples.

    Status

    Open · disputed claim

    abc retains open_disputed_claim. Joshi's February 2025 revision is not adjudicated by the older Scholze–Stix objection to Mochizuki; the July 2026 abc-type proposal is another conjecture. Current acceptance of a full proof was not established.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    abc and Szpiro conjectures · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 9–13.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  10. Rank010Generalized Riemann Hypothesis for Dirichlet L-functionsDirichlet L-functions extend the zeta-function setting by using periodic arithmetic weights. For every Dirichlet character, do all zeros of its L-function with real part between zero and one lie on the middle line, with real part one-half?Number theory & arithmetic geometryRiemann hypotheses and exceptional zeros · 3 problems

    Exact statement

    For every integer q>=1 and every Dirichlet character chi modulo q, including principal and imprimitive characters, every zero s of the meromorphic continuation of L(s,chi)=sum_{n>=1}chi(n)n^(-s) in 0<Re(s)<1 satisfies Re(s)=1/2. The continuation is from Re(s)>1. Only open-strip zeros are quantified: principal-character poles and boundary zeros of imprimitive Euler factors are not counterexamples. No simplicity or central nonvanishing requirement is added.

    Why it matters

    A proof would bound how far one must search for a prime in an arithmetic progression. For q > 3 and a coprime to q, the least prime congruent to a modulo q would be at most (φ(q) log q)², where φ(q) counts the residue classes coprime to q.

    Status

    Open

    Earlier note (2026-09-14): Dirichlet GRH remains open. No independently established full-target resolution was obtained; neighboring Riemann-family claims and source-access limitations remain qualified. October 7, 2026: OpenAI’s October 6 release reports partial progress. The theorem includes every Dirichlet L-function and allows the principal pole, but its zero-free boundary is 7/8 rather than 1/2. Thus its character quantification agrees broadly with the target while its zero-location conclusion is strictly weaker. The GRH target remains unresolved. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Related problems

    Problem family

    Riemann hypotheses and exceptional zeros · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 9–14.

    Source-fit opponent count: 50.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  11. Rank011NP versus coNP ProblemWhenever yes answers to a decision problem have polynomial-length certificates checkable in polynomial time, do its no answers always have such certificates as well?Theoretical computer scienceClassical complexity separations · 6 problems

    Exact statement

    Determine whether NP = coNP, that is, whether the class NP is closed under complementation. Equivalently, by the Cook-Reckhow criterion, decide whether there exists a polynomially bounded propositional proof system.

    Why it matters

    Equality would make NP closed under exchanging yes and no answers and provide a polynomially bounded propositional proof system. Separation would rule out every such proof system.

    Status

    Open

    NP versus coNP remains open with an explicit claim warning. Lin's August 2026 v28 and Gordeev–Haeusler's May 2026 revision make opposing claims; older independent critiques do not automatically adjudicate those revisions. Restricted proof-system lower bounds are not the full separation.

    Status reviewed

    Sources

    Problem family

    Classical complexity separations · 6 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 9–14.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  12. Rank012Extended Riemann hypothesis for Dedekind zeta functionsEvery number field—a finite algebraic extension of the rationals—has a Dedekind zeta function. Must all its zeros with real part between zero and one lie on the middle line, with real part one-half?Number theory & arithmetic geometryRiemann hypotheses and exceptional zeros · 3 problemsOpen · disputed claim

    Exact statement

    For every algebraic number field K (a finite extension of Q, including K=Q), does every zero s of its Dedekind zeta function ζ_K(s) in the critical strip 0<Re(s)<1 satisfy Re(s)=1/2?

    Why it matters

    A proof would supply the zeta-function hypothesis used in effective Chebotarev estimates for how primes split in number fields, with their other conditions retained. Including the rational field would also establish the ordinary Riemann Hypothesis.

    Status

    Open · disputed claim

    Earlier note (2026-09-14): Dedekind ERH remains open, with a disputed proof claim. Zhang's posting was revised to v19 on September 14, 2026; objections or prior assessments of v18 do not review v19. Independent acceptance of the newest claim was not established. October 7, 2026: OpenAI’s October 6 release reports partial progress. The stated half-plane theorem applies to zeta_Q and, by the usual abelian L-function factorization, to Dedekind zeta functions in the abelian settings represented by its Dirichlet/finite-order Hecke factors. It supplies neither the critical line 1/2 nor all number fields, so it is only restricted zero-free progress for the universal Dedekind-zeta target. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Related problems

    Problem family

    Riemann hypotheses and exceptional zeros · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 9–15.

    Source-fit opponent count: 45.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  13. Rank013NP versus P/polyCan every yes-or-no problem with short, efficiently checkable evidence for yes answers be solved by Boolean circuits of polynomial size, if a separate circuit may be chosen for each input length? No efficient method of constructing those circuits is required.Theoretical computer science

    Exact statement

    Determine the truth of NP subseteq P/poly. Here NP consists of languages over finite binary strings accepted by nondeterministic Turing machines in polynomial time. P/poly consists of languages L for which there exist constants C>=1 and integer k>=1 and Boolean circuits C_n for all n>=0, of size at most C(n+1)^k, deciding membership in L on every length-n input. Circuits are finite acyclic single-output circuits with fan-in-two AND/OR, fan-in-one NOT, input bits and hardwired constants0/1; count all nodes. Each language has its own bound and circuit family, with no uniform generation requirement.

    Why it matters

    Containment would collapse the polynomial hierarchy to its second level by the Karp–Lipton theorem. Noncontainment would establish a general circuit-size barrier for an NP language and imply P ≠ NP.

    Status

    Open

    NP containment in nonuniform P/poly remains open; February 2026 notes pose the matching general-circuit question. Restricted circuits and Karp–Lipton implications do not decide it.

    Status reviewed

    Sources

    Problem family

    NP versus P/poly · 1 problem in this edition. Family membership alone does not assert an implication.

    • #13 · NP versus P/poly (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 10–15.

    Source-fit opponent count: 34.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  14. Rank014P versus PSPACECan every yes-or-no problem solvable by a deterministic computer using polynomially bounded memory also be solved deterministically in time polynomial in the input length?Theoretical computer scienceClassical complexity separations · 6 problems

    Exact statement

    Determine whether P = PSPACE, that is, whether every language decidable by a deterministic Turing machine in polynomial space is decidable by a deterministic Turing machine in polynomial time.

    Why it matters

    Equality would turn every polynomial-space decision procedure into a polynomial-time one. Separation would establish that some problems admit modest memory requirements while resisting every deterministic polynomial-time algorithm.

    Status

    Open

    P versus PSPACE remains open. Recent space simulations and fixed-exponent time lower bounds do not separate polynomial time from polynomial space; conditional equality implications are not unconditional resolutions.

    Status reviewed

    Sources

    Problem family

    Classical complexity separations · 6 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 10–15.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  15. Rank015Tate Conjecture for Algebraic CyclesAlgebraic subvarieties defined over the base field produce l-adic cohomology classes fixed by arithmetic symmetries after a normalization called a Tate twist. For smooth projective varieties over finitely generated fields, are all even-degree classes with this symmetry property generated by algebraic cycles over the same field, for every prime l different from the field’s characteristic?Geometry & topology

    Exact statement

    For every smooth geometrically irreducible projective variety X over a field k finitely generated over its prime field, every codimension i, and every prime l different from char(k), prove that the l-adic cycle-class map from codimension-i algebraic cycle classes defined over k, tensored with Q_l, is surjective onto H^(2i)(X_bar,Q_l(i))^Gal(k_bar/k).

    Why it matters

    A proof would recover every invariant class in the specified cohomology from algebraic cycles defined over the base field. A counterexample would identify an invariant class with no such algebraic description.

    Status

    Open · subcases solved

    Earlier note (2026-09-14): The rational Tate conjecture remains open, with solved subcases. Integral counterexamples do not refute the stated rational cycle map. Kahn's unsuccessful surface approach does not prove the universal target; its HTML/history date discrepancy was not authenticated as a new revision. October 7, 2026: OpenAI’s October 6 release reports partial progress. The release reports algebraicity for specializations of rational Hodge classes on good-reduction abelian varieties and, separately, the Tate conjecture for abelian varieties over finite fields. The specialization theorem concerns the geometric special fiber; descent of an invariant class to a chosen finite field uses a further trace argument. These results do not establish cycle-class surjectivity for arbitrary smooth projective varieties over every field finitely generated over its prime field. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Tate Conjecture for Algebraic Cycles · 1 problem in this edition. Family membership alone does not assert an implication.

    • #15 · Tate Conjecture for Algebraic Cycles (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 11–15.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  16. Rank016BQP Versus NP: Complete Language-Class RelationHow do efficient quantum solving and efficient classical verification compare for yes-or-no problems? Does quantum computation with bounded error solve every problem with short, efficiently checkable evidence for yes answers—and do all problems it solves have such evidence?Theoretical computer science

    Exact statement

    Determine the truth values of BOTH NP subseteq BQP and BQP subseteq NP for unrelativized, advice-free classes of total languages over {0,1}. NP consists of languages with polynomial-length binary witnesses and a deterministic polynomial-time verifier. BQP consists of languages decided by polynomial-time uniformly generated polynomial-size quantum circuits (a deterministic Turing machine outputs Q_n on 1^n), using Watrous's fixed Toffoli, Hadamard and phase-i gates with |0> ancillas and discarding, followed by a standard-basis output measurement. On every yes input acceptance is at least 2/3, and on every no input it is at most 1/3. No arbitrary real gate constants, advice or oracle are available. Establish equality, either proper containment, or incomparability.

    Why it matters

    Resolving both inclusions would determine whether these computational classes coincide, one strictly contains the other, or neither contains the other. Each outcome gives a different relationship between efficient verification and bounded-error quantum computation.

    Status

    Open

    Both unrelativized BQP/NP inclusions remain unresolved on the checked evidence. Random-permutation oracle separations do not settle the stated uniform total-language relationship.

    Status reviewed

    Sources

    Problem family

    BQP Versus NP: Complete Language-Class Relation · 1 problem in this edition. Family membership alone does not assert an implication.

    • #16 · BQP Versus NP: Complete Language-Class Relation (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 16–18.

    Source-fit opponent count: 37.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  17. Rank017Polynomial-time classical integer factorizationIs there a deterministic classical algorithm that finds a nontrivial factor of every composite integer in time polynomial in the number of its binary digits?Theoretical computer science

    Exact statement

    Determine whether there is a deterministic classical algorithm that, on input a composite integer N in binary, outputs a nontrivial factor of N in time polynomial in log N.

    Why it matters

    An algorithm would guarantee efficient factor finding for every composite input in this model. A negative result would establish that no deterministic classical algorithm can satisfy that polynomial-time bound.

    Status

    Open

    Deterministic classical polynomial-bit-time factorization remains open. Conditional improvements to exponents in the integer being factored are not polynomial time in its bit length, and quantum factoring is outside this target.

    Status reviewed

    Sources

    Problem family

    Polynomial-time classical integer factorization · 1 problem in this edition. Family membership alone does not assert an implication.

    • #17 · Polynomial-time classical integer factorization (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 16–18.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  18. Rank018Hardy-Littlewood Prime k-Tuple ConjectureTake a fixed finite pattern of integer offsets. If no prime divides the product of the shifted entries for every integer shift, do shifts making every entry prime occur with exactly the asymptotic frequency predicted by the Hardy–Littlewood formula, including its correction for divisibility?Number theory & arithmetic geometrySimultaneous prime values · 5 problems

    Exact statement

    For every finite admissible set H={h_1,...,h_k} of distinct integers, prove that the number of n <= x for which every n+h_i is prime is asymptotic to S(H)x/(log x)^k, where S(H) is the Hardy-Littlewood singular series, or disprove this formula.

    Why it matters

    A proof would give quantitative frequencies for every admissible finite prime pattern. A counterexample would identify a pattern whose simultaneous-prime count departs from the proposed asymptotic law.

    Status

    Open

    The all-fixed-admissible-tuples Hardy–Littlewood asymptotic remains open. Published short-average progress does not establish every fixed tuple. The located bitstring-sieve full-proof claim lacks independent acceptance.

    Status reviewed

    Sources

    Related problems

    Problem family

    Simultaneous prime values · 5 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 17–21.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  19. Rank019Unconditional Separation of Quantum and Classical Bounded-Error Polynomial Time (BQP != BPP)Can quantum computers solve some decision problem in polynomial time with bounded error that every classical randomized polynomial-time algorithm fails to solve with bounded error?Quantum information & computation

    Exact statement

    Prove unconditionally that bounded-error quantum polynomial time (BQP) strictly contains bounded-error probabilistic polynomial time (BPP).

    Why it matters

    A proof would establish an unconditional separation between these quantum and classical computational resources. A negative resolution would equate their polynomial-time decision power when both are allowed bounded error.

    Status

    Open

    Unconditional unrelativized BQP versus BPP remains open. The September Fourier-hierarchy result is oracle-relative and does not establish an ordinary language-class separation.

    Status reviewed

    Sources

    Problem family

    Unconditional Separation of Quantum and Classical Bounded-Error Polynomial Time (BQP != BPP) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #19 · Unconditional Separation of Quantum and Classical Bounded-Error Polynomial Time (BQP != BPP) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 17–21.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  20. Rank020Twin Prime ConjectureAre there infinitely many pairs of prime numbers differing by exactly two, so that such pairs continue to occur however far along the number line we look?Number theory & arithmetic geometrySimultaneous prime values · 5 problems

    Exact statement

    Prove that there are infinitely many primes p such that p+2 is also prime.

    Why it matters

    A proof would show that the exact gap of two recurs at arbitrarily large primes. It would turn this smallest positive gap between odd primes into an endlessly recurring feature of prime spacing.

    Status

    Open

    Twin primes remains open with uncorroborated 2026 proof-claim warnings. Repository hosting and proof titles are not independent mathematical acceptance; bounded gaps, finite computations and prime-pair analogues are not exact gap-two infinitude.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    Simultaneous prime values · 5 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 17–22.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  21. Rank021Smooth four-dimensional Poincare conjectureA homotopy equivalence identifies spaces up to continuous deformation, without requiring a smooth match point by point. If a smooth, closed four-dimensional manifold has the homotopy type of the four-dimensional sphere, must there be a smooth bijection between them with a smooth inverse?Geometry & topologySmooth four-dimensional spheres · 2 problems

    Exact statement

    Determine whether every smooth closed 4-manifold homotopy equivalent to the standard 4-sphere S^4 is diffeomorphic to S^4.

    Why it matters

    A proof would show that the stated homotopy condition determines the sphere’s smooth structure. A counterexample would exhibit an exotic smooth four-sphere with a different diffeomorphism type.

    Status

    Open

    Smooth four-dimensional Poincare remains open. The reviewed X(41,189,73) project explicitly disclaims a peer-accepted counterexample and leaves geometric obligations outside its formal work. Restricted knot-route evidence does not decide the whole conjecture.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    Smooth four-dimensional spheres · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 20–32.

    Source-fit opponent count: 34.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  22. Rank022The NC versus P Problem for Efficient Parallel ComputationCan every decision problem solvable in polynomial time be solved with polynomially many parallel processors in only a power of log n sequential steps, where n is the input length? The parallel computations must come from a uniform algorithmic construction.Theoretical computer science

    Exact statement

    Determine whether NC equals P, where NC consists of decision problems solvable by uniform polynomial-size circuit families of polylogarithmic depth and P consists of decision problems solvable in deterministic polynomial time.

    Why it matters

    Equality would allow every polynomial-time decision problem to be organized into shallow uniform circuits. Separation would identify a problem whose polynomial-time computation cannot be parallelized to that depth within polynomial size.

    Status

    Open

    NC versus P remains open with a dated exact-status-source warning. Recent cellular-automaton or restricted-model classifications do not separate the same ordinary language classes.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    The NC versus P Problem for Efficient Parallel Computation · 1 problem in this edition. Family membership alone does not assert an implication.

    • #22 · The NC versus P Problem for Efficient Parallel Computation (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 21–31.

    Source-fit opponent count: 49.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  23. Rank023Grothendiecks Generalized Hodge ConjectureHodge coniveau measures how far Hodge theory suggests cohomology classes can be concentrated on a smaller subset. On a smooth complex projective variety, must every rational Hodge substructure of coniveau at least c be supported on an algebraic subset of codimension at least c?Number theory & arithmetic geometryHodge conjectures · 3 problems

    Exact statement

    For every smooth projective complex variety X and integers k,c, prove that the largest rational Hodge substructure of H^k(X,Q) with Hodge coniveau at least c equals the subspace of classes that vanish on X minus some closed algebraic subset of codimension at least c.

    Why it matters

    A proof would identify a condition on Hodge structures with a geometric support condition in every specified degree and codimension. A counterexample would show that the Hodge condition can predict support that no suitable algebraic subset provides.

    Status

    Open · subcases solved

    Earlier note (2026-09-14): Generalized Hodge remains open, with solved subcases. July 2026 treatment retains an unresolved general conjecture; moduli-space and specified coniveau cases do not cover every smooth projective variety. October 7, 2026: OpenAI’s October 6 release reports partial progress. The CM manuscript explicitly gives the generalized Hodge conjecture for CM abelian varieties. A companion also states it for powers of K3 surfaces on a specified quadratic locus and gives a Kuga–Satake implication. These are subclasses of the target, which requires the coniveau equality for all smooth projective complex varieties and all degrees; no global resolution follows. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Related problems

    Problem family

    Hodge conjectures · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 21–34.

    Source-fit opponent count: 32.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  24. Rank024Binary Goldbach Problem for Even IntegersCan every even integer greater than two be written as the sum of two prime numbers, allowing the same prime to be used twice?Number theory & arithmetic geometry

    Exact statement

    Prove or disprove that every even integer n greater than 2 is the sum of two primes.

    Why it matters

    A proof would guarantee that two prime summands suffice to represent every even integer above two. It would give a universal additive representation across all those integers, however large.

    Status

    Open

    Goldbach remains open. July/August 2026 sources distinguish the conjecture from almost-prime results; the revised uniformity preprint explicitly does not yield pointwise Goldbach.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Binary Goldbach Problem for Even Integers · 1 problem in this edition. Family membership alone does not assert an implication.

    • #24 · Binary Goldbach Problem for Even Integers (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 21–38.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  25. Rank025P versus BPP (Does BPP = P?)Can every decision problem solved by a randomized polynomial-time algorithm with bounded error on both answers also be solved deterministically in polynomial time?Cryptography, coding, information & optimization

    Exact statement

    Determine whether every language computable by a probabilistic polynomial-time algorithm with bounded two-sided error (BPP) can be computed by a deterministic polynomial-time algorithm, i.e., whether BPP = P.

    Why it matters

    Equality would remove randomness from every computation in this class while preserving polynomial-time solvability. Separation would identify a decision problem for which bounded-error randomization provides power beyond deterministic polynomial time.

    Status

    Open

    Earlier note (2026-09-14): P versus BPP remains open. Recent pseudorandom-generator and derandomization advances retain additional assumptions, not a deterministic simulation of every BPP language. October 7, 2026: OpenAI’s October 6 release reports partial progress. The theorem derandomizes the logarithmic-workspace subclass of polynomial-time bounded-error computation. Such machines form a restricted subcase of the full BPP=P question. The publisher gives no deterministic simulation for arbitrary polynomial-space work tapes used by polynomial-time BPP machines, so it does not establish the unrestricted equality BPP=P. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    P versus BPP (Does BPP = P?) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #25 · P versus BPP (Does BPP = P?) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 22–38.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  26. Rank026Schanuel's conjectureIf n complex numbers have no nonzero rational linear relation, must at least n of those numbers and their exponentials be algebraically independent, meaning that no nonzero polynomial with rational coefficients relates them?Number theory & arithmetic geometryExponential transcendence · 3 problems

    Exact statement

    Prove or disprove that if z_1,...,z_n in C are linearly independent over Q, then the field Q(z_1,...,z_n,e^{z_1},...,e^{z_n}) has transcendence degree at least n over Q.

    Why it matters

    A proof would force at least n algebraically independent quantities into every field generated this way. A counterexample would reveal an exponential tuple with more algebraic dependence than the bound allows.

    Status

    Open

    Schanuel remains open. Reformulations and almost-everywhere Schanuel properties do not establish its universal algebraic-independence conclusion.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    Exponential transcendence · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 22–41.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  27. Rank027Matrix-Multiplication Exponent Equals TwoCan two n-by-n rational matrices be multiplied exactly, without division, with a number of arithmetic operations arbitrarily close to quadratic: O(n^(2 + ε)) for every positive ε? Each rational arithmetic operation counts as one step, regardless of the number of bits in its operands.Theoretical computer science

    Exact statement

    Let R_Q(n) be the tensor rank over Q of the bilinear map multiplying two n by n rational matrices: the least r for which AB is a sum of r products alpha_j(A) beta_j(B) C_j, with rational linear forms alpha_j, beta_j and rational output matrices C_j. Set omega_Q = inf{tau in R : R_Q(n) = O(n^tau)}. Prove omega_Q = 2. Equivalently, for every epsilon > 0, exact multiplication admits division-free arithmetic circuits over Q of size O(n^(2+epsilon)), counting rational additions, subtractions, multiplications and constant scalar operations at unit cost. Circuits need not be uniformly generated and rational coefficient bit lengths are not charged. Constants and thresholds may depend on epsilon. Literal O(n^2) attainment is not required.

    Why it matters

    A proof would allow exact matrix multiplication with arithmetic exponents arbitrarily close to two and carry through applicable algebraic reductions to other computations. The statement measures arithmetic operations; it leaves coefficient bit costs and literal quadratic attainment outside its guarantee.

    Status

    Open

    Earlier note (2026-09-14): The exact rational-field matrix multiplication exponent target remains open. The August 2026 claimed exponent below 2.371177 is not exponent 2; complex-field results require field/scope review before transfer to Q. October 7, 2026: OpenAI’s October 6 release reports partial progress. The primary manuscript gives omega<=9/4 over C; the secondary manuscript explicitly gives omega<2.258 over every characteristic-zero field, including the full field Q. The all-field companion supplies a weaker square bound and the rectangular exponent improvement concerns a different shape. These are arithmetic upper-bound improvements but none gives omega_Q=2. In particular the strongest complex-only headline is not silently restated as an explicit rational-field theorem. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Matrix-Multiplication Exponent Equals Two · 1 problem in this edition. Family membership alone does not assert an implication.

    • #27 · Matrix-Multiplication Exponent Equals Two (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 23–37.

    Source-fit opponent count: 49.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  28. Rank028Generalized Ramanujan Conjecture for GL_nAutomorphic representations connect arithmetic with highly symmetric functions. For every cuspidal representation of GL_n over a number field with unitary central character, must its component at every place satisfy the sharp local condition called temperedness?Number theory & arithmetic geometry

    Exact statement

    Prove that for every number field F and every cuspidal automorphic representation pi of GL_n(A_F) with unitary central character, each local component pi_v is tempered at every place v.

    Why it matters

    A proof would force the global cuspidal and unitary hypotheses to constrain local behavior at every place. It would rule out non-tempered local components throughout the specified automorphic family.

    Status

    Open · subcases solved

    Generalized Ramanujan remains open, with solved subcases. Recent analytic consequences assume the conjecture and restricted automorphic cases do not prove all GL(n) targets; an inaccessible purported full proof was not accepted.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Generalized Ramanujan Conjecture for GL_n · 1 problem in this edition. Family membership alone does not assert an implication.

    • #28 · Generalized Ramanujan Conjecture for GL_n (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 24–38.

    Source-fit opponent count: 42.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  29. Rank029Polynomial Hierarchy Noncollapse ConjectureDoes each additional level of alternating polynomial-time verification give a strictly larger class of decision problems, so that the polynomial hierarchy never collapses at a finite level?Theoretical computer scienceClassical complexity separations · 6 problems

    Exact statement

    Resolve the assertion that the ordinary unrelativized polynomial hierarchy is infinite: for every integer k>=0, Sigma_k^P is a proper subclass of Sigma_{k+1}^P, equivalently PH != Sigma_k^P for every k>=0, where Sigma_0^P=P, Sigma_{k+1}^P=NP^{Sigma_k^P}, and PH is the union of these classes over constant finite k; all are decision languages on finite binary strings in the standard uniform Turing-machine model.

    Also known as

    PH noncollapse; Infinitude of the polynomial hierarchy; Strictness of every finite level of PH

    Why it matters

    This asks whether the successive levels of the polynomial hierarchy describe genuinely different computational powers. Full noncollapse would imply P≠NP; a collapse at a higher level would not by itself settle P versus NP.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Related problems

    • #1 · P versus NPRelated problemP versus NP

      Scope qualification: Full PH noncollapse has independent complexity-theory standing. Its affirmative branch implies P!=NP; a higher-level collapse does not itself answer P versus NP. This is a branch-dependent relation, not an unconditional implication between complete resolution events. No unconditional both-outcome resolution implication is adopted.

    Problem family

    Classical complexity separations · 6 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 21–45.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  30. Rank030Anderson Model Extended States in Dimension at Least ThreeIn a lattice of dimension at least three, can a quantum particle in a weak random potential have a conducting regime, separated from localized behavior by a mobility edge? The mathematical target is an absolutely continuous part of the Anderson model’s energy spectrum.Mathematical physics

    Exact statement

    Prove that the Anderson model, the discrete random Schroedinger operator on Z^d with d >= 3, has a regime of absolutely continuous spectrum, equivalently that a conducting phase and a mobility edge exist at weak disorder, so that the metal-insulator transition is established.

    Why it matters

    A proof would establish the specified spectral regime and the associated transition sought in this random-operator model. It would connect the model’s weak-disorder behavior to the conducting phase described in the target.

    Status

    Open

    Earlier note (2026-09-14): The lattice Anderson extended-states target remains open, explicitly so in a September 7 primary paper. Bethe-lattice theorems and finite-timescale lattice diffusion do not establish absolutely continuous spectrum on Z^d at fixed positive disorder. The claimed equivalence with conduction/mobility edges remains unverified. October 7, 2026: OpenAI’s October 6 release reports partial progress. For the nearest-neighbor lattice model with independent uniform variables on [-1,1], the manuscript reports nonzero, purely absolutely continuous spectrum on a deterministic interval in each fixed dimension d≥3 at every fixed sufficiently small positive coupling. It also reports a disjoint interval of localized spectrum at that coupling. Diffusion and the location of the full mobility edge remain untreated. The broader formulation here does not fix a single-site distribution and also uses conduction and mobility-edge language whose equivalence with the spectral assertion has not been established; that complete formulation remains open. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Anderson Model Extended States in Dimension at Least Three · 1 problem in this edition. Family membership alone does not assert an implication.

    • #30 · Anderson Model Extended States in Dimension at Least Three (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 24–37.

    Source-fit opponent count: 53.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  31. Rank031Valiants Algebraic P versus NP Conjecture (VP != VNP)The permanent uses the determinant’s sum over permutations but gives every term a plus sign. Over fields of characteristic zero, must arithmetic circuits computing n-by-n permanents exactly require more than polynomially many operations as n grows?Theoretical computer scienceAlgebraic computation lower bounds · 3 problems

    Exact statement

    Prove that the permanent polynomial family cannot be computed by polynomial-size algebraic circuits over fields of characteristic zero, equivalently that VP is not equal to VNP.

    Why it matters

    A proof would separate the algebraic complexity classes VP and VNP through a circuit lower bound for the permanent. Polynomial-size circuits for the permanent would refute that separation.

    Status

    Open

    VP versus VNP remains open over the stated characteristic-zero model. January 2026 primary work still poses the unrestricted question; symmetry-restricted circuits and unconstructed elusive-function reductions do not resolve it.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    • #287 · Koiran's Real Tau-Conjecture on Real Roots of Sums of Products of Sparse PolynomialsStronger assertionKoiran's Real Tau-Conjecture on Real Roots of Sums of Products of Sparse Polynomials

      Koiran's real-root bound would imply the separation VP ≠ VNP over C. Establishing that separation does not, by the relation recorded here, establish the root bound.

      Condition: The stronger assertion is proved; a negative answer to it need not settle the weaker assertion.

      Scope qualification: This records a positive implication over the complex numbers, without asserting a converse or an implication from a negative answer. Only the positive-answer implication is recorded. No converse or equivalence of resolution outcomes is established here.

    • #59 · Permanent versus Determinant over CVariantPermanent versus Determinant over C

      A proof of this assertion would also establish #59.

    Problem family

    Algebraic computation lower bounds · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 26–38.

    Source-fit opponent count: 48.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  32. Rank032Basing One-Way Functions on NP-HardnessCan worst-case NP-hardness provide a rigorous foundation for one-way functions—functions that are easy to compute but hard to reverse on random inputs—through a reduction that makes P ≠ NP imply their existence?Cryptography, coding, information & optimization

    Exact statement

    Determine whether one-way functions can be based on NP-hardness, i.e., whether there is a reduction from a worst-case NP-complete (or NP-hard) decision problem to the task of average-case inverting a polynomial-time computable function, so that P != NP would imply the existence of one-way functions.

    Why it matters

    Such a reduction would derive average-case inversion hardness from the stated worst-case complexity separation. It would connect the existence of one-way functions to the assumption P ≠ NP through a precise computational implication.

    Status

    Open

    Constructing one-way functions from worst-case NP hardness alone remains open. Located meta-complexity, zero-knowledge and derandomization characterizations add unproved hypotheses beyond P != NP.

    Status reviewed

    Sources

    Problem family

    Basing One-Way Functions on NP-Hardness · 1 problem in this edition. Family membership alone does not assert an implication.

    • #32 · Basing One-Way Functions on NP-Hardness (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 27–41.

    Source-fit opponent count: 43.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  33. Rank033P versus NP ∩ coNPIf both yes and no answers to a decision problem have short evidence that can be checked efficiently, must a deterministic computer be able to solve the problem in polynomial time?Theoretical computer scienceClassical complexity separations · 6 problems

    Exact statement

    Determine whether P = NP ∩ coNP for ordinary uniform deterministic and nondeterministic polynomial-time Turing-machine classes of total languages over a finite alphabet; equivalently, whether every language with polynomially checkable certificates for both membership and nonmembership has a deterministic polynomial-time decision algorithm.

    Why it matters

    Equality would turn efficient certification on both sides into efficient decision-making for every such language. A separation would exhibit a language whose two kinds of certificates are easy to check but whose answers are not deterministically computable in polynomial time.

    Status

    Open

    The ordinary uniform P = NP intersect coNP question remains open on the checked evidence. The November 2025 random-permutation separation is relativized and does not settle the ordinary question.

    Status reviewed

    Sources

    Related problems

    • #1 · P versus NPRelated problemP versus NP

      Related questions with different scopes.

      Scope qualification: These are related complexity questions. No implication between these exact statements is established here; their shared family does not assert one.

    Problem family

    Classical complexity separations · 6 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 25–45.

    Source-fit opponent count: 41.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  34. Rank034Deterministic Polynomial-Time Solvability of PPAD (P versus PPAD)Can every total search problem in PPAD be solved by a deterministic algorithm in polynomial time? Here the algorithm must find a valid solution, not merely decide whether one exists.Theoretical computer scienceTotal search complexity · 1 problem

    Exact statement

    Resolve whether every total search relation R in PPAD has a deterministic polynomial-time selector f_R: for every finite binary input x, R(x,f_R(x)) holds and f_R runs in time polynomial in |x|, with the polynomial allowed to depend on R; PPAD consists of polynomially balanced, polynomial-time-verifiable total search relations reducible by polynomial-time search reductions to End-of-Line on a succinctly represented directed graph of indegree and outdegree at most one with a given source, seeking a different source or a sink; this is PPAD containment in FP in search-class notation, not equality with the decision class P.

    Also known as

    P versus PPAD; FP versus PPAD; PPAD containment in deterministic polynomial-time search; Polynomial-time End-of-Line; Polynomial-time bimatrix Nash equilibrium

    Why it matters

    PPAD captures search problems whose solutions are guaranteed by a directed-graph argument, including important equilibrium computations. The question asks whether those existence guarantees can always be turned into efficient deterministic algorithms.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Related problems

    • #1 · P versus NPRelated problemP versus NP

      Scope qualification: Deterministic total-search PPAD solvability has independent standing through End-of-Line and bimatrix equilibrium. P=NP implies polynomial selectors; failure of a PPAD selector implies P!=NP. The opposite outcomes do not provide converse implications. No unconditional both-outcome resolution implication is adopted.

    Problem family

    Total search complexity · 1 problem in this edition. Family membership alone does not assert an implication.

    • #34 · Deterministic Polynomial-Time Solvability of PPAD (P versus PPAD) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 22–51.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  35. Rank035Extension of Kroneckers Theorem on Abelian Fields to Any Algebraic Base FieldCan all finite abelian extensions of any number field be generated by values of explicit analytic or arithmetic functions? These are extensions whose field symmetries commute; the goal is an explicit recipe for their generators.Number theory & arithmetic geometryExplicit class field theory and Stark conjectures · 3 problems

    Exact statement

    For each number field K, construct generators for its finite abelian extensions as special values of explicitly defined analytic or arithmetic functions associated with K, extending roots of unity over Q and complex multiplication over imaginary quadratic fields.

    Why it matters

    A construction would give explicit generators for every extension in scope through a common special-value description. An obstruction for one number field would show a limit to that proposed universal construction.

    Status

    Open · subcases solved

    Hilbert's twelfth problem over arbitrary number fields remains open, with solved subcases. Dasgupta–Kakde and the independent ICMS account cover a totally real p-adic construction, not every number field or the original complex-analytic formulation. October 6, 2026: The October 4 manuscript Stark units for real quadratic fields and reciprocity laws claims a real-quadratic advance using explicit units and reciprocity. At a fixed modulus, its Proposition 6 identifies a particular fixed subfield; the units need not generate the entire ray class field for every modulus. This claimed specialized progress does not settle Hilbert’s twelfth problem over arbitrary number fields. The manuscript discloses extensive LLM generation followed by human checking and editing; full independent validation was not established in the evidence reviewed.

    Status reviewed

    Sources

    Related problems

    Problem family

    Explicit class field theory and Stark conjectures · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 31–46.

    Source-fit opponent count: 51.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  36. Rank036Prove the Penrose inequality or present a counterexampleFor isolated gravitational initial data in three spatial dimensions obeying the dominant energy condition, must an outermost apparent horizon force the total mass to satisfy m ≥ √(A/(16π)), where A is the least area enclosing it, with equality characterized by Schwarzschild geometry?Mathematical physics

    Exact statement

    Let (M,g,k) be a three-dimensional asymptotically flat initial data set for the Einstein equations satisfying the dominant energy condition and containing an outermost apparent horizon. If A is the least area needed to enclose that horizon and m_ADM is the ADM mass, prove the sharp spacetime Penrose inequality m_ADM >= sqrt(A/(16*pi)), with the appropriate Schwarzschild rigidity statement, or construct initial data satisfying the hypotheses that violate the inequality.

    Why it matters

    A proof would give a sharp lower bound on total mass from the least area enclosing the horizon and identify the equality geometry. A counterexample would show that the stated geometric and energy conditions permit a mass below the proposed area-based bound.

    Status

    Open · subcases solved

    Earlier note (2026-09-14): The sharp spacetime Penrose inequality remains open, with solved subcases. Located MOTS and geometric extensions retain extra hypotheses or a weaker constant; no full sharp unconditional proof was verified. October 7, 2026: OpenAI’s October 6 release reports partial progress. The release reports the three-dimensional enclosing-area inequality for complete, asymptotically flat initial data satisfying the dominant energy condition and the stated weakly future trapped boundary assumptions, with arbitrary second fundamental form and momentum. Its rigidity statements require additional outer area-minimizing, outermostness and decay conditions; these do not follow just from the apparent-horizon condition in the full formulation here. The numerical inequality is substantial progress in the stated class, while the complete inequality-plus-rigidity target remains unresolved. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Prove the Penrose inequality or present a counterexample · 1 problem in this edition. Family membership alone does not assert an implication.

    • #36 · Prove the Penrose inequality or present a counterexample (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 31–48.

    Source-fit opponent count: 40.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  37. Rank037Modularity of Elliptic Curves over Number Fields (End_K(E)=Z)For any number field K, does every elliptic curve whose K-defined endomorphisms are just integer multiplications have the predicted automorphic realization: an embedding of its finite-place representation into the cohomology of the associated arithmetic spaces?Number theory & arithmetic geometryLanglands functoriality and modularity · 2 problems

    Exact statement

    For every number field K and every elliptic curve E/K with End_K(E)=Z, prove that pi(E) admits a GL_2(A_K^infinity)-equivariant injection into A tensor_Q C, exactly as in Thorne's Conjecture 2.6. Here A is the direct limit over open compact U of H^*(Y_U,Q) for the adelic locally symmetric spaces defined on page 647, and pi(E) is the restricted tensor product at all finite places v of rec_{K_v}^{-1}(r_v tensor |.|^{1/2},N_v), with (r_v,N_v) attached to E in the source's geometric-Frobenius normalization on pages 646 and 649.

    Why it matters

    A proof would realize the arithmetic data encoded by the curve’s specified finite-place representation inside the cohomology of those spaces. The injection would respect the adelic group action, connecting the curve’s local representations with this geometric construction.

    Status

    Open · subcases solved

    Earlier note (2026-09-14): Modularity over arbitrary number fields remains open, with solved subcases for the stated End_K(E)=Z scope. Totally real tower and odd-degree uniformization results are restricted or conditional; End_K(E)=Z is not silently replaced by geometrically non-CM. October 7, 2026: OpenAI’s October 6 release reports partial progress. The manuscript reports modularity of every elliptic curve over every imaginary quadratic field, with all finite local parameters in the stated cohomological normalization. For the End_K(E)=Z formulation here, the eligible non-CM and induced geometrically CM cases supply the imaginary-quadratic subcase through the usual cuspidal cohomological realization. The universal target over all number fields remains unresolved. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Langlands functoriality and modularity · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 30–50.

    Source-fit opponent count: 45.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  38. Rank038Higher-dimensional Euclidean Kakeya conjectureIn four or more dimensions, must a compact set containing a unit line segment in every direction have full Hausdorff dimension? This asks for full fractal dimension, even when the set need not have positive volume.Analysis & PDEKakeya, restriction and smoothing estimates · 4 problems

    Exact statement

    Prove or disprove that for every integer n >= 4, every compact set E in R^n that contains a unit line segment in every direction has Hausdorff dimension n.

    Why it matters

    A proof would show that accommodating all directions forces full Hausdorff dimension in these spaces. A counterexample would contain every required segment within a compact set of smaller dimension.

    Status

    Open · subcases solved

    Earlier note (2026-09-14): The stated higher-dimensional Euclidean Kakeya target (n >= 4) remains open, with solved subcases. The solved three-dimensional theorem is outside that range. A February positive-volume preprint is unaccepted and does not overcome the arbitrary zero-measure Kakeya-set issue; September lifting results remain conditional. October 7, 2026: OpenAI’s October 6 release reports partial progress. The full Kakeya target requires full Hausdorff dimension in every dimension n at least four. The second manuscript proves precisely the n=4 subcase, even for noncompact and nonregular sets; dimensions n at least five remain outside the family. The separate three-dimensional maximal estimate does not turn this into an all-dimensional theorem. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Kakeya, restriction and smoothing estimates · 4 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 34–48.

    Source-fit opponent count: 53.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  39. Rank039Conformal Field Theory Limit of the Critical Three-Dimensional Ising ModelAt criticality, does the three-dimensional Ising model have a conformally invariant continuum limit with non-Gaussian spin fluctuations? The question asks for both the symmetry of the limit and spin correlations that do not factorize as they would in a Gaussian field.Mathematical physics3D Ising · 3 problems

    Exact statement

    Prove that the scaling limit of the critical three-dimensional Ising model exists and is conformally invariant, with a non-Gaussian spin field whose limiting spin correlations do not factorize according to Wick's rule.

    Why it matters

    A proof would connect the critical lattice model to a conformally invariant continuum theory with non-Gaussian spin fluctuations. It would also establish the separately listed non-triviality target.

    Status

    Open

    The critical three-dimensional lattice Ising CFT/scaling-limit target remains open with verification limits. September fuzzy-sphere numerical certification concerns another model and does not prove convergence or conformal invariance of the frozen lattice limit.

    Status reviewed

    Sources

    Related problems

    Problem family

    3D Ising · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 30–49.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  40. Rank040Hilbert's Sixteenth Problem, Second PartFor planar differential equations whose right-hand sides are polynomials of a fixed degree at least two, what is the maximum number of isolated closed trajectories, and how can they be arranged? Is that maximum finite for every degree?Probability, ergodic theory & dynamics

    Exact statement

    For every degree n >= 2, prove finiteness and determine the maximal number and possible configurations of limit cycles of planar polynomial vector fields of degree n.

    Why it matters

    A resolution would determine how polynomial degree limits the number and arrangement of isolated periodic motions in planar systems. It would provide a complete degree-by-degree description of the possibilities across all vector fields in scope.

    Status

    Open

    Earlier note (2026-09-14): Hilbert's sixteenth problem, part II, remains open. The checked purported quadratic bound faces a published critique; no universal bound for each degree was established. This is not a claim of a formal retraction absent such evidence. October 7, 2026: OpenAI’s October 6 release reports partial progress. The exact Hilbert XVI target has three requirements for every degree: uniform finiteness, the maximal number, and possible configurations of limit cycles. The first paper establishes a finite degree-dependent bound but states no exact maxima or configuration classification. The second settles a quintic Lienard subclass with maximum two, not all degree-five planar fields. Thus uniform boundedness is resolved while the full compound target remains only partially covered. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Hilbert's Sixteenth Problem, Second Part · 1 problem in this edition. Family membership alone does not assert an implication.

    • #40 · Hilbert's Sixteenth Problem, Second Part (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 33–51.

    Source-fit opponent count: 44.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  41. Rank041The Quantum PCP ConjectureEven allowing a fixed small error per interaction, is estimating the lowest energy of a locally interacting quantum system as hard as every problem whose yes answers have a quantum proof that a quantum computer can check efficiently?Quantum information & computation

    Exact statement

    Prove or disprove that approximating the ground-state energy of a local Hamiltonian to within a constant additive fraction of its number of local terms is QMA-hard.

    Why it matters

    A proof would preserve QMA-hardness even at this coarse energy precision. A disproof would rule out the proposed constant-fraction hardness statement for the specified local-Hamiltonian problem.

    Status

    Open

    Quantum PCP remains open. The August 2026 classical private-PCP advance explicitly distinguishes the unresolved QMA problem and is not a quantum-PCP proof.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    The Quantum PCP Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #41 · The Quantum PCP Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 35–50.

    Source-fit opponent count: 47.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  42. Rank042Unconditional Separation of BQP from the Polynomial HierarchyCan an efficient quantum computer solve a yes-or-no problem beyond every fixed level of the polynomial hierarchy, which allows successively nested layers of existential and universal verification? The quantum algorithm must have bounded error and use no oracle.Quantum information & computation

    Exact statement

    Prove in the standard unrelativized model that bounded-error quantum polynomial time BQP is not contained in the polynomial hierarchy PH.

    Why it matters

    A proof would place some efficient quantum computation beyond the entire polynomial hierarchy. The opposite conclusion would contain all bounded-error quantum polynomial-time decision problems within that hierarchy.

    Status

    Open

    Unrelativized BQP versus PH remains open with a dated exact-status-source warning. Recent oracle, query and promise-language results do not exhibit the required ordinary total language outside PH.

    Status reviewed

    Sources

    Problem family

    Unconditional Separation of BQP from the Polynomial Hierarchy · 1 problem in this edition. Family membership alone does not assert an implication.

    • #42 · Unconditional Separation of BQP from the Polynomial Hierarchy (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 30–55.

    Source-fit opponent count: 47.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  43. Rank043The Exponential Time Hypothesis for 3-SATThe 3-SAT problem asks whether Boolean variables can satisfy constraints with three literals per clause. Is there a positive constant c such that no deterministic algorithm solves all n-variable instances in time 2^(cn)?Theoretical computer science

    Exact statement

    Prove or disprove the Exponential Time Hypothesis: there exists a constant c > 0 such that 3-SAT cannot be solved in deterministic time 2^{c n}.

    Why it matters

    A proof would establish an exponential barrier to deterministic 3-SAT algorithms measured by variable count. A negative resolution would remove the asserted positive lower bound on the exponential rate.

    Status

    Open

    ETH remains open. The 2026 source treats positive-exponent hardness as a conjecture; conditional algorithmic lower bounds do not establish unrestricted exponential hardness.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    The Exponential Time Hypothesis for 3-SAT · 1 problem in this edition. Family membership alone does not assert an implication.

    • #43 · The Exponential Time Hypothesis for 3-SAT (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 36–55.

    Source-fit opponent count: 45.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  44. Rank044The Fontaine-Mazur Conjecture on Geometric Galois RepresentationsA Galois representation records arithmetic symmetries as matrices. Must every irreducible l-adic representation over a number field, unramified outside finitely many places and de Rham at primes above l—a local condition describing behavior expected from geometry—come from a smooth projective variety’s cohomology, allowing subquotients and Tate twists?Number theory & arithmetic geometry

    Exact statement

    Let K be a number field and l a prime. Every irreducible continuous finite-dimensional l-adic representation rho of Gal(Kbar/K), unramified outside finitely many places and de Rham at every place above l, occurs as a subquotient of H^i_et(X_Kbar)(m) for some smooth projective variety X over K, integer i >= 0, and integer Tate twist m. Etale cohomology is taken with the same l-adic coefficient field as rho. There is no nonnegative Hodge-Tate weight restriction.

    Why it matters

    A proof would make these arithmetic conditions sufficient for geometric realization across all number fields and finite dimensions. The realization may use any integer Tate twist, including negative twists, and imposes no nonnegative Hodge–Tate weight restriction.

    Status

    Open · subcases solved

    Earlier note (2026-09-16): The full conjecture remains open, with solved subcases. The one-dimensional case is known; Thorne’s August 2026 preprint treats restricted two-dimensional cases and does not establish the assertion for every number field and dimension. This status is inferred from the reviewed scopes, not a certification of the proofs. October 7, 2026: OpenAI’s October 6 release reports partial progress. The classical-modularity companion treats odd, regular, irreducible two-dimensional representations of G_Q at l=2, de Rham at 2 and finitely ramified. Modular forms supply geometric realizations in that setting, but Fontaine-Mazur quantifies over all number fields, dimensions and primes without oddness or distinct-weight restrictions. The separate completed-Hecke pro-modularity result explicitly does not assert classical modularity and must not be used to remove the de Rham condition. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    The Fontaine-Mazur Conjecture on Geometric Galois Representations · 1 problem in this edition. Family membership alone does not assert an implication.

    • #44 · The Fontaine-Mazur Conjecture on Geometric Galois Representations (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 37–54.

    Source-fit opponent count: 35.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  45. Rank045BQP versus QMAIf a quantum computer can efficiently check a supplied quantum proof for a promised yes-or-no problem, can it also solve that problem efficiently without receiving the proof? Both verification and solving allow bounded error.Quantum information & computation

    Exact statement

    For promise problems A=(A_yes,A_no) with disjoint sets of finite binary strings, use polynomial-time uniform quantum circuits over a fixed effectively specified finite universal gate set, with polynomially many gates and |0> ancillas and a measured output bit. BQP has such a circuit accepting each yes input with probability>=2/3 and each no input with probability<=1/3. QMA permits a polynomial-length quantum witness: each yes input has a witness accepted with probability>=2/3, and every witness on each no input is accepted with probability<=1/3. Outside the promise no condition is imposed. Is QMA contained in BQP, equivalently BQP=QMA?

    Why it matters

    Equality would give efficient quantum decision procedures for QMA-complete promise problems, including the standard local-Hamiltonian energy-threshold problem. Separation would show that quantum witnesses add decision power beyond efficient quantum computation alone.

    Status

    Open

    BQP versus QMA remains open for unrelativized uniform promise classes. Unitary/classical oracle results and August qIO work with an assumed average-case QMA gap do not decide that equality.

    Status reviewed

    Sources

    Problem family

    BQP versus QMA · 1 problem in this edition. Family membership alone does not assert an implication.

    • #45 · BQP versus QMA (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 39–55.

    Source-fit opponent count: 71.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  46. Rank046Fourier restriction conjectureDo all the conjectured sharp norm bounds hold when a Fourier transform is restricted to the unit sphere, in every dimension at least two? These bounds control the transformed function’s size on the sphere using the original function’s size in space.Analysis & PDEKakeya, restriction and smoothing estimates · 4 problems

    Exact statement

    Prove the conjectured sharp L^p(R^d)-to-L^q(S^{d-1}) boundedness range for restriction of the Fourier transform to the unit sphere in every dimension d at least 2, including the non-endpoint range specified by the standard Knapp necessary conditions.

    Why it matters

    A proof would establish the stated sphere-restriction estimates across the proposed range of exponents. A counterexample at an admissible pair would reveal an additional obstruction beyond the stated necessary conditions.

    Status

    Open · subcases solved

    Earlier note (2026-09-14): Fourier restriction remains open, with solved subcases. Sawyer's earlier full-proof version was withdrawn, while the current 2026 version treats averaged probabilistic estimates. Those do not settle the deterministic restriction conjecture. October 7, 2026: OpenAI’s October 6 release reports partial progress. The family proves bounded-data sphere extension for p>3 and diagonal extension L^p(Sigma) to L^p(R^3) for p>3 on compact smooth positively curved surfaces. These directly address three-dimensional restriction/extension estimates, with the stated boundary and curvature scope. The full restriction target requires the entire Knapp-admissible exponent range in every dimension at least two; the abstracts do not claim that all-dimensional conclusion. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Kakeya, restriction and smoothing estimates · 4 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 39–55.

    Source-fit opponent count: 42.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  47. Rank047Grothendieck's Standard Conjecture DOn a smooth projective variety, do rational algebraic cycles that give the same intersection numbers against every complementary cycle necessarily have the same class in every available Weil cohomology theory?Number theory & arithmetic geometryGrothendieck’s standard conjectures · 2 problems

    Exact statement

    For every field k, every Weil cohomology theory H* available over k, every smooth projective k-variety X of pure dimension n, every integer r with 0<=r<=n, and every rational algebraic cycle z of codimension r on X, if deg(z.w)=0 for every rational algebraic cycle w of codimension n-r on X, then the H* cycle class of z is zero. Equivalently, numerical and H*-homological equivalence of rational algebraic cycles agree in every codimension.

    Why it matters

    A proof would make intersection numbers sufficient to detect homological vanishing in every codimension and cohomology theory in scope. A counterexample would separate numerical equivalence from homological equivalence.

    Status

    Open · subcases solved

    Standard Conjecture D remains open, with solved subcases. The August 2026 abelian-fourfold claim does not cover all smooth projective varieties; earlier conditional uses neither prove the universal target nor corroborate the newest claim.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Grothendieck’s standard conjectures · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 40–59.

    Source-fit opponent count: 44.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  48. Rank048Grothendieck Period ConjecturePeriods encode integration data from algebraic geometry. For every Nori motive over the rationals, does the maximum number of algebraically independent periods equal the dimension of its motivic Galois group, which describes its algebraic symmetries?Number theory & arithmetic geometry

    Exact statement

    Prove that for every Nori motive M over Q, trdeg_Q Q(periods of M) equals dim G_mot(M), the dimension of its motivic Galois group.

    Why it matters

    A proof would match the algebraic independence measured by the period field with the dimension of the motive’s symmetry group. A counterexample would show that those two quantities can differ.

    Status

    Open

    Grothendieck's arithmetic period conjecture remains open. Geometric period theorems and selected CM Kummer-surface results do not prove the all-motives algebraic-relations assertion.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Grothendieck Period Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #48 · Grothendieck Period Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 42–60.

    Source-fit opponent count: 46.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  49. Rank049Non-Perturbative Construction and Non-Triviality of 4D Euclidean Phi4 Quantum Field TheoryCan a genuinely non-Gaussian scalar quantum field with quartic interaction be constructed in four Euclidean spacetime dimensions, satisfying the Osterwalder–Schrader axioms that make the Euclidean construction a rigorous quantum field theory?Mathematical physicsReviewed hold · conditional rank

    Exact statement

    Construct a non-trivial, non-Gaussian Euclidean scalar quantum field theory with quartic interaction (phi^4)_4 in four spacetime dimensions satisfying the Osterwalder-Schrader axioms.

    Why it matters

    A construction would realize the specified interacting field theory within those axioms. A general triviality theorem would instead exclude every such non-Gaussian construction in four dimensions.

    Status

    Reviewed hold · conditional rank

    Recorded hold reason. Retain open only with the inherited phi4 claim/scope warning. Accepted lattice-cutoff triviality is not a universal no-go theorem for every OS formulation. The separate nontrivial/OS-positivity construction claim lacks located independent acceptance; August mean-field-sector triviality is narrower. No positive or universal negative resolution is certified.

    The phi4 target remains open, with a claim and scope warning. Accepted lattice-cutoff triviality is not a universal no-go theorem for every OS formulation. Independent acceptance of the separate nontrivial/OS-positivity construction claim was not established; August mean-field-sector triviality is narrower. No positive or universal negative resolution is certified.

    Status reviewed

    Sources

    Problem family

    Non-Perturbative Construction and Non-Triviality of 4D Euclidean Phi4 Quantum Field Theory · 1 problem in this edition. Family membership alone does not assert an implication.

    • #49 · Non-Perturbative Construction and Non-Triviality of 4D Euclidean Phi4 Quantum Field Theory (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 46–64.

    Source-fit opponent count: 45.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  50. Rank050Equivariant Tamagawa number conjecture: Burns-Flach Conjecture4 packageFor a motive over a number field with extra symmetries, possibly noncommuting, does its equivariant L-function continue to s = 0, allowing poles, with the predicted zero or pole orders and rational and integral leading-term identities from its arithmetic cohomology? This is conditional on the Burns–Flach hypotheses on motivic cohomology, comparison maps, coefficient compatibility and coherence.Number theory & arithmetic geometry

    Exact statement

    For every motive M over a number field K in Burns-Flach Section3.1's motivic-structure formalism, with an action of a finite-dimensional semisimple Q-algebra A, and every order mathcal A in A admitting the compatible projective mathcal A-structure of Section3.3, prove Burns-Flach Conjecture4(i)-(iv) in its stated conditional setting. The external defining hypotheses are the finite motivic cohomology and real comparison exact sequence of Conjecture1, the motivic-to-p-adic comparison of Conjecture2 for M and Mdual(1), coefficient compatibility giving well-defined equivariant local factors as in4.1 (Conjecture3 or the weaker Euler-factor compatibility of Remark7), and the stated Coherence hypothesis. A projective structure consists of full projective lattices in the archimedean realizations whose l-adic comparison images agree and are Galois-stable, not merely an arbitrary maximal-order choice. The INCLUDED assertions are: (i) the finite equivariant L-function L(AM,s) has meromorphic continuation to0; (ii) its componentwise integer order r equals rr_A(H_f^1(K,Mdual(1))^*)-rr_A(H_f^0(K,Mdual(1))^*), with reduced rank as in2.6; (iii) for the nonzero componentwise leading Laurent coefficient L*(AM,0)=lim_(s->0) s^(-r)L(AM,s), the class T_Omega=delta_hat^1_(mathcal A,R)(L*(AM,0))+R_Omega(M,mathcal A) lies in Cl(mathcal A,Q); and (iv) T_Omega=0 in Cl(mathcal A,R), hence in K_0(mathcal A,R). The R_Omega term is the class of the fundamental virtual object with its integral compact-support cohomology structures and real comparison trivialization in3.4; delta_hat is Lemma9's canonical extended reduced-norm boundary, not an inverse to a surjective real reduced norm. Equivalently the final identity is delta_hat(L*)=-R_Omega. Preserve graded determinant conventions and all suitable nonmaximal/noncommutative orders. Coherence-free Remark8's Conjecture5 and Conjecture6 for EVERY prime are an alternative to the rational/integral clauses in the source's stated sense, not removal of(i)-(ii) or replacement by a single p-local or epsilon assertion. The external hypotheses are not presumed universally proved or restricted to currently established cases, and proving all of them is not added as a separate conclusion.

    Why it matters

    Under its stated hypotheses, the conjecture relates an equivariant L-function’s behavior at zero to arithmetic cohomology, including an integral leading-term identity. This relation must hold for every suitable order, including nonmaximal and noncommutative orders.

    Status

    Open · subcases solved

    The full Burns–Flach ETNC package remains open, with solved subcases. Recent conditional p-parts, modular motives and restricted Tate motives do not prove all four clauses for every motive and suitable order; specialist package-level currency remains qualified.

    Status reviewed

    Sources

    Problem family

    Equivariant Tamagawa number conjecture: Burns-Flach Conjecture4 package · 1 problem in this edition. Family membership alone does not assert an implication.

    • #50 · Equivariant Tamagawa number conjecture: Burns-Flach Conjecture4 package (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 45–64.

    Source-fit opponent count: 66.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  51. Rank051Equivalence of Worst-Case and Average-Case Hardness for NP (Excluding Heuristica)Suppose some efficiently verifiable decision problem resists polynomial-size circuits even when a separate circuit is chosen for each input length. Must some distributional NP problem—a problem together with its input distribution—also be hard on average, admitting no average-polynomial-time algorithm?Theoretical computer science

    Exact statement

    Prove or disprove that if NP is hard in the worst case (NP not in P/poly), then NP is hard on average (DistNP not in AvgP).

    Why it matters

    A proof would turn the stated worst-case circuit hardness assumption into average-case hardness. It would rule out the combination of that worst-case barrier with average-polynomial-time algorithms for every distributional NP problem.

    Status

    Open

    The exact NP not in P/poly implies DistNP not in AvgP target remains provisionally open. Located reductions add iO or quantitative hypotheses and do not recertify every stated nonuniform/AvgP convention; exact-formulation evidence remains limited.

    Status reviewed

    Sources

    Problem family

    Equivalence of Worst-Case and Average-Case Hardness for NP (Excluding Heuristica) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #51 · Equivalence of Worst-Case and Average-Case Hardness for NP (Excluding Heuristica) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 47–64.

    Source-fit opponent count: 38.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  52. Rank052Dickson's Conjecture on Simultaneous Prime Values of Linear FormsFor any finite family of integer linear forms with positive slopes, must infinitely many positive inputs make all values prime if no prime divides their product at every input?Number theory & arithmetic geometrySimultaneous prime values · 5 problems

    Exact statement

    For every integer k>=1 and every family f_i(n)=a_i*n+b_i (1<=i<=k) with integers a_i>=1 and b_i, if for each prime p some integer n has p not dividing the product of the f_i(n), then infinitely many positive integers n make every f_i(n) prime.

    Why it matters

    A proof would make the absence of these divisibility obstructions sufficient for infinitely many simultaneous prime values. A counterexample would reveal an additional obstruction for a particular family of linear forms.

    Status

    Open · subcases solved

    Dickson remains open, with solved subcases. May 2026 exposition explicitly retains the conjecture; Xiao's unaccepted Hypothesis H claim would imply the linear case, but no independent acceptance was established. Other inaccessible certificate/proof leads were not adopted.

    Status reviewed

    Sources

    Related problems

    Problem family

    Simultaneous prime values · 5 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 45–68.

    Source-fit opponent count: 50.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  53. Rank053Existence of One-Way PermutationsCan one polynomial-time algorithm permute binary strings bijectively at every length n, while every randomized polynomial-time algorithm reverses a random output with success smaller than every inverse power of n for sufficiently large n?Cryptography, coding, information & optimizationOne-way functions and permutations · 2 problems

    Exact statement

    There exists a family f_n:{0,1}^n->{0,1}^n, n>=1, of bijections computed by one deterministic polynomial-time algorithm on (1^n,x), such that for every probabilistic polynomial-time algorithm A its inversion success Pr[A(1^n,f_n(x))=x], over uniform x and A's coins, is negligible in n: for every c>0 it is at most n^(-c) for all sufficiently large n. This is unconditional existence in the classical unrelativized model, with no key sampling or variable domains.

    Why it matters

    A proof would establish computationally hard inversion even when every output has exactly one preimage and the domain is all n-bit strings. The guarantee would hold uniformly across lengths in the classical model without oracles or sampled keys.

    Status

    Open

    Unconditional one-way permutations remain open. Full-domain trapdoor constructions require iO and one-way functions; query inversion taking square-root domain size is exponential in bit length and does not refute classical polynomial-time security.

    Status reviewed

    Sources

    Related problems

    Problem family

    One-way functions and permutations · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 45–68.

    Source-fit opponent count: 42.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  54. Rank054The Bombieri-Lang Conjecture for Varieties of General TypeFor every positive-dimensional variety of general type over a number field K, must its K-rational points lie in a proper algebraic subset, rather than being Zariski dense in the whole variety?Number theory & arithmetic geometry

    Exact statement

    Prove that for every variety X of general type defined over a number field K, the set of K-rational points X(K) is not Zariski dense in X.

    Scope qualification. Interpret the inherited Bombieri-Lang target as the standard positive-dimensional general-type non-density conjecture over number fields. The zero-dimensional point is excluded; it is not a claimed counterexample to that canonical problem. The Campana positive implication is expressly limited to this standard scope.

    Why it matters

    A proof would give a broad geometric reason that rational points cannot fill a variety. It would connect the general-type condition with arithmetic scarcity across all positive dimensions.

    Status

    Open

    Bombieri–Lang remains open for arithmetic non-density over number fields. The July 2026 specialist survey says so; the geometric function-field finite-to-abelian theorem does not resolve the arithmetic target.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    • #84 · Campana's Conjecture Characterizing Special Varieties by Potential Density of Rational PointsStronger assertionCampana's Conjecture Characterizing Special Varieties by Potential Density of Rational Points

      For positive-dimensional varieties of general type, this non-density assertion follows from Campana's implication that potential density requires specialness. No converse to Campana's full conjecture is established here.

      Condition: The stronger assertion is proved; a negative answer to it need not settle the weaker assertion.

      Scope qualification: The Bombieri–Lang statement carries its recorded positive-dimensional scope qualification. Only the positive-answer implication is recorded. No converse or equivalence of resolution outcomes is established here.

    Problem family

    The Bombieri-Lang Conjecture for Varieties of General Type · 1 problem in this edition. Family membership alone does not assert an implication.

    • #54 · The Bombieri-Lang Conjecture for Varieties of General Type (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 48–65.

    Source-fit opponent count: 54.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  55. Rank055Schinzel's Hypothesis H on Simultaneous Prime Values of Irreducible PolynomialsCan any fixed collection of irreducible integer polynomials with positive leading coefficients take prime values together at infinitely many positive inputs, provided no prime divides their product at every input? Irreducible means the polynomial cannot be factored further over the integers.Number theory & arithmetic geometrySimultaneous prime values · 5 problems

    Exact statement

    Let f1,...,fr in Z[x] be irreducible with positive leading coefficients and no prime dividing f1(n)...fr(n) for every n. Prove that infinitely many positive integers n make every fi(n) prime.

    Why it matters

    A proof would make the absence of the stated divisibility obstruction sufficient for infinitely many simultaneous prime values. A counterexample would reveal a polynomial family with another obstruction to that conclusion.

    Status

    Open · subcases solved

    Schinzel H remains open, with solved subcases, with an unaccepted full-proof claim warning. Xiao's June 2025 posting matches the polynomial target but is explicitly unrefereed; later conditional use does not independently adjudicate it.

    Status reviewed

    Sources

    Related problems

    Problem family

    Simultaneous prime values · 5 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 48–67.

    Source-fit opponent count: 44.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  56. Rank056Novikov Conjecture on homotopy invariance of higher signaturesHigher signatures are numbers combining a manifold’s geometry with information about its fundamental group. For closed oriented smooth manifolds, do all rational higher signatures stay unchanged under orientation-preserving homotopy equivalence?Geometry & topology

    Exact statement

    For every closed oriented smooth manifold M, every u in H^*(B pi_1(M); Q), and every classifying map r, prove that the higher signature <L(M) cup r^*u,[M]> is invariant under orientation-preserving homotopy equivalence.

    Why it matters

    A proof would make these higher signatures invariants of oriented homotopy type throughout the stated class of manifolds. A counterexample would give equivalent homotopy types with different higher-signature values.

    Status

    Open · subcases solved

    Novikov remains open, with solved subcases. Finite decomposition complexity and the August 2026 finite-complexity Banach-embedding criterion retain group restrictions. Embedding all groups in a Banach space does not remove the finite-complexity hypothesis.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Novikov Conjecture on homotopy invariance of higher signatures · 1 problem in this edition. Family membership alone does not assert an implication.

    • #56 · Novikov Conjecture on homotopy invariance of higher signatures (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 50–65.

    Source-fit opponent count: 50.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  57. Rank057L = P ProblemCan every decision problem solvable by a deterministic polynomial-time algorithm also be solved deterministically using only logarithmically many working-memory bits relative to its input length?Theoretical computer science

    Exact statement

    Prove or disprove that every language decidable by a deterministic polynomial-time Turing machine is decidable using O(log n) work space; equivalently, determine whether the complexity classes L and P are equal.

    Why it matters

    Equality would give logarithmic-workspace algorithms for every polynomial-time language. Separation would establish that some efficiently solvable decision problems require more working memory than that bound permits.

    Status

    Open

    L versus P remains open with source-quality and recency limits. Fixed polynomial time-space tradeoffs and square-root-space simulation do not give logarithmic space for every polynomial-time language.

    Status reviewed

    Sources

    Problem family

    L = P Problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #57 · L = P Problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 51–71.

    Source-fit opponent count: 40.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  58. Rank058Inverse Galois problem over the rational numbersCan every finite group occur as the Galois group—the group of field symmetries—of some finite Galois extension of the rational numbers?Algebra, representation & category theoryInverse Galois over Q · 2 problems

    Exact statement

    Prove or disprove that every finite group G is isomorphic to the Galois group Gal(L/Q) of some finite Galois extension L of the rational numbers Q.

    Why it matters

    A proof would realize every finite abstract group as arithmetic symmetries over the rationals. A disproof would identify a finite group excluded from all such Galois extensions.

    Status

    Open · subcases solved

    The inverse Galois problem over Q remains open, with solved subcases. The August 2026 M23 preprint reports one finite group, not realization of every finite group; favorable informal commentary is not journal-proof certification.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    • #111 · Regular Inverse Galois Problem over QStronger assertionRegular Inverse Galois Problem over Q

      An affirmative solution of the regular problem over Q(T) would prove this assertion over Q. A converse implication is not established here.

      Condition: The stronger assertion is proved; a negative answer to it need not settle the weaker assertion.

      Scope qualification: Only the positive-answer implication is recorded. No converse or equivalence of resolution outcomes is established here.

    Problem family

    Inverse Galois over Q · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 53–70.

    Source-fit opponent count: 48.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  59. Rank059Permanent versus Determinant over CThe permanent is the determinant-like sum with all permutation terms positive. If an n-by-n permanent is represented exactly as a determinant with affine-linear expressions in the original entries, must the required matrix size escape every polynomial bound in n, over the complex numbers?Theoretical computer scienceAlgebraic computation lower bounds · 3 problems

    Exact statement

    Over C, let dc(per_n) be the least m for which the n×n permanent per_n(X) equals the determinant of an m×m matrix whose entries are affine-linear forms in the n^2 variables X, as an exact polynomial identity. Assert that dc(per_n) is not O(n^c) for any real constant c>0: explicitly, for every c>0, C>0 and integer N≥1, some n≥N satisfies dc(per_n)>C*n^c. No restrictions of equivariance, sparsity or symmetry are imposed on the representing matrix. This is exact, not border, determinantal complexity over C; the selected assertion is the source’s VP_ws != VNP formulation, not an assertion that it is equivalent to VP != VNP.

    Why it matters

    A proof would separate the permanent from polynomial-size weakly-skew algebraic computation. A polynomial-size determinant representation would refute this separation, even if no uniform algorithm for producing those representations were supplied.

    Status

    Open

    Earlier note (2026-09-14): Permanent versus determinant remains open for unrestricted exact affine determinantal complexity over C. Quadratic lower bounds, border results and power-sum polynomials do not prove superpolynomial permanent growth; exact current-status evidence remains partly dated. October 7, 2026: OpenAI’s October 6 release reports partial progress. The border bound is over C and allows arbitrary affine-linear determinant representations, so it implies the same Omega(n^3) lower bound for exact determinantal complexity in the full model. The full assertion requires growth beyond every fixed polynomial n^c. A cubic bound remains polynomial and is therefore partial lower-bound progress, not permanent-versus-determinant separation. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Related problems

    Problem family

    Algebraic computation lower bounds · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 54–75.

    Source-fit opponent count: 57.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  60. Rank060NEXP versus P/polyIf a decision problem’s yes answers have certificates whose length and verification time are at most exponential in a polynomial of input length, must polynomial-size Boolean circuits decide it, with a separate circuit chosen for every input length?Theoretical computer science

    Exact statement

    Determine whether NEXP (nondeterministic exponential time) is contained in P/poly (the class of languages computable by polynomial-size Boolean circuit families).

    Why it matters

    Containment would give small circuits for every language in NEXP. Separation would establish that at least one nondeterministic exponential-time language requires circuit sizes beyond every polynomial bound.

    Status

    Open

    NEXP versus P/poly remains open on bounded evidence. Consistency of lower bounds with bounded arithmetic does not prove them in the standard model; the independent exact-status notes have no authenticated recent update date.

    Status reviewed

    Sources

    Problem family

    NEXP versus P/poly · 1 problem in this edition. Family membership alone does not assert an implication.

    • #60 · NEXP versus P/poly (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 53–77.

    Source-fit opponent count: 43.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  61. Rank061Vojtas Main Conjecture on Height Inequalities in Diophantine ApproximationHeights measure the arithmetic size of rational points. For every positive ε, can one remove a proper algebraic subset of a smooth projective variety over a number field so that proximity to a simple-normal-crossings divisor plus canonical height is bounded by ε times an ample height, up to an additive constant?Number theory & arithmetic geometry

    Exact statement

    Let k be a number field, X a smooth projective variety over k, K_X a canonical divisor, A an ample divisor, D an effective simple normal-crossings divisor, and S a finite set of places containing the archimedean places. Prove that for every epsilon > 0 there are a proper Zariski-closed subset Z of X and a constant C such that m_{D,S}(P)+h_{K_X}(P) < epsilon h_A(P)+C for every P in X(k) outside Z.

    Why it matters

    A proof would uniformly control the stated combination of proximity and canonical height outside a proper algebraic exceptional set. A counterexample would require violations that cannot be confined by any permitted exceptional set and constant.

    Status

    Open

    Vojta's number-field height inequality remains open. Special integral-point results and function-field analogues change the quantifiers or ground field and do not establish every stated SNC-divisor inequality.

    Status reviewed

    Sources

    Problem family

    Vojtas Main Conjecture on Height Inequalities in Diophantine Approximation · 1 problem in this edition. Family membership alone does not assert an implication.

    • #61 · Vojtas Main Conjecture on Height Inequalities in Diophantine Approximation (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 59–77.

    Source-fit opponent count: 44.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  62. Rank062Modularity of Elliptic Curves over Totally Real FieldsFor an elliptic curve over a totally real number field, must there be a Hilbert cuspidal newform of parallel weight two with exactly the same L-function? Totally real means every embedding of the field into the complex numbers has real image.Number theory & arithmetic geometry

    Exact statement

    For every totally real number field K and every elliptic curve E over K, prove that there is a Hilbert cuspidal newform f over K of parallel weight 2 satisfying L(E,s)=L(f,s).

    Why it matters

    A proof would match each elliptic curve in scope to an automorphic form through equality of their L-functions. A counterexample would identify a curve for which no form of the specified kind provides that match.

    Status

    Open · subcases solved

    Totally real modularity remains open, with solved subcases. The accessible theorem covers restricted towers with additional conditions, and finite database verification is not all curves over all fields. The final 2026 journal text could not be verified, so source currency remains qualified.

    Status reviewed

    Sources

    Problem family

    Modularity of Elliptic Curves over Totally Real Fields · 1 problem in this edition. Family membership alone does not assert an implication.

    • #62 · Modularity of Elliptic Curves over Totally Real Fields (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 60–78.

    Source-fit opponent count: 55.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  63. Rank063Strong Exponential Time Hypothesis (SETH)As the allowed number of literals per Boolean clause grows, must the best satisfiability algorithms approach the brute-force exponent in 2^n, where n counts variables? The claim is that no fixed improvement of the exponent below one works for every fixed clause width.Theoretical computer science

    Exact statement

    Prove or disprove that the optimal exponent s_k for k-SAT satisfies lim_{k to infinity} s_k = 1.

    Why it matters

    A proof would show that improvements in the exponential rate cannot stay uniformly below one as clause width increases. A disproof would establish a gap below that proposed limiting exponent.

    Status

    Open

    SETH remains open. Restricted resolution-over-parities lower bounds are not bounds against every deterministic k-SAT algorithm; 2026 scheduling work uses SETH as an assumption.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Strong Exponential Time Hypothesis (SETH) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #63 · Strong Exponential Time Hypothesis (SETH) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 63–78.

    Source-fit opponent count: 48.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  64. Rank064The Farrell-Jones Conjecture in Algebraic K- and L-TheoryCan the algebraic K- and L-theory of an integer group ring always be reconstructed from the data attached to its virtually cyclic subgroups—those containing a cyclic subgroup of finite index—through the Farrell–Jones assembly maps?Algebra, representation & category theory

    Exact statement

    Prove the Farrell-Jones conjecture for every group G: the assembly maps from the equivariant homology of the classifying space for the family of virtually cyclic subgroups to the algebraic K-theory, respectively L-theory, of the group ring ℤG are isomorphisms.

    Why it matters

    A proof would express both theories through the stated equivariant homology and assembly construction. A counterexample would locate a group where at least one map loses information or misses classes.

    Status

    Open · subcases solved

    Farrell–Jones remains open, with solved subcases. June 2026 primary work explicitly retains the general conjecture while proving specified group classes; broader coefficient frameworks do not remove group restrictions.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    The Farrell-Jones Conjecture in Algebraic K- and L-Theory · 1 problem in this edition. Family membership alone does not assert an implication.

    • #64 · The Farrell-Jones Conjecture in Algebraic K- and L-Theory (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 55–93.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  65. Rank065Artins Conjecture on the Holomorphy of Non-Abelian Artin L-FunctionsDoes every Artin L-function attached to a nontrivial irreducible Galois representation extend to a holomorphic function on the entire complex plane, with no poles anywhere?Number theory & arithmetic geometry

    Exact statement

    Prove that every non-trivial irreducible Artin L-function L(s, rho) associated to a Galois representation rho is an entire holomorphic function on the whole complex plane.

    Why it matters

    A proof would establish entire continuation uniformly across the stated representation family. A counterexample would identify a nontrivial irreducible representation whose L-function fails that analytic property.

    Status

    Open

    Artin holomorphy remains open for all nontrivial irreducible Artin representations. The 2025 primary source and independent expert exposition retain the general question; special groups and criteria do not establish universal entireness.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Artins Conjecture on the Holomorphy of Non-Abelian Artin L-Functions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #65 · Artins Conjecture on the Holomorphy of Non-Abelian Artin L-Functions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 62–82.

    Source-fit opponent count: 46.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  66. Rank066Finiteness of the Tate-Shafarevich Group for Elliptic CurvesFor every elliptic curve over a number field, are there only finitely many classes in its Tate–Shafarevich group, which measures obstructions to turning local solutions into global ones?Number theory & arithmetic geometry

    Exact statement

    Prove that the Tate-Shafarevich group Sha(E/K) is finite for every elliptic curve E defined over a number field K.

    Why it matters

    A proof would establish finiteness separately for each elliptic curve and number field in scope. A counterexample would exhibit one with infinitely many Tate–Shafarevich classes; no uniform bound on all group sizes is requested.

    Status

    Open

    Earlier note (2026-09-14): Sha finiteness for every elliptic curve over every number field remains open. The September 8 rank-two CM p-primary criterion and five-curve computations are not whole-group universal finiteness. A separate inaccessible proof-claim lead remains unverified. October 7, 2026: OpenAI’s October 6 release reports partial progress. Finiteness of the entire Tate-Shafarevich group is explicitly asserted for E/Q of q-power Selmer corank zero or one at some q. The target quantifies over every elliptic curve over every number field. Neither higher Selmer corank nor general number fields is covered. The companion explicitly proves finiteness of Sha for rational-two-torsion curves over Q of 2-power Selmer corank zero or one. The universal Sha target allows every elliptic curve over every number field; neither the torsion hypothesis, low corank, nor the base field restriction can be dropped. The low-corank 2-converse includes finiteness of the entire Tate-Shafarevich group in its stated E/Q subcase. Even in combination with the twist-density statements, exceptional twists and curves over general number fields are not universally covered by this family. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Finiteness of the Tate-Shafarevich Group for Elliptic Curves · 1 problem in this edition. Family membership alone does not assert an implication.

    • #66 · Finiteness of the Tate-Shafarevich Group for Elliptic Curves (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 58–89.

    Source-fit opponent count: 67.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  67. Rank067P versus NC^1Can we exhibit a polynomial-time decision problem whose Boolean formulas built from AND, OR and NOT cannot have size bounded by any polynomial in the input length? Each length may have its own formula, but formulas cannot reuse intermediate results as circuits can.Theoretical computer science

    Exact statement

    Exhibit a Boolean function family computable in deterministic polynomial time whose nonuniform De Morgan formula size is superpolynomial; equivalently, prove that P is not contained in nonuniform NC^1.

    Why it matters

    A proof would show that polynomial-time computation cannot always be compressed into polynomial-size nonuniform formulas. It would separate P from containment in nonuniform NC¹ through an explicit kind of formula-size barrier.

    Status

    Open

    P versus nonuniform NC^1 remains open with dated-source qualification. Formula-composition progress does not establish the required superpolynomial De Morgan formula lower bound and is not the separate uniform P versus union-NC problem.

    Status reviewed

    Sources

    Problem family

    P versus NC^1 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #67 · P versus NC^1 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 59–91.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  68. Rank068Derandomization of Polynomial Identity TestingCan a deterministic polynomial-time algorithm decide whether any given arithmetic circuit computes the zero polynomial, so that its output vanishes identically rather than just at selected inputs?Theoretical computer science

    Exact statement

    Determine whether polynomial identity testing for arithmetic circuits (deciding whether a given arithmetic circuit computes the identically zero polynomial) can be derandomized into deterministic polynomial time.

    Why it matters

    A positive answer would give an efficient identity test without random choices for every arithmetic circuit in scope. A negative answer would exclude every deterministic polynomial-time algorithm for this testing task.

    Status

    Open

    General arithmetic-circuit PIT derandomization remains open. July 2026 read-4 formula algorithms and nonassociative-algebra results do not cover arbitrary ordinary associative arithmetic circuits.

    Status reviewed

    Sources

    Problem family

    Derandomization of Polynomial Identity Testing · 1 problem in this edition. Family membership alone does not assert an implication.

    • #68 · Derandomization of Polynomial Identity Testing (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 61–87.

    Source-fit opponent count: 37.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  69. Rank069Deligne conjecture on coefficient-equivariant critical motivic L-valuesFor a pure motive over the rationals, do its critical L-values, divided by the predicted periods and powers of 2πi, come from one element of its coefficient number field across every complex embedding? This asks whether geometric periods account for these values, assuming compatible realizations and analytic continuation giving finite values at those integers.Number theory & arithmetic geometry

    Exact statement

    For every pure motive M over Q with coefficients in a number field E, in Deligne's geometric realization formalism (Betti, de Rham and compatible l-adic realizations with comparison, dual, tensor and Tate-twist structures), and every critical integer n, the normalized finite L-value vector L(M,n)/((2*pi*i)^(n*d_sign)*c^sign(M)) lies in the diagonal E inside E tensor C, where sign=(-1)^n and d_sign=dim_E H_B^sign(M). Here H_B^sign is the sign eigenspace of complex conjugation, and c^sign is the determinant of the Betti/de Rham comparison to the corresponding filtration quotient, computed in E-rational bases and defined modulo E-units as in Deligne1.7/2.6. Critical means that L_infinity(sigma,M,s) and L_infinity(sigma,Mdual,1-s) have no pole at s=n for every embedding sigma:E->C (equivalently use Mdual(1) at -s). Use finite, not completed, L-functions with the compatible E-valued local factors of2.2. Analytic continuation giving finite values at the selected critical integer and the source's well-defined compatible realizations/local factors are defining prerequisites, not assertions proved by this period law and not a restriction to cases whose continuation is already known. The simultaneous embedding vector must come from one element of E; unrelated componentwise algebraicity is insufficient. All-zero values are allowed, but mixed zero/nonzero components do not satisfy the diagonal membership. No general nonvanishing or order-of-zero assertion from Deligne2.7 is appended.

    Why it matters

    A proof would relate critical L-values to Betti–de Rham periods compatibly across all coefficient embeddings. The requirement is one shared algebraic element; separate algebraicity of unrelated components would not suffice.

    Status

    Open · subcases solved

    Deligne critical values remain open, with solved subcases for the coefficient-equivariant universal motive target. Recent CM-automorphic results retain regularity/rationality hypotheses; numerical checks on curves are not universal proofs.

    Status reviewed

    Sources

    Problem family

    Deligne conjecture on coefficient-equivariant critical motivic L-values · 1 problem in this edition. Family membership alone does not assert an implication.

    • #69 · Deligne conjecture on coefficient-equivariant critical motivic L-values (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 64–82.

    Source-fit opponent count: 57.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  70. Rank070FPT versus W[1]Can we deterministically decide whether an n-vertex graph contains k mutually connected vertices in time f(k)n^c, for some computable function f and fixed constant c independent of k? This would confine any non-polynomial growth in the bound to the chosen clique size k.Theoretical computer science

    Exact statement

    Prove or disprove FPT!=W[1] in classical uniform parameterized complexity. Equivalently, determine whether there is a deterministic algorithm, a total computable function f:N->N and an absolute constant c such that on every finite simple n-vertex graph G and every integer 1<=k<=n it decides the existence of a k-clique within f(k)n^c steps. The conjecture asserts no such algorithm. The function and exponent are fixed with the algorithm; no input-length advice, oracle or nonuniform circuit family is allowed.

    Scope qualification. FPT != W[1] would imply P != NP: P = NP would give a polynomial-time Clique decider, which is an allowed FPT decider. The equality outcome FPT = W[1] allows arbitrary computable parameter dependence and does not itself supply a polynomial-time Clique algorithm in the full input size. The completion criterion remains the stated uniform parameterized class question; no additional theorem is a separate requirement.

    Why it matters

    Proving that no such uniform algorithm exists would separate FPT from W[1] and imply P ≠ NP. An FPT algorithm would allow arbitrary computable dependence on k, so it would not by itself give polynomial running time in the full input size.

    Status

    Open

    FPT versus W[1] remains open. Yi's counting-collapse claim was explicitly withdrawn May 10, 2026; independent 2026 work retains the separation as an assumption. The ordinary uniform k-Clique target is unchanged.

    Status reviewed

    Sources

    Problem family

    FPT versus W[1] · 1 problem in this edition. Family membership alone does not assert an implication.

    • #70 · FPT versus W[1] (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 62–91.

    Source-fit opponent count: 43.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  71. Rank071Termination of Flips Conjecture in the Minimal Model ProgramThe minimal model program simplifies algebraic varieties through controlled transformations called flips. For any projective log canonical pair over the complex numbers, must every sequence of flips end after finitely many steps, in every dimension?Geometry & topologyMinimal model conjectures · 1 problem

    Exact statement

    Prove that there is no infinite sequence of flips in the minimal model program: for a projective log canonical (or klt) pair over the complex numbers, every sequence of (K_X + Delta)-flips terminates after finitely many steps, in every dimension.

    Why it matters

    A proof would guarantee that the flip stage cannot continue indefinitely, regardless of which permitted sequence is chosen. This would remove infinite flipping as an obstruction in the stated minimal-model setting in every dimension.

    Status

    Open · subcases solved

    Earlier note (2026-09-14): Termination of flips remains open, with solved subcases. The 2026 Enriques theorem is restricted, and other all-sequence or dimension-four implications retain unproved hypotheses; none proves arbitrary flip-sequence termination in all dimensions. October 7, 2026: OpenAI’s October 6 release reports partial progress. The projective manuscript proves termination of arbitrary permitted MMPs for lc fourfolds with rational boundary over algebraically closed characteristic-zero fields, allowing arbitrary rays and mixed steps. This includes the complex fourfold rational-boundary flip-sequence subcase. The full target requires termination in every dimension and has no fourfold limitation. The Kähler companions broaden geometry but remain dimension four and concern existing diagrams, so they do not close the universal target. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Minimal model conjectures · 1 problem in this edition. Family membership alone does not assert an implication.

    • #71 · Termination of Flips Conjecture in the Minimal Model Program (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 66–91.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  72. Rank072Polynomial-Time Solvability of Mean-Payoff GamesCan a deterministic algorithm decide in polynomial time whether Max can force a nonnegative long-run average payoff in a finite two-player game with binary-encoded integer edge weights?Theoretical computer sciencePolynomial-time algorithms for finite games · 3 problems

    Exact statement

    Does there exist a deterministic Turing-machine algorithm that, given a finite directed graph with at least one outgoing edge per vertex, a partition of its vertices between Max and Min, signed integer edge weights encoded in binary, and a designated starting vertex, decides in time polynomial in the complete input bit length whether Max has a strategy forcing liminf as T tends to infinity of (1/T) times the sum of the first T edge weights to be at least zero against every Min strategy?

    Also known as

    Mean-payoff games in P; Deterministic polynomial-time mean-payoff threshold problem; Polynomial-time binary-weight mean-payoff games

    Why it matters

    The running-time bound must depend polynomially on the full input length, including the number of bits used for each weight. Such an algorithm would also give a polynomial-time algorithm for parity games.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The deterministic ordinary-game manuscript computes the full zero-threshold winning set for arbitrary signed binary weights, exactly covering the full input and liminf objective, and also computes values and positional strategies. Its stated bit complexity is 2^{O((log(L+2))^2)}, however, which is quasipolynomial rather than polynomial in the complete input length. This is an algorithmic bound toward the target, not a positive solution of polynomial-time solvability. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Related problems

    • #360 · Polynomial-Time Solvability of Condon's Simple Stochastic GamesConditional implicationPolynomial-Time Solvability of Condon's Simple Stochastic Games

      A proof of #360 would also establish this assertion.

      Condition: An affirmative simple-stochastic-games-in-P theorem implies deterministic-mean-payoff-in-P. Its contrapositive transfers a negative mean-payoff answer upward; a negative simple-stochastic-games answer need not resolve deterministic mean payoff.

      Scope qualification: An affirmative simple-stochastic-games-in-P theorem implies deterministic-mean-payoff-in-P. Its contrapositive transfers a negative mean-payoff answer upward; a negative simple-stochastic-games answer need not resolve deterministic mean payoff.

    • #73 · Polynomial-Time Solvability of Parity GamesConditional implicationPolynomial-Time Solvability of Parity Games

      A proof of this assertion would also establish #73.

      Condition: An affirmative mean-payoff-in-P theorem implies parity-in-P; its contrapositive transfers a negative parity answer upward. A negative mean-payoff answer need not resolve parity.

      Scope qualification: An affirmative mean-payoff-in-P theorem implies parity-in-P; its contrapositive transfers a negative parity answer upward. A negative mean-payoff answer need not resolve parity.

    Problem family

    Polynomial-time algorithms for finite games · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 60–97.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  73. Rank073Polynomial-Time Solvability of Parity GamesCan a deterministic algorithm find the winner from every vertex of a finite parity game in polynomial time, when the winner is determined by the greatest priority visited infinitely often?Theoretical computer sciencePolynomial-time algorithms for finite games · 3 problems

    Exact statement

    Does there exist a deterministic Turing-machine algorithm that, for every finite directed graph with at least one outgoing edge per vertex, a partition of its vertices between Even and Odd, and nonnegative integer vertex priorities encoded in binary, returns the winning player at every vertex in time polynomial in the complete input bit length, where Even wins exactly when the greatest priority occurring infinitely often is even?

    Also known as

    Parity games in P; Deterministic polynomial-time parity-game solving; Polynomial-time parity winner problem

    Why it matters

    Parity games connect infinite play on finite graphs with logical verification. The question asks for a polynomial bound in the complete encoded input, rather than a bound that becomes polynomial only when a game parameter is fixed.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Related problems

    • #72 · Polynomial-Time Solvability of Mean-Payoff GamesConditional implicationPolynomial-Time Solvability of Mean-Payoff Games

      A proof of #72 would also establish this assertion.

      Condition: An affirmative mean-payoff-in-P theorem implies parity-in-P; its contrapositive transfers a negative parity answer upward. A negative mean-payoff answer need not resolve parity.

      Scope qualification: An affirmative mean-payoff-in-P theorem implies parity-in-P; its contrapositive transfers a negative parity answer upward. A negative mean-payoff answer need not resolve parity.

    Problem family

    Polynomial-time algorithms for finite games · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 63–96.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  74. Rank074Non-Triviality of the Three-Dimensional Ising ModelDoes the large-scale limit of the critical three-dimensional Ising spin model remain non-Gaussian, with correlations that do not reduce to sums of products of pair correlations as Wick’s rule would require?Mathematical physics3D Ising · 3 problems

    Exact statement

    Prove that the critical three-dimensional Ising model is non-trivial, that is, that its scaling limit is not Gaussian and its critical spin correlations do not factorize according to Wick's rule.

    Why it matters

    A proof would establish that the limiting correlations retain behavior beyond a Gaussian theory. A Gaussian limit with Wick factorization would refute the stated non-triviality claim.

    Status

    Open

    Three-dimensional short-range Ising nontriviality remains provisionally open with an exact-proof warning. Long-range effective-dimension-four Gaussianity is a different model. The accessible Kai exponent-certificate abstract and demo/surrogate artifact do not establish the frozen non-Wick lattice scaling-limit theorem; full article/artifact correspondence was not verified.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    3D Ising · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 67–89.

    Source-fit opponent count: 32.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  75. Rank075Montgomerys Pair Correlation Conjecture for Zeros of the Riemann Zeta FunctionIf vertical separations between pairs of nontrivial zeta zeros are rescaled by the local mean spacing of zeros, does their limiting pair density at every fixed separation u > 0 equal 1 − (sin(πu)/(πu))²?Number theory & arithmetic geometry

    Exact statement

    Prove that normalized vertical gaps between pairs of nontrivial zeros of the Riemann zeta function have limiting pair density 1 - (sin(pi u)/(pi u))^2 at fixed positive normalized separation.

    Why it matters

    A proof would give the specified quantitative law for two-point spacing statistics of zeta zeros. A counterexample would show a departure from that density at a positive normalized separation.

    Status

    Open

    Montgomery pair correlation remains open. Jerby's December 2025 proof claim assumes RH, which the stated target does not assume, and lacks established independent acceptance. It is recorded as an unverified conditional claim, not an accepted conditional theorem or an unconditional resolution.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Montgomerys Pair Correlation Conjecture for Zeros of the Riemann Zeta Function · 1 problem in this edition. Family membership alone does not assert an implication.

    • #75 · Montgomerys Pair Correlation Conjecture for Zeros of the Riemann Zeta Function (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 66–101.

    Source-fit opponent count: 34.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  76. Rank076Woodins Ultimate L Conjecture for Supercompact CardinalsIf an extendible cardinal δ exists, must there be an inner model of set theory inside HOD—the hereditarily ordinal-definable sets—that satisfies the Ultimate-L axiom and is a weak extender model for δ’s supercompactness?Logic, foundations & set theory

    Exact statement

    Assuming delta is an extendible cardinal, prove that there is an inner model N contained in HOD such that N is a weak extender model for the supercompactness of delta and N satisfies V=Ultimate-L.

    Why it matters

    An inner model is an internal universe of sets; HOD consists of hereditarily ordinal-definable sets. The conjecture would build such a universe inside HOD that retains the specified supercompactness structure and satisfies Ultimate-L under the ambient extendibility assumption.

    Status

    Open

    Ultimate-L remains provisionally open with a formulation/consistency warning. The exacting-cardinal obstruction is conditional, not an unconditional disproof. Woodin's August 26, 2026 formulation remains conjectural but differs from the stated weak-extender-inside-HOD wording; equivalence requires specialist adjudication. The unrefereed broad proof essay was not accepted.

    Status reviewed

    Sources

    Problem family

    Woodins Ultimate L Conjecture for Supercompact Cardinals · 1 problem in this edition. Family membership alone does not assert an implication.

    • #76 · Woodins Ultimate L Conjecture for Supercompact Cardinals (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 66–101.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  77. Rank077Chowla conjecture on correlations of the Liouville functionThe Liouville function is +1 or −1 according to whether an integer has an even or odd number of prime factors, counted with multiplicity. For k ≥ 2 linear forms a_i n + b_i with natural coefficients a_i, distinct nonnegative offsets b_i and no proportional pair, do products of these signs average to zero?Probability, ergodic theory & dynamics

    Exact statement

    Prove that for the Liouville function lambda, for every k >= 2, distinct nonnegative integers b_1,...,b_k and natural numbers a_1,...,a_k with a_i b_j - a_j b_i != 0 for i<j, sum_{n<=x} lambda(a_1 n + b_1) ... lambda(a_k n + b_k) = o(x) as x -> infinity.

    Why it matters

    The Liouville function is +1 or −1 according to whether the number of prime factors, counted with multiplicity, is even or odd. A proof would show that positive and negative products balance in the limit for every fixed admissible family of affine forms.

    Status

    Open

    Earlier note (2026-09-14): Chowla remains open for every stated nonproportional affine-form tuple. Xiao's unrefereed fixed-shift claim is overlapping but not explicitly all affine forms; logarithmic and averaged August results do not establish every ordinary Cesaro correlation. October 7, 2026: OpenAI’s October 6 release reports partial progress. The target requires cancellation for every number k>=2 of pairwise nonproportional affine forms. This theorem supplies ordinary cutoff cancellation, even with a logarithmic saving, for exactly two factors. It covers the k=2 target subcase with matching affine-form scope but leaves k>=3 open. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Chowla conjecture on correlations of the Liouville function · 1 problem in this edition. Family membership alone does not assert an implication.

    • #77 · Chowla conjecture on correlations of the Liouville function (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 72–99.

    Source-fit opponent count: 35.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  78. Rank078Weight–Monodromy ConjectureMonodromy organizes a variety’s l-adic cohomology into layers. Are the Frobenius eigenvalues on each layer algebraic, with the absolute sizes predicted by its degree and layer index in every complex embedding, for every smooth proper variety over a nonarchimedean local field and every prime l different from the residue characteristic?Number theory & arithmetic geometry

    Exact statement

    Let K be ANY nonarchimedean local field with finite residue field F_q of characteristic p, X ANY proper smooth variety over K, i>=0, and l ANY prime different from p. Put V=H^i_et(X_over_Kbar,Qbar_l), retaining the original K. On an open inertia subgroup the action is unipotent and defines nilpotent monodromy N:V->V(-1), with its Galois-equivariant Tate twist. Let M be the unique increasing monodromy filtration centred at zero: N(M_j) is contained in M_(j-2)(-1), and N^a:gr^M_a(V)->gr^M_(-a)(V)(-a) is an isomorphism for every a>=0. For EVERY integer j and ANY lift Phi of geometric Frobenius of F_q, every eigenvalue alpha on gr^M_j(V) is conjectured to be algebraic and have absolute value q^((i+j)/2) under ALL complex embeddings. This is purity of weight i+j. Defining N on an open inertia subgroup does not replace K or q. Finite-extension proof reductions must change cardinality and Frobenius consistently. Extending l-adic coefficients does not change the target. The equivalent source convention compares zero-centred M_j with W_(i+j); weight-filtration existence is not an added unresolved assertion. No restriction to projective, semistable, geometrically connected or special complete-intersection varieties is imposed. l=p, crystalline, torsion and arbitrary rigid-analytic analogues are not substituted.

    Why it matters

    A proof would force every such eigenvalue to have absolute value q^((i + j)/2) under every complex embedding. It would link local monodromy to prescribed Frobenius weights without imposing projectivity or semistability.

    Status

    Open · subcases solved

    Weight–monodromy remains open, with solved subcases. Complete intersections in abelian varieties are a subcase, and 2026 formality work assumes weight–monodromy. Rigid-analytic counterexamples or l=p analogues are not the stated smooth proper algebraic l != p statement.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Weight–Monodromy Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #78 · Weight–Monodromy Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 73–97.

    Source-fit opponent count: 54.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  79. Rank079Graph Isomorphism in Polynomial TimeCan a deterministic classical algorithm decide in polynomial time whether any two finite graphs describe the same connections after their vertices are relabeled?Theoretical computer science

    Exact statement

    Determine whether the graph isomorphism problem for general finite graphs is decidable by a deterministic classical Turing machine in polynomial time.

    Why it matters

    A positive answer would give an efficient general procedure for recognizing identical graph structure despite different labels. A negative answer would establish a polynomial-time barrier even for this precisely defined comparison task.

    Status

    Open

    General graph isomorphism in deterministic polynomial time remains open. Heated-space methods explicitly lack all-input guarantees; the older Xue claim was withdrawn after an author-reported counterexample. A separate inaccessible 2026 claim was not accepted.

    Status reviewed

    Sources

    Problem family

    Graph Isomorphism in Polynomial Time · 1 problem in this edition. Family membership alone does not assert an implication.

    • #79 · Graph Isomorphism in Polynomial Time (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 70–110.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  80. Rank080The Lindeloef Hypothesis on the Growth of the Riemann Zeta FunctionAs we move up or down the line with real part one-half, is the magnitude of the Riemann zeta function eventually bounded by every positive power of the height above or below the real axis, up to a constant depending on that power?Number theory & arithmetic geometry

    Exact statement

    Prove that for every epsilon>0 there is a constant C_epsilon such that |zeta(1/2+it)| <= C_epsilon |t|^epsilon for all sufficiently large real |t|.

    Why it matters

    A proof would rule out growth at any fixed positive power along the critical line, controlling even its unusually large values. A disproof would demonstrate repeated growth exceeding at least one such power bound.

    Status

    Open

    Lindelof remains open with a renewed full-proof claim warning. Unterberger's v48 was withdrawn August 12, then v49 and September 4 v50 restored the claim. No independent assessment of the repair was established; it must not be called simply withdrawn, accepted or refuted. A separate formula argument explicitly lacks convergence justification.

    Status reviewed

    Sources

    Problem family

    The Lindeloef Hypothesis on the Growth of the Riemann Zeta Function · 1 problem in this edition. Family membership alone does not assert an implication.

    • #80 · The Lindeloef Hypothesis on the Growth of the Riemann Zeta Function (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 72–103.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  81. Rank081Sarnak's Mobius randomness conjecture for zero-entropy systemsThe Möbius function is zero on integers divisible by a prime square, and otherwise is +1 or −1 according to the parity of the number of prime factors. Does it have zero average correlation with every continuous measurement along every orbit of a compact metric dynamical system with zero topological entropy?Probability, ergodic theory & dynamics

    Exact statement

    Prove that for every compact metric topological dynamical system (X,T) of zero topological entropy, every continuous f on X, and every x in X, the average N^(-1) sum_{n<=N} mu(n) f(T^n x) tends to zero as N tends to infinity.

    Why it matters

    A proof would exclude persistent correlation between the Möbius function and any observable orbit in this entire zero-entropy class. A counterexample would locate structured dynamical behavior with a surviving arithmetic correlation.

    Status

    Open

    Sarnak remains open for ordinary n-iterate Cesaro averages. Polynomial-iterate counterexamples and a special infinite-torus theorem do not resolve it. An author-site disproof assertion lacks a verified exact counterexample or independent acceptance.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Sarnak's Mobius randomness conjecture for zero-entropy systems · 1 problem in this edition. Family membership alone does not assert an implication.

    • #81 · Sarnak's Mobius randomness conjecture for zero-entropy systems (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 68–106.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  82. Rank082Bateman-Horn conjectureFor a fixed collection of distinct irreducible integer polynomials with positive leading coefficients and no prime dividing their product at every integer, does the Bateman–Horn formula correctly predict how often they are all prime at the same positive input, including the leading constant from their divisibility patterns?Number theory & arithmetic geometrySimultaneous prime values · 5 problems

    Exact statement

    Prove or disprove the Bateman-Horn asymptotic formula for the number of positive integers n up to x for which any fixed admissible finite family of distinct irreducible integer polynomials with positive leading coefficients are simultaneously prime.

    Why it matters

    A proof would quantify how frequently simultaneous polynomial prime values occur, including the constant prescribed for each family. It would supply counting predictions more precise than merely asserting infinitely many successful inputs.

    Status

    Open · subcases solved

    Bateman–Horn remains open, with solved subcases with an unaccepted full-proof claim warning. Xiao's posting invokes hypothesis F and is unrefereed; a 2026 random-set model is not a theorem for the actual primes.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    Simultaneous prime values · 5 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 70–109.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  83. Rank083Zilber-Pink Conjecture on Unlikely IntersectionsMust every irreducible complex subvariety of a mixed Shimura variety contain only finitely many maximal pieces where intersection with a special subvariety is larger than dimension counting predicts? These oversized pieces are the conjecture’s atypical intersections.Number theory & arithmetic geometry

    Exact statement

    Let S be a mixed Shimura variety and W an irreducible subvariety of S defined over C. Call a subvariety A of W atypical if there is a special subvariety T of S with A contained in the intersection of T and W and dim A > dim W + dim T - dim S. Prove that the union of all atypical subvarieties of W is a finite union of atypical subvarieties, equivalently that W contains only finitely many maximal atypical subvarieties.

    Why it matters

    A proof would organize all these unexpectedly large intersections into finitely many maximal exceptional pieces. This would give a finite geometric description even when the variety meets infinitely many special subvarieties.

    Status

    Open · subcases solved

    Earlier note (2026-09-14): Zilber–Pink remains open, with solved subcases for the full mixed-Shimura target. Strong non-Qbar/Hodge-generic hypotheses and pure-Shimura cases do not cover it; current universal mixed-scope certification remains limited. October 7, 2026: OpenAI’s October 6 release reports partial progress. The release reports finiteness of intersections of closed Hodge-generic curves in A_2 over Qbar with special subvarieties of dimension at most one. This is a subcase of the mixed-Shimura target. A separate abelian-variety result concerns torsion cosets over algebraic closures of number fields; it does not identify that setting with every mixed Shimura variety. Neither result covers every complex subvariety in the full target. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Zilber-Pink Conjecture on Unlikely Intersections · 1 problem in this edition. Family membership alone does not assert an implication.

    • #83 · Zilber-Pink Conjecture on Unlikely Intersections (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 73–107.

    Source-fit opponent count: 43.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  84. Rank084Campana's Conjecture Characterizing Special Varieties by Potential Density of Rational PointsFor a smooth projective geometrically integral variety over a number field, is Campana's geometric condition of specialness exactly what allows its rational points to become Zariski dense after a finite extension of the field? Specialness is tested over the algebraic closure.Number theory & arithmetic geometryCampana specialness and rational points · 1 problem

    Exact statement

    For every number field K and every smooth projective geometrically integral variety X/K, prove or disprove that X over an algebraic closure of K is special in Campana's sense if and only if X(L) is Zariski dense in X for some finite extension L/K. Here specialness has its usual projective meaning: for each integer 1<=p<=dim X and rank-one coherent subsheaf F of the sheaf of algebraic p-forms on X over the algebraic closure, the Kodaira dimension kappa(F) is less than p (equivalently, the Campana core is a point). A proof must establish the biconditional for all such X; one counterexample to either implication refutes it. This target excludes nonproper integral-point, weakly-special, analytic and function-field variants.

    Why it matters

    This would characterize an arithmetic property—potential density of rational points—by geometry. A proof would also give potential density for K3 surfaces and the Bombieri–Lang non-density assertion for positive-dimensional varieties of general type.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Related problems

    • #54 · The Bombieri-Lang Conjecture for Varieties of General TypeWeaker assertionThe Bombieri-Lang Conjecture for Varieties of General Type

      The direction from potential density to specialness would imply Bombieri–Lang non-density for positive-dimensional varieties of general type, by contraposition. A disproof of Campana's biconditional need not refute Bombieri–Lang.

      Condition: The stronger assertion is proved; a negative answer to it need not settle the weaker assertion.

      Scope qualification: The Bombieri–Lang target is read with its recorded positive-dimensional scope qualification. Campana, Bombieri–Lang and K3 potential density retain separate family identities. Only the positive-answer implication is recorded. No converse or equivalence of resolution outcomes is established here.

    • #393 · Potential Density of Rational Points on K3 SurfacesWeaker assertionPotential Density of Rational Points on K3 Surfaces

      The direction from specialness to potential density would give potential density for every K3 surface over a number field. A counterexample to Campana's full biconditional need not refute the K3 assertion.

      Condition: The stronger assertion is proved; a negative answer to it need not settle the weaker assertion.

      Scope qualification: Only the positive-answer implication is recorded. No converse or equivalence of resolution outcomes is established here.

    Problem family

    Campana specialness and rational points · 1 problem in this edition. Family membership alone does not assert an implication.

    • #84 · Campana's Conjecture Characterizing Special Varieties by Potential Density of Rational Points (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 74–105.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  85. Rank085Deligne's Absolute Hodge ConjectureHodge classes are cohomology classes singled out by a variety's complex geometry. On every smooth projective complex variety, must each rational Hodge class remain one after any automorphism of the complex number field?Number theory & arithmetic geometryHodge conjectures · 3 problems

    Exact statement

    For every smooth projective variety X over C, every p>=0 and every rational Hodge class alpha in H^(2p)(X,Q(p)), alpha is absolute Hodge: for every automorphism sigma of C, its sigma-conjugate under algebraic de Rham comparison on X^sigma is again a rational Hodge class, with the same Tate-twist convention.

    Why it matters

    A proof would make the Hodge property invariant under all these field automorphisms, using the specified de Rham comparison and Tate twist. A counterexample would identify a class whose conjugate loses that property.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. Algebraic cycle classes remain rational Hodge classes under conjugation, so algebraicity supplies the absolute-Hodge assertion on the complex projective K3 products and other explicitly covered varieties. This is a mathematical consequence restricted to those classes of varieties, not a proof that every Hodge class on every smooth projective complex variety is absolute. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Related problems

    • #4 · Hodge ConjectureStronger assertionHodge Conjecture

      A proof of #4 would also establish this assertion.

      Scope qualification: The recorded direction concerns positive assertions only; no converse or whole binary-completion equivalence is asserted.

    Problem family

    Hodge conjectures · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 78–105.

    Source-fit opponent count: 40.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  86. Rank086Grothendieck–Teichmüller ConjectureThe absolute Galois group of the rationals describes the field symmetries of algebraic numbers. Does its standard Ihara embedding account for every element of Drinfeld’s original profinite Grothendieck–Teichmüller group, which encodes compatible symmetries of fundamental-group data?Number theory & arithmetic geometryGrothendieck–Teichmüller theory · 1 problem

    Exact statement

    Let G_Q=Gal(Qbar/Q). Let GT-hat be Drinfeld's original profinite Grothendieck-Teichmuller group, including its standard commutator and invertibility conditions, not merely the pair-equation monoid, gentle GT, a pro-l or prounipotent version, or a higher-genus subgroup. Fix the standard Ihara embedding Ih:G_Q -> GT-hat specified by the Ihara splitting in Bortnovskyi-Dolgushev-Holikov-Pashkovskyi Section1.3, equations1.5-1.6: Ih(g)=((chi(g)-1)/2,f_g) in that source's coordinates, with chi the cyclotomic character, and associated action x -> x^chi(g), y -> f_g^(-1)y^chi(g)f_g on the profinite free group. The conjecture asserts that this actual continuous injective homomorphism is surjective, hence an isomorphism of profinite groups. The source's group law and fixed splitting are retained; formulas from an opposite composition convention are not mixed into this definition.

    Why it matters

    Surjectivity would identify that entire profinite group with the arithmetic Galois group through the actual embedding. Failure would exhibit elements satisfying the group’s defining conditions that do not arise from Galois symmetries.

    Status

    Open

    Grothendieck–Teichmuller remains open for surjectivity of the fixed Ihara embedding into the original full profinite GT. August 2026 quotient results and underlying-space homeomorphisms do not prove this group-homomorphism surjectivity.

    Status reviewed

    Sources

    Problem family

    Grothendieck–Teichmüller theory · 1 problem in this edition. Family membership alone does not assert an implication.

    • #86 · Grothendieck–Teichmüller Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 81–107.

    Source-fit opponent count: 64.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  87. Rank087Elliott-Halberstam Conjecture on the Level of Distribution of PrimesWhen counting primes in arithmetic progressions, can their distribution errors be controlled for step sizes nearly as large as the counting limit x? Elliott–Halberstam predicts arbitrarily strong logarithmic savings in the sum of the largest errors over residue classes and shorter counting ranges, for step sizes up to x^θ with any 0 < θ < 1.Number theory & arithmetic geometry

    Exact statement

    Prove that for every A > 0 and every 0 < theta < 1, the sum over q <= x^theta of the maximal Chebyshev error |psi(y; q, a) - y/phi(q)| over y <= x and (a,q)=1 is O(x / (log x)^A), extending Bombieri-Vinogradov from theta < 1/2 to every theta < 1.

    Why it matters

    A proof would control the combined error in prime distribution across arithmetic progressions for moduli extending to every fixed power below x. The estimate controls the sum of worst errors for each modulus, retaining uniformity over shorter cutoffs.

    Status

    Open

    Elliott–Halberstam remains open unconditionally. A recent implication assumes both GRH and Dirichlet-L pair correlation, while restricted-modulus results do not prove the stated all-moduli maximal-error estimate.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Elliott-Halberstam Conjecture on the Level of Distribution of Primes · 1 problem in this edition. Family membership alone does not assert an implication.

    • #87 · Elliott-Halberstam Conjecture on the Level of Distribution of Primes (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 80–113.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  88. Rank088The Mumford-Tate Conjecture for Abelian Varieties over Number FieldsAn abelian variety has symmetries seen through both number theory and complex geometry. Over any number field, do the Lie algebras of its l-adic Galois image and its Mumford–Tate group agree after passing to l-adic coefficients, so that the two descriptions give the same infinitesimal symmetries?Number theory & arithmetic geometry

    Exact statement

    Prove that for every abelian variety A over a number field K, the Lie algebra of the l-adic Galois image Lie(Gal(Kbar/K)) equals the Lie algebra of the Mumford-Tate group MT(A) tensor Q_l.

    Why it matters

    A proof would identify the specified arithmetic Galois symmetries with the Mumford–Tate description at the Lie-algebra level. A counterexample would expose a mismatch between these two symmetry measurements for an abelian variety.

    Status

    Open

    Mumford–Tate remains open for every number-field abelian variety. Frobenius-field progress does not prove equality of the two Lie algebras universally; a rendered 2026 HTML date is not an authenticated revision of the 2024 source.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    The Mumford-Tate Conjecture for Abelian Varieties over Number Fields · 1 problem in this edition. Family membership alone does not assert an implication.

    • #88 · The Mumford-Tate Conjecture for Abelian Varieties over Number Fields (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 85–110.

    Source-fit opponent count: 35.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  89. Rank089PSPACE versus EXPTIMEIf a deterministic algorithm solves a yes-or-no problem in time 2^(n^k) for some fixed k, must another deterministic algorithm solve it using only polynomially many working-memory bits in the input length n?Theoretical computer science

    Exact statement

    Determine whether PSPACE=EXPTIME for languages over finite binary strings in the standard deterministic multitape Turing-machine model with a read-only input tape and work tapes. PSPACE is the union over integers k>=1 of DSPACE(n^k), with asymptotic constant factors allowed and read-only input cells not counted as work space. EXPTIME is the union over integers k>=1 of DTIME(2^(n^k)), with asymptotic constant factors allowed. Machines are total deciders that halt on every input. This is an unrelativized equality of the full classes, not of one fixed resource exponent.

    Why it matters

    An equality would show that polynomial memory suffices for every computation in the full exponential-time class. A separation would identify a decision problem whose time bound can be exponential but whose memory requirements exceed every polynomial.

    Status

    Open

    PSPACE versus EXPTIME remains open with bounded recency evidence. The Czerwinski separation was withdrawn in March 2024 with an independent critique, which defeats that claim but does not certify absence of every later one.

    Status reviewed

    Sources

    Problem family

    PSPACE versus EXPTIME · 1 problem in this edition. Family membership alone does not assert an implication.

    • #89 · PSPACE versus EXPTIME (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 82–112.

    Source-fit opponent count: 58.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  90. Rank090Classical Szpiro Discriminant–Conductor Conjecture over QCan the magnitude of an elliptic curve's minimal discriminant be controlled by its conductor—two measures of its behavior at bad primes—using every power greater than six, with a bound uniform over all curves defined over the rationals?Number theory & arithmetic geometryabc and Szpiro conjectures · 2 problems

    Exact statement

    Prove or disprove that for every epsilon>0 there is C(epsilon)>0 such that every elliptic curve E over Q with minimal discriminant Delta_E and conductor N_E satisfies |Delta_E| <= C(epsilon)*N_E^(6+epsilon).

    Why it matters

    A proof would uniformly control the discriminant in terms of the conductor, with the constant depending only on ε. Failure would mean that some positive ε admits no such uniform constant.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Related problems

    Problem family

    abc and Szpiro conjectures · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 80–119.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  91. Rank091Effective Mordell Conjecture for Curves of Genus at Least TwoCan we give a guaranteed terminating algorithm listing every point defined over a number field on any smooth projective curve of genus at least two, or an explicit bound on the points’ arithmetic size, called height, that makes such a search finite?Number theory & arithmetic geometry

    Exact statement

    For every smooth projective curve X of genus at least two over a number field K, give an explicit height bound for X(K), or an algorithm guaranteed to terminate that computes all points of X(K).

    Why it matters

    A resolution would make the curve’s rational points accessible through a provably complete search. An explicit height bound or guaranteed terminating enumeration would provide a stopping criterion, so a list could be known to omit no points.

    Status

    Open · subcases solved

    Effective Mordell remains open, with solved subcases for one uniform algorithm taking both field and curve as input. The 2025 automorphism/rank-dependent height theorem is restricted, not an all-curves terminating enumeration.

    Status reviewed

    Sources

    Problem family

    Effective Mordell Conjecture for Curves of Genus at Least Two · 1 problem in this edition. Family membership alone does not assert an implication.

    • #91 · Effective Mordell Conjecture for Curves of Genus at Least Two (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 82–122.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  92. Rank092Littlewood ConjectureFor every pair of real numbers α and β, can q times the product of the distances from qα and qβ to their nearest integers become arbitrarily small for arbitrarily large positive integers q?Probability, ergodic theory & dynamics

    Exact statement

    Prove or disprove that for every pair of real numbers alpha and beta, liminf_{q to infinity} q ||q alpha|| ||q beta|| = 0.

    Why it matters

    A proof would guarantee this multiplicative simultaneous approximation for every real pair. The product allows an exceptionally close approximation to one coordinate to compensate for a less close approximation to the other.

    Status

    Open

    Classical Littlewood remains open. The March 2026 disproof and August positive-dimensional counterexamples concern the different uniform Littlewood conjecture, not the stated real-pair liminf assertion.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Littlewood Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #92 · Littlewood Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 75–137.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  93. Rank093Grothendiecks Section Conjecture in Anabelian GeometryDoes the arithmetic fundamental group, which combines geometric loops with field symmetries, encode the rational points of every smooth complete curve of genus at least two over a number field exactly as ways to split its projection to field symmetries, counted up to conjugacy?Number theory & arithmetic geometry

    Exact statement

    Prove that for every smooth, complete hyperbolic curve C of genus g >= 2 over a number field K, the map sending K-rational points C(K) to conjugacy classes of splittings of the core exact sequence 1 -> pi_1(C_{Kbar}) -> pi_1(C) -> Gal(Kbar/K) -> 1 is a bijection.

    Why it matters

    A proof would recover exactly the rational points from group-theoretic splittings. Every splitting class would come from one rational point, turning this fundamental-group structure into a precise description of the curve’s arithmetic points.

    Status

    Open

    Earlier note (2026-09-14): The number-field section conjecture remains open with explicit current-status uncertainty. Recent conditional finite-type consequences and restricted affine S-Selmer sections do not establish all sections for every smooth complete hyperbolic number-field curve. October 7, 2026: OpenAI’s October 6 release reports partial progress. The manuscript reports the section bijection over finite extensions of Q_p and a global consequence conditional on equality of the finite-cover descent locus with the diagonal rational points in the modified adelic space. That arithmetic equality is an additional hypothesis. Stated global subcases include the specified genus-at-least-two modular curves and criteria involving abelian varieties or Jacobians with finitely many rational points and trivial divisible Tate–Shafarevich subgroup. The required descent equality is not established for all number-field curves, so the universal global conjecture remains open. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Grothendiecks Section Conjecture in Anabelian Geometry · 1 problem in this edition. Family membership alone does not assert an implication.

    • #93 · Grothendiecks Section Conjecture in Anabelian Geometry (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 82–130.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  94. Rank094Kannan-Lovasz-Simonovits conjecture on isoperimetry of log-concave measuresFor a log-concave probability distribution, can a hyperplane cut achieve a boundary-to-mass ratio within a universal constant factor of the best possible cut, in every dimension? Each cut’s boundary is normalized by the smaller of the two masses it separates.Applied & computational mathematicsResolution claim under review · historical rank

    Exact statement

    Prove that there is a universal constant C > 0 such that, for every log-concave probability measure mu on R^n, its Cheeger isoperimetric constant is at least C times the infimum of mu^+(H)/min(mu(H), 1-mu(H)) over halfspaces H. Equivalently, prove the dimension-free KLS lower bound psi_mu >= C/sqrt(||Cov(mu)||_op), with a universal constant C.

    Why it matters

    A proof would make simple halfspace cuts sufficient, up to a universal factor, to describe the measure’s worst isoperimetric bottleneck. The bound would remain independent of dimension and admit the stated covariance formulation.

    Status

    Resolution claim under review

    Earlier note (2026-10-06): Claimed pending refereeing — full KLS proof claims. Song and Zhang v2 and Bizeul, Klartag and Lehec v1 claim a universal Poincaré bound for isotropic log-concave measures, giving the dimension-free Cheeger target stated here. Song–Zhang v1 only gave an unbounded log-star bound. Bizeul–Klartag–Lehec build on the earlier Song–Zhang criterion, so these are not wholly independent confirmations. The pinned October 6 manuscript by Balasubramanian and Kasiviswanathan makes another full claim. No public Lean formalization or kernel receipt was located in the reviewed sources; this was a source-and-scope review, not a complete proof audit. October 7, 2026: OpenAI’s October 6 release reports partial progress. The logarithmic Sobolev estimate supplies a dimension-free Poincare-type bound for log-concave measures with a controlled linear subgaussian parameter, hence a KLS-type subcase when that parameter is uniformly controlled after isotropic normalization. The full KLS target covers every log-concave measure using only its covariance scale. Uniform subgaussian marginals are an extra assumption, and the abstract does not bound their parameter by a universal multiple of the covariance scale for all log-concave measures. It cannot upgrade the full full-KLS status. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Kannan-Lovasz-Simonovits conjecture on isoperimetry of log-concave measures · 1 problem in this edition. Family membership alone does not assert an implication.

    • #94 · Kannan-Lovasz-Simonovits conjecture on isoperimetry of log-concave measures (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 86–122.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  95. Rank095L versus NLCan a deterministic algorithm decide whether a directed graph has a path from one chosen vertex to another using only logarithmically many bits of working memory relative to input length? This is equivalent to asking whether nondeterminism adds power when working memory is logarithmic.Theoretical computer science

    Exact statement

    Determine whether every language decidable by a nondeterministic Turing machine using O(log n) work space is decidable by a deterministic Turing machine using O(log n) work space; equivalently, decide whether directed s-t reachability belongs to L.

    Why it matters

    Equality would remove nondeterminism without increasing the logarithmic workspace bound, yielding such an algorithm for directed reachability. Separation would show that nondeterministic choices increase computational power at this memory scale.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    L versus NL · 1 problem in this edition. Family membership alone does not assert an implication.

    • #95 · L versus NL (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 87–120.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  96. Rank096Grothendieck Standard Conjecture of Hodge TypeAfter identifying algebraic cycles with the same intersection numbers, does the sign-corrected intersection form give a positive value on every nonzero primitive class? This is asked in codimensions up to half the dimension, for every ample class on smooth projective irreducible varieties over algebraically closed fields of any characteristic.Algebra, representation & category theoryGrothendieck’s standard conjectures · 2 problems

    Exact statement

    For every smooth projective irreducible variety X of dimension n over an algebraically closed field k of any characteristic and every ample divisor class L, put N^i(X)_Q equal to codimension-i algebraic cycles with rational coefficients modulo numerical equivalence, with N^j=0 outside 0<=j<=n. For each integer 0<=i<=n/2 let P_L^i=ker(L^(n-2i+1):N^i(X)_Q -> N^(n-i+1)(X)_Q). The numerical standard conjecture of Hodge type asserts that q_L^i(a,b)=(-1)^i deg(a.b.L^(n-2i)) is positive definite on P_L^i after extending scalars to R. This is numerical primitive positivity, not an unconditional identification with primitive algebraic cohomology for an arbitrary Weil theory; it does not assume or require proving the other standard conjectures separately.

    Why it matters

    A proof would give a positive inner-product structure to these primitive numerical cycle spaces. The conclusion concerns cycles modulo numerical equivalence and does not itself identify them with primitive classes in an arbitrary cohomology theory.

    Status

    Open · subcases solved

    Earlier note (2026-09-14): Standard Conjecture of Hodge type remains open, with solved subcases. Fourfolds, powers of threefolds and other specified examples do not establish numerical primitive positivity in every dimension and characteristic; newer preprint proofs were not independently certified. October 7, 2026: OpenAI’s October 6 release reports partial progress. The stated consequence is the Hodge standard conjecture for abelian varieties in arbitrary characteristic. This supplies the abelian-variety part of the full numerical primitive-positivity question. Its universal quantifier over smooth projective irreducible varieties remains wider, even though the characteristic range of the abelian consequence is unrestricted. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Grothendieck’s standard conjectures · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 87–120.

    Source-fit opponent count: 53.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  97. Rank097Green-Griffiths-Lang Conjecture on entire curves in varieties of general typeAn entire curve is a holomorphic map from the whole complex plane. On a projective variety of general type, must each such curve stay inside some proper algebraic subvariety?Geometry & topologyReviewed hold · conditional rank

    Exact statement

    Prove the Green-Griffiths-Lang conjecture: for a projective variety X of general type, every entire holomorphic curve f: C -> X factors through a proper closed subvariety (equivalently, jet differentials define an algebraic differential equation for non-constant entire curves).

    Why it matters

    A proof would prevent any entire curve from spreading densely through a general-type variety in the Zariski sense. It would connect the variety’s algebraic type to a restriction on globally defined complex-analytic maps.

    Status

    Reviewed hold · conditional rank

    Recorded hold reason. Green–Griffiths–Lang remains open with its inherited scope warning. The standard uniform exceptional locus and the frozen per-curve wording must be distinguished; the asserted equivalence with jet-differential equations is not certified. High-degree generic hypersurfaces and plane-curve complements do not resolve all projective general-type varieties.

    Green–Griffiths–Lang remains open with its inherited scope warning. The standard uniform exceptional locus and the stated per-curve wording must be distinguished; the asserted equivalence with jet-differential equations is not certified. High-degree generic hypersurfaces and plane-curve complements do not resolve all projective general-type varieties.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Green-Griffiths-Lang Conjecture on entire curves in varieties of general type · 1 problem in this edition. Family membership alone does not assert an implication.

    • #97 · Green-Griffiths-Lang Conjecture on entire curves in varieties of general type (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 86–127.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  98. Rank098Bunyakovsky conjectureIf a polynomial with integer coefficients cannot split into lower-degree integer polynomials, eventually takes positive values, and has no prime dividing every value, must it produce primes at infinitely many positive integer inputs?Number theory & arithmetic geometryOpen · disputed claim

    Exact statement

    Prove or disprove that every irreducible polynomial f in Z[x] with positive leading coefficient and no fixed prime divisor takes prime values for infinitely many positive integer inputs.

    Why it matters

    A proof would establish infinitely many prime values for every polynomial satisfying these conditions, including x² + 1. A counterexample would reveal an obstruction beyond factorization, sign and fixed prime divisors.

    Status

    Open · disputed claim

    Status review date not recorded in this edition

    Sources

    Problem family

    Bunyakovsky conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #98 · Bunyakovsky conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 78–136.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  99. Rank099Strongly Polynomial Linear ProgrammingCan every rational linear program be solved deterministically using a number of exact arithmetic operations bounded by a polynomial in the numbers of variables and constraints, independently of the coefficient lengths?Theoretical computer scienceStrongly polynomial linear programming and simplex methods · 2 problems

    Exact statement

    Does a uniform deterministic algorithm solve every rational LP max c^T x, Ax <= b, x >= 0, returning an exact optimum or correctly reporting infeasibility or unboundedness, with poly(m,n) exact arithmetic operations and comparisons independent of coefficient encoding lengths and polynomial intermediate representation size in the binary input length? Operations on input and intermediate rationals are addition, subtraction, multiplication, division and comparison. For a stored p/q with q positive, size is ceil(log2(abs(p)+1)) + ceil(log2(q+1)); reduction to coprime terms, floor and bit extraction are not free operations. No nonuniform advice or arbitrary real constants are allowed. The displayed form is without loss under strongly polynomial slack-variable and variable-splitting reductions. This is the deterministic formulation; a randomized algorithm alone does not settle it.

    Also known as

    TOPP Problem8; Megiddo strongly polynomial LP question; Smale's ninth problem (rational strongly polynomial formulation)

    Why it matters

    Linear programming is central to optimization. A strongly polynomial algorithm would separate its arithmetic complexity from the size of its coefficients, while also keeping intermediate number representations polynomial in the binary input length.

    Status

    Open

    The deterministic strongly polynomial problem for rational linear programming remains open. An exact local audit of Awoniyi’s July 4, 2026 proof claim (arXiv:2503.12041v10) found a reachable two-variable counterexample to Lemma 6.1 and Corollary 1(c), invalidating the submitted iteration-bound argument. The checks were reproduced by the ranking coordinator and reviewed by a separate Claude model. This is a reviewed local audit, not an accepted external rebuttal or formal verification; it does not disprove eventual success of the complete algorithm. A randomized pivot-count bound for bounded nondegenerate simplex instances would not, by itself, settle this deterministic general rational-LP target. Smale’s real-BSS formulation is a separate model variant.

    Status reviewed

    Sources

    Related problems

    • #141 · Polynomial pivot rule for the simplex methodVariantPolynomial pivot rule for the simplex method

      Related questions with different algorithmic scopes. The LP target requires a deterministic algorithm for all rational instances, with arithmetic-operation and intermediate-size bounds. The simplex target permits randomization and bounds pivots on bounded nondegenerate instances from a feasible basis.

      Scope qualification: No unconditional implication between these complete targets is asserted. A general LP algorithm need not supply a simplex pivot rule, and this randomized special-case pivot bound alone does not yield the required deterministic LP algorithm.

    Problem family

    Strongly polynomial linear programming and simplex methods · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 79–141.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  100. Rank100Two-Dimensional Jacobian Conjecture in Characteristic ZeroIf a polynomial transformation of a plane has a nonzero constant determinant of derivatives, must it have an inverse that is also polynomial, over every field of characteristic zero?Algebra, representation & category theory

    Exact statement

    Let k be a field of characteristic zero and let F:k^2 -> k^2 be a polynomial map whose Jacobian determinant is a nonzero constant. Determine whether F must have a polynomial inverse, equivalently whether every two-variable Keller map is a polynomial automorphism of the affine plane.

    Why it matters

    A proof would turn this differential condition into global polynomial reversibility for every planar map in scope. A counterexample would show that constant nonzero Jacobian determinant does not guarantee an algebraic inverse.

    Status

    Open

    Gao's July 31, 2026 preprint reports counterexamples only in dimensions greater than two and explicitly leaves the two-dimensional case open. This higher-dimensional claim does not resolve this entry. October 6, 2026: Van den Essen explains the three-variable Jacobian counterexample. The current ranked entity explicitly concerns two variables. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Problem family

    Two-Dimensional Jacobian Conjecture in Characteristic Zero · 1 problem in this edition. Family membership alone does not assert an implication.

    • #100 · Two-Dimensional Jacobian Conjecture in Characteristic Zero (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 88–135.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  101. Rank101Superpolynomial Lower Bounds for Frege Propositional Proof SystemsIn standard Frege systems for reasoning with Boolean formulas, can we exhibit an explicit family of always-true formulas whose proofs exceed every polynomial bound in the formula length, measured by total proof size or by the number of proof lines?Theoretical computer scienceClassical complexity separations · 6 problems

    Exact statement

    Prove a superpolynomial lower bound on the size, or number of lines, of proofs in standard Frege propositional proof systems for an explicit family of propositional tautologies.

    Why it matters

    A proof would give concrete logical truths for which every proof in these systems must become very large. It would establish an intrinsic limitation of the proof system, beyond difficulties in choosing a proof-search method.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Classical complexity separations · 6 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 93–135.

    Source-fit opponent count: 35.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  102. Rank102Polignac's conjectureDoes every positive even number occur infinitely often as the gap between two consecutive primes?Number theory & arithmetic geometry

    Exact statement

    For every positive even integer 2k, there are infinitely many pairs of consecutive primes whose difference is 2k.

    Why it matters

    A proof would show that every positive even gap recurs indefinitely among neighboring primes. A disproof would identify an even gap that eventually disappears or never occurs in consecutive-prime spacings.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Polignac's conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #102 · Polignac's conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 90–140.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  103. Rank103Exact Kissing Numbers in Higher DimensionsIn every dimension greater than four, what is the maximum number of equal unit spheres that can touch one central unit sphere without overlapping one another?Cryptography, coding, information & optimization

    Exact statement

    For every integer n > 4, determine the exact kissing number tau_n: the maximum number of pairwise nonoverlapping unit balls in Euclidean R^n that can touch a central unit ball of the same radius.

    Why it matters

    Exact kissing numbers would determine the largest possible local contact arrangement in each dimension. A complete answer needs both an arrangement attaining the number and a proof that more contacts are impossible.

    Status

    Open · subcases solved

    Open with solved instances n=8 and n=24 within the target range; the March 2026 primary source still records a gap between 40 and 44 in dimension five. No full-sequence resolution was located. October 6, 2026: A four-dimensional packing-density/24-cell claim does not determine exact kissing numbers in every dimension n>4. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Problem family

    Exact Kissing Numbers in Higher Dimensions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #103 · Exact Kissing Numbers in Higher Dimensions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 96–132.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  104. Rank104Global Regularity of the 3D Incompressible Magnetohydrodynamic EquationsFor an incompressible conducting fluid filling three-dimensional space, do all sufficiently smooth divergence-free velocity and magnetic fields evolve uniquely and smoothly forever when both viscosity and magnetic diffusion are positive?Analysis & PDEIncompressible fluid regularity · 2 problems

    Exact statement

    Prove or disprove that every sufficiently smooth divergence-free initial velocity and magnetic field on R^3 generates a unique global smooth solution of the standard incompressible three-dimensional MHD system with positive viscosity and magnetic diffusivity.

    Why it matters

    A proof would ensure that the coupled fluid and magnetic equations preserve smooth evolution for every datum in the stated class. A disproof would reveal a breakdown of regularity or uniqueness despite both forms of dissipation.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Related problems

    Problem family

    Incompressible fluid regularity · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 84–140.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  105. Rank105Selberg 1/4 ConjectureOn every congruence quotient of the upper half-plane, an arithmetic hyperbolic space, is the first nonzero Laplacian eigenvalue at least one-quarter? These eigenvalues describe how functions oscillate on the space.Number theory & arithmetic geometry

    Exact statement

    Prove that the first nonzero eigenvalue lambda_1 of the Laplace-Beltrami operator on every congruence quotient of the upper half-plane is at least 1/4.

    Why it matters

    A proof would exclude all nonzero Laplace eigenvalues below one-quarter throughout this arithmetic family of hyperbolic surfaces. It would provide one uniform spectral gap despite the different geometries of the quotients.

    Status

    Open

    Selberg 1/4 remains open with recency and claim-verification limits. Random covers, high-energy statistics and small-level checks do not cover every congruence quotient; the private formal-route assertion lacks an independently verified exact-target transfer.

    Status reviewed

    Sources

    Problem family

    Selberg 1/4 Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #105 · Selberg 1/4 Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 89–147.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  106. Rank106Invariant Subspace Problem for Hilbert Space OperatorsMust every bounded linear operator on a separable infinite-dimensional complex Hilbert space map some closed linear subspace into itself, other than the zero subspace or the whole space?Analysis & PDE

    Exact statement

    Does every bounded linear operator T on an infinite-dimensional separable complex Hilbert space H have a nontrivial closed invariant subspace, that is, a closed subspace M with {0} != M != H such that T(M) is a subset of M?

    Why it matters

    A proof would guarantee that every such operator has a smaller closed space on which its action can be studied. A counterexample would show that even this basic form of reduction can fail.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Invariant Subspace Problem for Hilbert Space Operators · 1 problem in this edition. Family membership alone does not assert an implication.

    • #106 · Invariant Subspace Problem for Hilbert Space Operators (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 99–139.

    Source-fit opponent count: 35.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  107. Rank107Unbounded Ranks of Elliptic Curves over QFor every nonnegative integer r, does some elliptic curve over the rational numbers have at least r independent rational points of infinite order?Number theory & arithmetic geometry

    Exact statement

    Prove that for EVERY nonnegative integer r there is SOME elliptic curve E/Q whose actual Mordell-Weil group E(Q) has Z-rank at least r. The base field is exactly Q and rank means the free abelian rank in Mordell's theorem, not analytic rank, a Selmer bound or random-matrix corank. Thus the set of actual ranks of ALL elliptic curves over Q is unbounded. Counting one representative per Q-isomorphism class does not alter this existence question. No effective construction, specified family, uniform curve-producing algorithm or predetermined rate of growth is required.

    Why it matters

    A proof would show that the number of independent generators needed for rational points has no universal ceiling across elliptic curves over the rationals. It would establish existence without requiring an algorithm that constructs the curves.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Unbounded Ranks of Elliptic Curves over Q · 1 problem in this edition. Family membership alone does not assert an implication.

    • #107 · Unbounded Ranks of Elliptic Curves over Q (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 103–137.

    Source-fit opponent count: 36.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  108. Rank108The Haldane Conjecture on Spectral Gaps of Antiferromagnetic Quantum Spin ChainsIn the one-dimensional isotropic antiferromagnetic Heisenberg spin chain, is there a positive energy gap above the ground state for every positive integer spin, but no gap for every half-integer spin?Mathematical physics

    Exact statement

    Prove that the one-dimensional isotropic antiferromagnetic Heisenberg chain has a positive spectral gap for every integer spin S >= 1 and is gapless for every half-integer spin.

    Why it matters

    This would establish a precise distinction in the low-energy behavior of the two spin classes: an energy threshold for excitations in integer-spin chains and excitations arbitrarily close to the ground-state energy in half-integer chains.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The periodic theorem establishes the positive even-periodic gap formulation for S=1. The canonical target quantifies over every integer spin S>=1 and also asks for gaplessness at every half-integer spin. Neither S>=2 nor the half-integer conclusion is established by these abstracts. The odd open-chain companion includes fixed endpoint fields and therefore cannot silently be substituted for the unrestricted chain statement. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    The Haldane Conjecture on Spectral Gaps of Antiferromagnetic Quantum Spin Chains · 1 problem in this edition. Family membership alone does not assert an implication.

    • #108 · The Haldane Conjecture on Spectral Gaps of Antiferromagnetic Quantum Spin Chains (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 99–143.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  109. Rank109Quantum Unique Ergodicity on Negatively Curved ManifoldsOn a compact negatively curved space, must every sequence of normalized Laplace eigenfunctions—standing-wave patterns—whose energies tend to infinity spread uniformly in both position and direction?Probability, ergodic theory & dynamics

    Exact statement

    Let (M,g) be a compact negatively curved Riemannian manifold. Prove that every complete high-energy sequence of L2-normalized Laplace eigenfunctions has semiclassical measures converging to Liouville measure on the unit cosphere bundle; consequently, the position densities converge weakly to normalized Riemannian volume.

    Why it matters

    A proof would exclude exceptional high-energy sequences that retain concentration in phase space. It would also force their spatial probability densities to approach normalized volume throughout the manifold.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Quantum Unique Ergodicity on Negatively Curved Manifolds · 1 problem in this edition. Family membership alone does not assert an implication.

    • #109 · Quantum Unique Ergodicity on Negatively Curved Manifolds (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 102–146.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  110. Rank110Prove the BKL locality conjecture in the general inhomogeneous contextNear a singularity in generic gravitational collapse, do different spatial points evolve increasingly independently, following the oscillations of the spatially homogeneous vacuum Bianchi IX model, with the singularity forming a spacelike boundary?Mathematical physics

    Exact statement

    Prove that the singularity in generic inhomogeneous gravitational collapse is spacelike, local, and oscillatory, and that the evolution at each spatial point approaches vacuum homogeneous Bianchi IX Mixmaster behavior.

    Why it matters

    A proof would justify describing the approach to these singularities through local oscillatory dynamics despite spatial inhomogeneity. Failure of locality or of the proposed limiting behavior would require a different description of generic collapse.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Prove the BKL locality conjecture in the general inhomogeneous context · 1 problem in this edition. Family membership alone does not assert an implication.

    • #110 · Prove the BKL locality conjecture in the general inhomogeneous context (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 89–162.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  111. Rank111Regular Inverse Galois Problem over QCan every finite group occur as the Galois group of a finite Galois extension of Q(T) whose only elements algebraic over Q are rational numbers? This asks for a realization over a rational function field without introducing new algebraic constants.Algebra, representation & category theoryInverse Galois over Q · 2 problems

    Exact statement

    For every finite group G, does there exist a finite Galois extension F/Q(T) with Gal(F/Q(T)) isomorphic to G and Q algebraically closed in F? Equivalently, require a geometrically integral Galois cover of the projective line over Q with group G. A positive resolution proves this for all finite groups; a negative resolution proves nonexistence for at least one finite group.

    Why it matters

    Such realizations would supply geometric covers for every finite group. Hilbert irreducibility would then give every finite group as a Galois group over Q, settling ordinary inverse Galois affirmatively.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Related problems

    • #58 · Inverse Galois problem over the rational numbersWeaker assertionInverse Galois problem over the rational numbers

      Realizing every finite group regularly over Q(T) would, by Hilbert irreducibility, realize every finite group over Q. A negative answer to the regular problem need not refute ordinary inverse Galois.

      Condition: The stronger assertion is proved; a negative answer to it need not settle the weaker assertion.

      Scope qualification: Only the positive-answer implication is recorded. No converse or equivalence of resolution outcomes is established here.

    Problem family

    Inverse Galois over Q · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 102–143.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  112. Rank112Grothendieck–Katz p-Curvature ConjectureAn integrable algebraic connection describes a compatible system of differential equations. If its p-curvature vanishes after reduction at almost every prime, must it have a complete set of algebraic solutions?Number theory & arithmetic geometry

    Exact statement

    Let X be a smooth connected complex variety and (V,∇) an algebraic vector bundle with integrable connection, spread out over a finitely generated Z-algebra. Prove that if the reductions have zero p-curvature for almost all primes, then (V,∇) has a full set of algebraic solutions, equivalently becomes trivial on a finite étale cover and has finite monodromy.

    Why it matters

    A proof would make arithmetic information from reductions modulo primes sufficient to guarantee algebraic solvability over the complex numbers. Equivalently, the connection would become trivial on a finite étale cover and have finite monodromy.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Grothendieck–Katz p-Curvature Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #112 · Grothendieck–Katz p-Curvature Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 101–149.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  113. Rank113Serre's congruence subgroup conjectureFor arithmetic groups arising from simply connected, absolutely almost simple algebraic groups, is the congruence kernel—the gap between finite-quotient and congruence information—finite in total rank at least two when all selected nonarchimedean local ranks are positive, but infinite in rank one?Algebra, representation & category theory

    Exact statement

    Let K be a global field (a number field or the function field of a curve over a finite field), G an absolutely almost simple simply connected algebraic K-group, and S a finite nonempty set of places of K containing every archimedean place when K is a number field. Put r_S(G) = sum over v in S of rank_{K_v}(G), and let C^S(G) be the kernel of the natural surjection from the profinite completion of G(O_S) (S-arithmetic topology) onto its S-congruence completion. Prove Serre's congruence subgroup conjecture: (a) if r_S(G) >= 2 and rank_{K_v}(G) > 0 for every nonarchimedean v in S, then C^S(G) is finite; (b) if r_S(G) = 1, then C^S(G) is infinite. The case r_S(G) = 0 (finite G(O_S), trivial kernel) is excluded. Finiteness of C^S(G) is the assertion; triviality is a stronger statement not required. Which specific groups remain unsettled is status information, not part of the target.

    Why it matters

    The kernel measures the discrepancy between two ways of recovering an arithmetic group from finite information. A proof would classify that discrepancy as finite or infinite under these rank hypotheses, without requiring finite kernels to be trivial.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Serre's congruence subgroup conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #113 · Serre's congruence subgroup conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 108–149.

    Source-fit opponent count: 34.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  114. Rank114Erdos-Szemeredi Sum-Product Conjecture for IntegersAs finite sets of integers grow, must either their distinct pairwise sums or their distinct pairwise products number almost quadratically many—at least a constant times the set size to the power 2−ε, for every ε>0, with the constant depending only on ε?Combinatorics & discrete geometry

    Exact statement

    Prove that for every epsilon greater than zero there is a constant c_epsilon greater than zero such that every finite set A of integers satisfies max(|A+A|, |A*A|) at least c_epsilon times |A|^(2-epsilon).

    Why it matters

    A proof would show that a large integer set cannot have both highly compressed addition and highly compressed multiplication. At least one operation would generate almost quadratically many distinct outcomes.

    Status

    Open

    October 6, 2026: Counterexamples using real algebraic integers do not refute the stated target for subsets of ordinary integers Z. The earlier arXiv:2605.28781 also concerns real numbers. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Problem family

    Erdos-Szemeredi Sum-Product Conjecture for Integers · 1 problem in this edition. Family membership alone does not assert an implication.

    • #114 · Erdos-Szemeredi Sum-Product Conjecture for Integers (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 108–145.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  115. Rank115No-Slip Vanishing-Viscosity Limit in the Energy NormAs viscosity tends to zero in a smooth bounded two- or three-dimensional container with no-slip walls, must every finite-energy Leray–Hopf flow approach the ideal Euler flow with mean-square velocity error tending uniformly to zero on every time interval where Euler stays smooth, when both start from the same smooth boundary-vanishing incompressible velocity?Analysis & PDE

    Exact statement

    For d in {2,3}, every smooth bounded domain Omega in R^d and every fixed smooth divergence-free initial velocity u0 vanishing on the boundary, prove or disprove that every no-slip Leray–Hopf Navier–Stokes family u^nu with viscosity nu and initial value u0 converges to the Euler solution u in L^infinity([0,T];L^2(Omega)) as nu tends to zero, for each T on which u is smooth and satisfies u.n=0.

    Why it matters

    A proof would justify this energy-norm limit on each interval where the impermeable-boundary Euler solution remains smooth. A counterexample would show that vanishing viscosity can retain a nonvanishing L² difference under these conditions.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    No-Slip Vanishing-Viscosity Limit in the Energy Norm · 1 problem in this edition. Family membership alone does not assert an implication.

    • #115 · No-Slip Vanishing-Viscosity Limit in the Energy Norm (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 107–150.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  116. Rank116QMA versus QCMA ProblemCan every decision problem checked efficiently by a quantum computer using a polynomial-size quantum certificate also be checked efficiently using only a polynomial-size classical certificate?Quantum information & computationQuantum verification classes · 2 problems

    Exact statement

    Determine whether QMA=QCMA: does every language decidable by a polynomial-time quantum verifier using a polynomial-size quantum witness also admit a polynomial-time quantum verification protocol using only a polynomial-size classical witness?

    Why it matters

    The answer would determine whether quantum states give witnesses more verification power than classical strings when the verifier is already quantum. Equality would permit classical witnesses for all languages in the quantum-witness class.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Quantum verification classes · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 99–164.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  117. Rank117Cohen-Lenstra class-group distributions for quadratic number fieldsClass groups measure how unique factorization can fail in number fields. Among quadratic fields ordered by discriminant size, do their odd-prime parts follow the Cohen–Lenstra frequency law, favoring groups with fewer symmetries and giving smaller groups extra weight in the real case?Number theory & arithmetic geometry

    Exact statement

    For u in {0,1} let K_u(X) be the set of isomorphism classes of quadratic number fields K with |disc(K)| < X that are imaginary (u = 0) or real (u = 1). For each fixed odd prime p and each finite abelian p-group A, prove that the limit as X tends to infinity of #{K in K_u(X) : Cl(K)[p^infinity] is isomorphic to A} / #K_u(X) exists and equals (1/Z_u(p)) * |A|^{-u} / |Aut(A)|, where Z_u(p) is the sum over isomorphism classes of finite abelian p-groups B of |B|^{-u}/|Aut(B)| (so Z_0(p) = prod_{i>=1}(1-p^{-i})^{-1} and Z_1(p) = prod_{i>=2}(1-p^{-i})^{-1}). Cl(K) is the ordinary ideal class group and Cl(K)[p^infinity] its Sylow p-subgroup. No assertion is made for p = 2, for narrow class groups, for a joint law over several primes, or for any non-quadratic family.

    Why it matters

    The formula would specify the limiting frequency of every finite abelian p-group in each quadratic family. It would turn a weighting rule into exact statistical predictions for ordinary class groups, separately for real and imaginary fields.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Cohen-Lenstra class-group distributions for quadratic number fields · 1 problem in this edition. Family membership alone does not assert an implication.

    • #117 · Cohen-Lenstra class-group distributions for quadratic number fields (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 112–154.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  118. Rank118Erdos-Hajnal ConjectureIf a graph avoids one fixed induced pattern, including its absent edges, must it contain vertices that are either all mutually connected or all mutually disconnected in a set of size at least a fixed positive power of the total number of vertices?Combinatorics & discrete geometry

    Exact statement

    Prove that for every fixed finite graph H there is a constant c_H>0 such that every n-vertex graph with no induced subgraph isomorphic to H contains a clique or an independent set of size at least n^{c_H}.

    Why it matters

    A proof would force a large completely ordered pattern of adjacency whenever one induced pattern is excluded. The guaranteed size would be a power of the total vertex count, with the exponent depending only on the forbidden graph.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Erdos-Hajnal Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #118 · Erdos-Hajnal Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 111–155.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  119. Rank119Volume conjecture for knotsA hyperbolic knot has both polynomial invariants and a geometric volume for the space around it. Does that volume equal 2π times the exponential growth rate of the absolute values of its normalized colored Jones polynomials, evaluated at e^(2πi/N) as the color N grows?Geometry & topology

    Exact statement

    Prove or disprove that for every hyperbolic knot K in S^3, the limit as N tends to infinity of (2*pi/N) log|J_{K,N}(exp(2*pi*i/N))| equals the hyperbolic volume of S^3 minus K, using the standard normalized N-colored Jones polynomial.

    Why it matters

    A proof would recover a geometric volume from a sequence of polynomial knot invariants. A counterexample would locate a failure of that proposed link, either through a different limit or a limit that does not exist.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Volume conjecture for knots · 1 problem in this edition. Family membership alone does not assert an implication.

    • #119 · Volume conjecture for knots (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 107–157.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  120. Rank120Density Hypothesis for Zeroes of the Riemann Zeta FunctionHow sparse are zeta zeros to the right of the critical line? For 1/2≤σ≤1, must their count with real part at least σ and imaginary part of magnitude at most T grow no faster than a constant times T^(2(1−σ)+ε), for every ε>0?Number theory & arithmetic geometryZeta zeros and large Dirichlet polynomial values · 2 problems

    Exact statement

    Prove that for every sigma with 1/2 <= sigma <= 1 and every epsilon > 0, the number N(sigma,T) of nontrivial zeros rho = beta + i gamma of the Riemann zeta function with beta >= sigma and |gamma| <= T satisfies N(sigma,T) <= C_{sigma,epsilon} T^{2(1-sigma)+epsilon} for all sufficiently large T.

    Why it matters

    The estimate would quantify how sparse zeros can be toward the right of the critical strip. Such control feeds estimates for the distribution of prime numbers, including primes in short intervals.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. For every sigma>7/8 the asserted zero-free half-plane makes N(sigma,T)=0, giving the target density estimate in that restricted sigma range. The density hypothesis also quantifies over every sigma from 1/2 through 7/8, which is not addressed by this conclusion. No full density-hypothesis resolution follows. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Zeta zeros and large Dirichlet polynomial values · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 96–190.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  121. Rank121Tarski's Exponential Function ProblemCan an algorithm always decide whether any first-order statement about real numbers, using arithmetic, inequalities and the exponential function, is true—even when it quantifies over all real numbers?Logic, foundations & set theory

    Exact statement

    Determine whether the first-order theory of the real exponential field (R; 0, 1, +, *, exp, <, =) is decidable.

    Why it matters

    A positive answer would provide a universal decision procedure for this precisely specified language, including nested quantifiers over real numbers. A negative answer would prove that no algorithm can decide all such sentences.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Tarski's Exponential Function Problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #121 · Tarski's Exponential Function Problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 110–159.

    Source-fit opponent count: 40.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  122. Rank122Polynomial Bound for Limit Cycles of Planar Polynomial Vector FieldsCan the number of isolated periodic motions in any polynomial differential system in the plane be bounded by a power of its degree, with one universal exponent?Probability, ergodic theory & dynamics

    Exact statement

    Determine whether the number K of limit cycles of every planar polynomial differential system of degree at most d satisfies K ≤ d^q for some universal exponent q.

    Why it matters

    Such a bound would control the number of isolated repeating motions directly from the degree of the equations. A negative answer would show that no fixed polynomial in the degree captures their possible number.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. A finite bound depending on degree does not imply K <= d^q for one universal exponent q. The first manuscript supplies no bound on how its degree-dependent constant grows, and the quintic Lienard theorem is a fixed subclass. The family establishes the broader finiteness prerequisite but neither proves nor refutes the full polynomial-in-degree estimate. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Polynomial Bound for Limit Cycles of Planar Polynomial Vector Fields · 1 problem in this edition. Family membership alone does not assert an implication.

    • #122 · Polynomial Bound for Limit Cycles of Planar Polynomial Vector Fields (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 91–177.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  123. Rank123Unconditional Single-Prover Classical Verification of BQP ComputationsCan an ordinary computer efficiently check any efficient quantum computation by exchanging classical messages with one quantum prover, reliably accepting correct answers and rejecting false ones even if a cheating prover has unlimited computing power?Quantum information & computation

    Exact statement

    Determine whether every language L in BQP admits a polynomial-round interactive proof with only classical messages between a probabilistic classical polynomial-time verifier and a single honest uniform quantum polynomial-time prover, with completeness at least 2/3 and soundness at most 1/3 against every computationally unbounded cheating prover.

    Why it matters

    An affirmative answer would give purely classical verification of every computation in BQP while requiring only an efficient honest quantum prover. The guarantee would resist computationally unbounded cheating without relying on cryptographic hardness assumptions.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Unconditional Single-Prover Classical Verification of BQP Computations · 1 problem in this edition. Family membership alone does not assert an implication.

    • #123 · Unconditional Single-Prover Classical Verification of BQP Computations (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 105–181.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  124. Rank124Local Connectivity of the Mandelbrot Set (MLC Conjecture)At every point of the Mandelbrot set—the parameters c for which iterating z ↦ z² + c from zero stays bounded—are there arbitrarily small connected neighborhoods within the set?Probability, ergodic theory & dynamicsMandelbrot set boundary · 2 problems

    Exact statement

    Determine whether the Mandelbrot set M = {c in C : the orbit of 0 under z -> z^2 + c is bounded} is locally connected at every point.

    Why it matters

    A proof would control the set's local topology even at its most intricate parameters. A counterexample would identify a point where connected neighborhoods cannot be made arbitrarily small.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Mandelbrot set boundary · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 112–171.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  125. Rank125Existence of a Length-Optimal Propositional Proof SystemIs there a Cook–Reckhow propositional proof system whose shortest proofs are at most polynomially longer than those in any other such system, with the polynomial allowed to depend on the other system? An efficient way to translate proofs is not required.Theoretical computer scienceOptimal propositional proof systems · 2 problems

    Exact statement

    Does there exist a polynomial-time computable function P from finite binary strings onto the encodings of all propositional tautologies such that, for every polynomial-time computable function Q onto the same tautologies, there is a polynomial q_Q for which every Q-proof pi has some P-proof sigma with P(sigma)=Q(pi) and |sigma| <= q_Q(|pi|)? The polynomial may depend on Q. No polynomial-time algorithm for finding sigma from pi is required. This is the ordinary unrelativized Cook–Reckhow setting without advice or randomness; resolve existence or nonexistence of such P.

    Also known as

    Optimal propositional proof system problem; Length-optimal Cook–Reckhow proof system; Existence of an optimal proof system for TAUT

    Why it matters

    This asks whether one proof system can match every other system in proof length. It separates the existence of short proofs from the algorithmic task of finding or translating them.

    Status

    Open

    Status reviewed

    Sources

    Related problems

    • #310 · Existence of a p-Optimal Propositional Proof SystemStronger assertionExistence of a p-Optimal Propositional Proof System

      Existence of a p-optimal system would establish length-optimality. The definition of length-optimality does not require the polynomial-time proof translation required by the p-optimality question.

      Scope qualification: No converse or proved strict separation is asserted. These are separately posed existence questions in the ordinary Cook–Reckhow setting.

    Problem family

    Optimal propositional proof systems · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 110–165.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  126. Rank126The Generalized Sato-Tate Conjecture for Higher-Dimensional Abelian VarietiesAs primes vary, do the normalized Frobenius data of every abelian variety over a number field, encoding its behavior modulo those primes, follow the probability distribution predicted by its Sato–Tate symmetry group?Number theory & arithmetic geometry

    Exact statement

    Prove that the normalized Frobenius conjugacy classes of an abelian variety A of dimension g over a number field K are equidistributed in the Sato-Tate group ST(A).

    Why it matters

    A proof would give a precise statistical law for these arithmetic data across primes. The Sato–Tate group would determine the limiting distribution rather than merely describe possible individual Frobenius classes.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    The Generalized Sato-Tate Conjecture for Higher-Dimensional Abelian Varieties · 1 problem in this edition. Family membership alone does not assert an implication.

    • #126 · The Generalized Sato-Tate Conjecture for Higher-Dimensional Abelian Varieties (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 108–184.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  127. Rank127NP-Completeness of the Minimum Circuit Size Problem (MCSP)Given a function's full table of binary inputs and outputs, is deciding whether it has a circuit of at most a chosen size as hard as every problem in NP, under deterministic polynomial-time many-one reductions?Theoretical computer science

    Exact statement

    Prove or disprove that MCSP is NP-hard under deterministic polynomial-time many-one reductions, where an input is the truth table of a Boolean function and a size bound and asks whether a circuit of at most that size computes the function.

    Why it matters

    A hardness proof would make deciding whether a compact circuit description exists a destination for reductions from every NP problem. A negative answer would distinguish this decision problem from that standard form of universal NP hardness.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    NP-Completeness of the Minimum Circuit Size Problem (MCSP) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #127 · NP-Completeness of the Minimum Circuit Size Problem (MCSP) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 125–159.

    Source-fit opponent count: 43.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  128. Rank128Borel normal number conjecture for irrational algebraic numbersMust every real irrational root of an integer polynomial have equally frequent digit blocks of each fixed length in every integer base of at least two?Number theory & arithmetic geometry

    Exact statement

    Prove or disprove that every real irrational algebraic number is normal to every integer base b >= 2.

    Why it matters

    A proof would impose precise long-run frequencies on all digit patterns of algebraic irrationals. It would connect numbers defined by polynomial equations to the same block frequencies as uniformly distributed random digits.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Borel normal number conjecture for irrational algebraic numbers · 1 problem in this edition. Family membership alone does not assert an implication.

    • #128 · Borel normal number conjecture for irrational algebraic numbers (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 111–179.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  129. Rank129Hyperlinearity of Every GroupCan every finite piece of any group's multiplication table be modeled arbitrarily accurately by unitary matrices, while nonidentity elements stay separated from the identity in the normalized Hilbert–Schmidt distance?Algebra, representation & category theoryGroup approximation and determinants · 1 problem

    Exact statement

    Determine whether every group G is hyperlinear. Precisely, for each finite F contained in G and each epsilon>0, must there be d>=1 and phi:F->U(d) such that ||phi(g)phi(h)-phi(gh)||_(2,d)<epsilon whenever g,h,gh belong to F, and ||phi(g)-I||_(2,d)>sqrt(2)-epsilon for every nonidentity g in F? Here ||X||_(2,d)=(Tr(X*X)/d)^(1/2) is the normalized Hilbert-Schmidt norm. A negative resolution is a group failing this condition.

    Why it matters

    A proof would give finite-dimensional unitary approximations for every group in the precise stated sense. A counterexample would identify a finite pattern and tolerance for which the required approximation and separation cannot coexist.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Group approximation and determinants · 1 problem in this edition. Family membership alone does not assert an implication.

    • #129 · Hyperlinearity of Every Group (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 113–179.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  130. Rank130Torsion Conjecture for Abelian Varieties of Dimension at Least TwoAfter fixing a number field and a dimension of at least two, is there one common bound on the number of finite-order points defined over that field on any abelian variety of that dimension?Number theory & arithmetic geometry

    Exact statement

    For every integer g>=2 and every number field K, there exists a finite constant B(g,K) such that every abelian variety A defined over K of dimension g satisfies #A(K)_tors <= B(g,K). Here A(K)_tors is the subgroup of K-rational torsion points. The bound is uniform over all such A, without dependence on A, its height, endomorphisms, polarization degree or reduction type; it may depend on g and K. No effective formula, explicit numerical value or classification of torsion groups is required.

    Why it matters

    A proof would prevent rational torsion from growing arbitrarily large while dimension and field remain fixed. The bound could exist without an explicit formula or a classification of the possible torsion groups.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Torsion Conjecture for Abelian Varieties of Dimension at Least Two · 1 problem in this edition. Family membership alone does not assert an implication.

    • #130 · Torsion Conjecture for Abelian Varieties of Dimension at Least Two (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 120–179.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  131. Rank131Indistinguishability Obfuscation from Constant-Degree Homomorphic AssumptionsCan we efficiently disguise programs so that programs doing the same computation are computationally indistinguishable, using only standard polynomial-time hardness assumptions and without a subexponential security loss?Cryptography, coding, information & optimization

    Exact statement

    Construct indistinguishability obfuscation based purely on standard polynomial-time assumptions (such as standard LWE, LPN, and degree-2 multilinear pairings) without sub-exponential loss.

    Why it matters

    Such a construction would connect the obfuscation guarantee to the specified assumptions without the stated security loss. Its security would still be conditional on those assumptions and would concern indistinguishability of equivalent programs.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Indistinguishability Obfuscation from Constant-Degree Homomorphic Assumptions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #131 · Indistinguishability Obfuscation from Constant-Degree Homomorphic Assumptions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 114–203.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  132. Rank132Arithmetic Quantum Unique ErgodicityOn arithmetic locally symmetric spaces, must high-energy wave patterns that are also Hecke eigenfunctions spread uniformly in the limit, leaving normalized volume as their only possible limiting distribution?Probability, ergodic theory & dynamics

    Exact statement

    Prove the arithmetic quantum unique ergodicity conjecture: for arithmetic locally symmetric spaces, the normalized Riemannian volume is the only arithmetic quantum limit of Hecke eigenfunctions.

    Why it matters

    A proof would force these arithmetically constrained eigenfunctions to spread uniformly in the limit. It would exclude alternative limiting distributions that retain concentration in particular regions of the space.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Arithmetic Quantum Unique Ergodicity · 1 problem in this edition. Family membership alone does not assert an implication.

    • #132 · Arithmetic Quantum Unique Ergodicity (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 119–184.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  133. Rank133Global Gan–Gross–Prasad Conjecture (Fixed-Representation Form)For a fixed tempered cuspidal automorphic representation in the Gan–Gross–Prasad setting of classical groups and their covers, with multiplicity one, is its global period nonzero exactly when every local compatibility test succeeds and its associated L-function has a nonzero central value?Number theory & arithmetic geometry

    Exact statement

    For every number field F, let E be F, a quadratic extension, or F×F, and let W⊂V be a nondegenerate relevant pair with form sign epsilon, split W-perp, and epsilon*(-1)^dim(W-perp)=-1; a two-dimensional orthogonal V or W is required to be nonsplit. Use exactly the groups, covers, Bessel/Fourier–Jacobi representation nu and auxiliary data of Gan–Gross–Prasad Section 23, and R of Section 24, including the skew-hermitian character mu and its matching inverse twist in R. For every irreducible tempered representation pi occurring with multiplicity one in the cuspidal automorphic spectrum, Conjecture 24.1 asserts: the global period F(nu) is nonzero on pi if and only if Hom_H(F_v)(pi_v,nu_v) is nonzero for every place v and L(pi,R,1/2) is nonzero. Quantification is over all data permitted there, including relevant orthogonal, hermitian, symplectic/metaplectic and skew-hermitian cases. This is the fixed-representation equivalence, not the refined period identity or packet-existence formulation.

    Why it matters

    The equivalence would decide global period nonvanishing from local representation data and one central L-value. It would connect these arithmetic and representation-theoretic tests for each fixed representation, without asserting a refined formula for the period's size.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Global Gan–Gross–Prasad Conjecture (Fixed-Representation Form) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #133 · Global Gan–Gross–Prasad Conjecture (Fixed-Representation Form) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 128–172.

    Source-fit opponent count: 42.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  134. Rank134Uniform Subconvexity for Standard Automorphic L-Functions on GL(n)An L-function’s analytic conductor measures its arithmetic and analytic complexity. For each fixed rank n and number field, are the magnitudes of the central values at s = 1/2 of all unitary cuspidal standard L-functions on GL(n) bounded by a constant times the conductor raised to an exponent strictly below one-quarter, uniformly as all parameters vary together?Number theory & arithmetic geometry

    Exact statement

    For every integer n≥1 and every number field F, there exist constants δ(n,F)>0 and A(n,F)>0 such that |L(1/2,π)|≤A(n,F) C(π)^(1/4−δ(n,F)) for every unitary cuspidal automorphic representation π of GL_n(A_F), where L(s,π) is the standard finite-part L-function and C(π) its analytic conductor. The bound must be uniform simultaneously in all finite and archimedean parameters of π; imaginary unitary twists are included. Any fixed positive exponent saving suffices; the optimal exponent is not requested.

    Why it matters

    A proof would give a uniform power-saving bound for central automorphic L-values at every fixed rank and number field, even when all arithmetic and analytic parameters vary together. A disproof would rule out any such saving in at least one of these families and contradict the corresponding generalized Lindelöf prediction.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Uniform Subconvexity for Standard Automorphic L-Functions on GL(n) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #134 · Uniform Subconvexity for Standard Automorphic L-Functions on GL(n) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 130–172.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  135. Rank135Murre's Chow-Kunneth ConjecturesDoes every smooth projective complex variety admit a Chow–Künneth splitting for rational algebraic cycles into cohomological degrees, with every such splitting placing codimension-j cycles only in degrees j through 2j and producing one intrinsic filtration whose first step consists exactly of cycles invisible in cohomology?Number theory & arithmetic geometry

    Exact statement

    For every smooth projective irreducible complex variety X of dimension d, let CH^j(X)_Q denote cycles modulo rational equivalence, with Betti Q-cohomology. A correspondence pi in CH^d(X x X)_Q acts by pi_*(a)=(p_2)_*(p_1^*(a).pi). Murre's full A-D package asserts: (A) there exists a Chow-Kunneth decomposition, namely mutually orthogonal composition-idempotents pi_0,...,pi_(2d) in CH^d(X x X)_Q summing to the diagonal and inducing projection onto H^i(X,Q) for pi_i; (B) for EVERY such decomposition and 0<=j<=d, pi_i acts as zero on CH^j when i<j or i>2j; (C) for EVERY such decomposition define F^r CH^j as the intersection of ker(pi_i)_* for max(0,2j-r+1)<=i<=2j, r>=0, with empty intersection F^0=CH^j; these filtrations are equal for ALL decompositions satisfying A; (D) for EVERY such decomposition, F^1 CH^j equals ker(CH^j(X)_Q -> H^(2j)(X,Q)). Under A and B, F^(j+1)=0. No multiplicative or self-dual decomposition, explicit construction algorithm, functorial-extra requirement, or arbitrary-base-field enlargement is included.

    Why it matters

    The full package would organize rational algebraic cycles into a filtration independent of the chosen projectors. Its first step would capture exactly the cycles invisible to Betti cohomology, with the prescribed vanishing properties throughout.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Murre's Chow-Kunneth Conjectures · 1 problem in this edition. Family membership alone does not assert an implication.

    • #135 · Murre's Chow-Kunneth Conjectures (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 129–174.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  136. Rank136Planted Clique ConjectureHide a complete subgraph on about n^(1/2−α) randomly chosen vertices in an otherwise random n-vertex graph whose edges appear independently with probability 1/2. For every fixed 0<α<1/2, is recovering all hidden vertices with probability tending to one impossible in randomized polynomial time?Theoretical computer science

    Exact statement

    For every fixed real alpha with 0<alpha<1/2, set k_n=floor(n^(1/2-alpha)). Sample a uniform k_n-subset S of n labeled vertices, put every edge inside S, and independently include each other edge with probability1/2. There is no randomized algorithm running in time polynomial in n that, given this graph, outputs exactly S with probability tending to1 as n tends to infinity, over S, remaining edges and its internal coins. Algorithms may depend on the fixed alpha but receive no advice depending on n.

    Why it matters

    A proof would establish a computational barrier to recovering the actual planted vertex set in this random-graph model. An algorithm refuting it would need exact recovery for some fixed exponent, rather than merely detecting that a clique was planted.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Planted Clique Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #136 · Planted Clique Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 122–190.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  137. Rank137Log-Rank Conjecture in Communication ComplexityTwo people each hold one input to a Boolean function and exchange bits to compute its value. Is their minimum deterministic communication always bounded by a polynomial in the logarithm of the real rank of the function's ±1 input-output matrix?Cryptography, coding, information & optimization

    Exact statement

    Prove that there is a universal constant C >= 1 such that, for every Boolean function f:X x Y -> {-1,1} on finite nonempty sets X,Y, D(f) <= C[log_2 max{2,rank_R(M_f)}]^C, where M_f(x,y)=f(x,y) is its sign communication matrix and D(f) is deterministic two-party communication complexity in bits.

    Wording clarified after assessment

    Earlier comparisons used the wording below. A recorded review treats the published wording above as a clarification of the same intended problem. The wording review itself adds no comparison judgments.

    Earlier assessed wording

    Prove that there is a universal constant C such that every Boolean function f has deterministic two-party communication complexity D(f) at most C(log rank_R(M_f))^C.

    Wording review recorded .

    Why it matters

    A proof would turn an algebraic property of the function's matrix into a bound on the information the parties must exchange. The bound would grow polynomially in logarithmic rank while allowing constant communication at small ranks.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Log-Rank Conjecture in Communication Complexity · 1 problem in this edition. Family membership alone does not assert an implication.

    • #137 · Log-Rank Conjecture in Communication Complexity (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 117–192.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  138. Rank138Superlinear Circuit Lower Bounds for Explicit Boolean FunctionsCan a family of yes-or-no functions be computable in polynomial time yet require circuits whose number of two-input gates grows faster than any constant times the input length, even with unrestricted depth and fan-out?Theoretical computer science

    Exact statement

    There exists a Boolean function family f_n:{0,1}^n->{0,1}, n>=1, evaluated by a single deterministic polynomial-time algorithm on (1^n,x), whose minimum gate count C_B2(f_n) over single-output fan-in-two circuits using arbitrary binary Boolean gates satisfies C_B2(f_n)=omega(n): for every c>0 there is N such that C_B2(f_n)>c*n for every n>=N. The circuits have unrestricted depth and fan-out; no fixed linear constant counts as superlinear.

    Why it matters

    A proof would exhibit an efficiently specified computation whose circuit size grows faster than every linear bound. The eventual growth requirement excludes merely improving a fixed constant in front of n or obtaining lower bounds at isolated input lengths.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Superlinear Circuit Lower Bounds for Explicit Boolean Functions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #138 · Superlinear Circuit Lower Bounds for Explicit Boolean Functions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 121–197.

    Source-fit opponent count: 32.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  139. Rank139Order of Magnitude of the Largest Three-Term-Progression-Free SetsAs N grows, what is the asymptotic order of the largest subset of {1,…,N} containing no three distinct numbers in arithmetic progression?Combinatorics & discrete geometry

    Exact statement

    Let r_3(N) be the largest cardinality of a subset of {1,...,N} containing no three distinct terms in arithmetic progression. Determine the asymptotic order of r_3(N) as N tends to infinity.

    Why it matters

    A matching asymptotic estimate would determine how large an integer set can be while avoiding the simplest nontrivial arithmetic progression. It would identify the density scale at which this additive pattern becomes unavoidable.

    Status

    Open

    Earlier note (null): Open. Bloom--Sisask 2023 and Raghavan 2026 improve upper bounds for progression-free sets but leave the matching asymptotic-order problem unresolved. October 7, 2026: OpenAI’s October 6 release reports partial progress. Specializing the all-fixed-k estimate to k=3 gives a sublinear upper bound of the form C N exp(-c(log N)^epsilon) for the largest three-term-progression-free subset. The full target asks for its asymptotic order, which requires matching upper and lower scales. No optimal exponent or matching lower bound is stated, so the theorem supplies an upper bound without determining that asymptotic order. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Order of Magnitude of the Largest Three-Term-Progression-Free Sets · 1 problem in this edition. Family membership alone does not assert an implication.

    • #139 · Order of Magnitude of the Largest Three-Term-Progression-Free Sets (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 131–184.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  140. Rank140Consistency of Reinhardt Cardinals without ChoiceCan ZF set theory, without the axiom of choice, consistently admit a nonidentity self-embedding of the entire set-theoretic universe that preserves statements in the original language of set theory, with the embedding included in the expanded theory's language?Logic, foundations & set theory

    Exact statement

    Determine whether ZFR is consistent, where ZFR is the first-order theory ZF(j) plus the assertion that a nontrivial function symbol j:V->V is Sigma_1-elementary and hence fully elementary as a theorem scheme.

    Why it matters

    The answer would determine whether this form of self-embedding can coexist with the specified set-theoretic axioms. An inconsistency proof would rule it out even when the axiom of choice is absent.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Consistency of Reinhardt Cardinals without Choice · 1 problem in this edition. Family membership alone does not assert an implication.

    • #140 · Consistency of Reinhardt Cardinals without Choice (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 115–221.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  141. Rank141Polynomial pivot rule for the simplex methodCan a rule for choosing the simplex method's next step solve every bounded nondegenerate rational linear program using polynomially many pivots in its numbers of variables and constraints, from any feasible starting basis? A randomized rule may meet this bound in expectation.Cryptography, coding, information & optimizationStrongly polynomial linear programming and simplex methods · 2 problems

    Exact statement

    Construct a deterministic or randomized simplex pivot rule and a polynomial p such that, for every bounded nondegenerate rational linear program with m equality constraints and n variables and every feasible starting basis, the rule reaches an optimal basis in at most p(m,n) pivots, in expectation when the rule is randomized; or prove that no such rule exists.

    Why it matters

    A construction would bound the number of steps along feasible edges using only the numbers of constraints and variables. It would control pivot count without, by itself, bounding the computational work needed to choose each pivot.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Related problems

    • #99 · Strongly Polynomial Linear ProgrammingVariantStrongly Polynomial Linear Programming

      Related questions with different algorithmic scopes. The LP target requires a deterministic algorithm for all rational instances, with arithmetic-operation and intermediate-size bounds. The simplex target permits randomization and bounds pivots on bounded nondegenerate instances from a feasible basis.

      Scope qualification: No unconditional implication between these complete targets is asserted. A general LP algorithm need not supply a simplex pivot rule, and this randomized special-case pivot bound alone does not yield the required deterministic LP algorithm.

    Problem family

    Strongly polynomial linear programming and simplex methods · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 120–211.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  142. Rank142Colliot-Thelene Conjecture: Brauer-Manin Obstruction Controls Rational Points on Rationally Connected VarietiesOn smooth proper geometrically rationally connected varieties over number fields, does passing the Brauer–Manin compatibility test ensure that local solutions can be approximated arbitrarily closely by rational points?Number theory & arithmetic geometry

    Exact statement

    Let X be a smooth, proper, geometrically irreducible variety over a number field k whose base change to an algebraic closure is rationally connected. Prove that X(k) is dense in the Brauer-Manin set X(A_k)^{Br(X)}; in particular the Brauer-Manin obstruction is the only obstruction to the Hasse principle and weak approximation for such varieties.

    Why it matters

    A proof would make the Brauer–Manin obstruction the complete obstruction to existence and weak approximation of rational points in this class. A counterexample would demonstrate an additional obstacle beyond that test.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Colliot-Thelene Conjecture: Brauer-Manin Obstruction Controls Rational Points on Rationally Connected Varieties · 1 problem in this edition. Family membership alone does not assert an implication.

    • #142 · Colliot-Thelene Conjecture: Brauer-Manin Obstruction Controls Rational Points on Rationally Connected Varieties (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 129–205.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  143. Rank143Asymptotic tightness of the binary Gilbert–Varshamov boundBinary codes separate bit strings so errors can be detected or corrected. For every fixed relative separation δ between zero and one-half, is the best asymptotic information rate exactly the Gilbert–Varshamov value 1−H₂(δ), where H₂ is binary entropy, even when nonlinear codes are allowed?Cryptography, coding, information & optimization

    Exact statement

    For integers n>=1 and d>=0, let A_2(n,d) be the maximum cardinality of a binary code C subset {0,1}^n, not necessarily linear, with Hamming distance at least d between distinct codewords. For each fixed real 0<delta<1/2 define R_2(delta)=limsup_{n->infinity} n^(-1) log_2 A_2(n,floor(delta n)) and H_2(delta)=-delta log_2(delta)-(1-delta)log_2(1-delta). Is R_2(delta)=1-H_2(delta) for every such delta?

    Why it matters

    Equality would determine the best exponential growth rate of binary code sizes at every such distance. A disproof would show a genuine rate improvement somewhere, without necessarily providing efficiently constructible or decodable codes.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Asymptotic tightness of the binary Gilbert–Varshamov bound · 1 problem in this edition. Family membership alone does not assert an implication.

    • #143 · Asymptotic tightness of the binary Gilbert–Varshamov bound (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 137–180.

    Source-fit opponent count: 46.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  144. Rank144Grothendieck's Variational Hodge ConjectureIn a smooth projective family over a smooth connected complex algebraic base, consider a cohomology class transported consistently across the fibres by a global algebraic horizontal section. If it represents an algebraic cycle with rational coefficients on one fibre, must it do so on every fibre?Geometry & topology

    Exact statement

    For every smooth projective morphism f:X->S of smooth connected complex algebraic varieties, every integer p>=0, and every global algebraic horizontal section xi of the relative algebraic de Rham cohomology bundle R^(2p)f_*Omega^._(X/S) for the Gauss-Manin connection, if xi(s0) is the de Rham cycle class of a rational algebraic codimension-p cycle on one fiber X_s0, then xi(s) is such a rational algebraic cycle class on every fiber X_s for s in S(C).

    Why it matters

    A proof would show that algebraicity persists under the specified global flat transport through a family. It would not automatically produce one cycle on the entire family or an effective method for constructing the fibrewise cycles.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Grothendieck's Variational Hodge Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #144 · Grothendieck's Variational Hodge Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 132–194.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  145. Rank145Congruent Number ProblemExactly which positive integers occur as the area of a right triangle whose three side lengths are rational numbers?Number theory & arithmetic geometryElliptic curve ranks and congruent numbers · 3 problems

    Exact statement

    Determine exactly which positive integers are congruent numbers, i.e. areas of right triangles with all three sides rational, equivalently for which squarefree n the elliptic curve y^2 = x^3 - n^2 x has a rational point with y != 0.

    Why it matters

    A complete answer would classify these rational triangles by their integer areas. For squarefree n, it would also determine exactly when y² = x³ − n²x has a rational point with nonzero y.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The rank-equality theorem applies in particular to congruent-number curves y^2=x^3-n^2 x when a full q-power Selmer group has corank at most one; it then identifies whether their Mordell-Weil rank is zero or one. There is no unconditional classification of every squarefree n or a guarantee that every congruent-number curve meets the low-corank hypothesis. Congruent-number curves have rational two-torsion, so the companion identifies their actual rank when their full 2-power Selmer corank is zero or one. This does not decide congruence for every squarefree n: higher-corank twists and an exhaustive criterion are not covered. The low-corank 2-converse applies to those quadratic twists of y^2=x^3-x whose full 2-power Selmer corank is zero or one, giving the corresponding actual rank. Analytic densities and a mean over signed squarefree twists do not by themselves classify every individual positive congruent number. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Related problems

    Problem family

    Elliptic curve ranks and congruent numbers · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 119–216.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  146. Rank146Global Regularity of Solutions to the Inviscid Surface Quasi-Geostrophic EquationIn the inviscid surface quasi-geostrophic model, a quantity transported across the plane also determines the velocity that carries it. Must every sufficiently smooth, decaying, finite-energy initial state remain smooth for all time?Analysis & PDE

    Exact statement

    For the inviscid SQG equation partial_t theta + u dot grad theta = 0 on R^2, with u = R-perp theta, prove global smooth existence for every sufficiently smooth decaying finite-energy initial datum, or construct such a datum whose solution develops a finite-time singularity.

    Why it matters

    A proof would exclude finite-time singularities throughout the specified class despite the absence of dissipation. A blow-up example would show that this scalar transport law can generate a singularity from smooth initial data.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Global Regularity of Solutions to the Inviscid Surface Quasi-Geostrophic Equation · 1 problem in this edition. Family membership alone does not assert an implication.

    • #146 · Global Regularity of Solutions to the Inviscid Surface Quasi-Geostrophic Equation (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 131–195.

    Source-fit opponent count: 38.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  147. Rank147Leopoldt's conjectureFor every number field and prime p, do the p-adic logarithms of its independent units retain the full expected number of independent directions, so the p-adic regulator never loses rank?Number theory & arithmetic geometry

    Exact statement

    Prove or disprove that for every number field K and every prime p, the Leopoldt defect of K at p is zero, equivalently that the p-adic regulator has the conjectured full rank.

    Why it matters

    A proof would rule out unexpected rank loss in these p-adic regulators uniformly across number fields and primes. A counterexample would identify a specific field and prime where the expected independence fails.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Leopoldt's conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #147 · Leopoldt's conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 132–194.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  148. Rank148Erdos-Rado Sunflower ConjectureFor every integer r≥3, is there a constant C depending only on r such that any family of more than C^k sets of size at most k contains r sets whose pairwise intersections are all the same?Combinatorics & discrete geometry

    Exact statement

    For every fixed integer r >= 3, prove or disprove that there is a constant C_r such that every family of more than C_r^k sets of size at most k contains an r-petal sunflower.

    Why it matters

    A proof would give a purely exponential threshold forcing a sunflower: sets sharing one common core and otherwise disjoint. It would quantify when sufficiently large set families must contain this reusable combinatorial pattern.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Erdos-Rado Sunflower Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #148 · Erdos-Rado Sunflower Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 142–200.

    Source-fit opponent count: 48.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  149. Rank149NP versus the Existential Theory of the RealsIf a system of polynomial equations and inequalities with integer coefficients has a real solution, must there be a certificate of polynomial size that an ordinary computer can check in polynomial time?Applied & computational mathematics

    Exact statement

    Determine whether every problem decidable in the existential theory of the reals (exists R) can be decided in nondeterministic polynomial time, i.e., decide whether NP = exists R.

    Why it matters

    Equality would give polynomial-size, efficiently checkable certificates for every yes-instance in this real-feasibility complexity class. A separation would establish that at least one such problem lies beyond NP.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The source places the existential theory of the reals, and even existential–universal sentences with circuit-encoded integer polynomials, in a fixed level of the counting hierarchy. The full question requires membership specifically in NP. Counting-hierarchy membership alone neither provides NP certificates nor separates exists R from NP, so this is an upper-bound contribution without resolution of the full equality. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    NP versus the Existential Theory of the Reals · 1 problem in this edition. Family membership alone does not assert an implication.

    • #149 · NP versus the Existential Theory of the Reals (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 134–210.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  150. Rank150Density of Hyperbolicity for Rational MapsFor each degree at least two, can every rational map of the Riemann sphere be approximated arbitrarily closely by maps whose critical points all approach attracting periodic cycles?Probability, ergodic theory & dynamicsDensity of hyperbolicity · 2 problems

    Exact statement

    Determine whether hyperbolic rational maps, whose critical points are all attracted to attracting periodic cycles, are dense in the space of rational maps of the Riemann sphere of each fixed degree d >= 2.

    Why it matters

    A positive answer would make this specific attracting-cycle behavior available arbitrarily close to every map of the same degree. A negative answer would reveal a region of parameter space containing no such maps.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Related problems

    Problem family

    Density of hyperbolicity · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 140–201.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  151. Rank151Berman-Hartmanis Isomorphism ConjectureAre all NP-complete decision problems equivalent by a reversible relabeling of binary-string inputs, where both the relabeling and its inverse are computable in polynomial time and the relabeling preserves both yes and no answers?Theoretical computer science

    Exact statement

    Prove or disprove that for every two languages A and B complete for NP under polynomial-time many-one reductions, there is a bijection h on binary strings such that h(A)=B and both h and its inverse are computable in polynomial time.

    Why it matters

    A proof would show that all such complete languages have the same structure up to efficient reversible relabeling. A counterexample would distinguish two complete languages despite their sharing the same reduction-based completeness property.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Berman-Hartmanis Isomorphism Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #151 · Berman-Hartmanis Isomorphism Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 125–233.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  152. Rank152Tate's rational leading-term Stark conjectureFor finite Galois extensions of number fields, do the ratios linking unit regulators to the first nonzero terms of Artin L-functions always lie in their corresponding character fields, compatibly with every automorphism of the complex number field?Number theory & arithmetic geometryExplicit class field theory and Stark conjectures · 3 problems

    Exact statement

    For every finite Galois extension L/K of number fields with group G, let S be any finite set of places of K containing the archimedean and ramified places. Let X be the augmentation kernel of the free Q-module on places of L above S, U=Q tensor_Z O_(L,S)^*, and f:X->U any Q[G]-isomorphism. Define lambda:R tensor U->R tensor X by lambda(u)=-sum_w log|u|_w w, using product-formula normalized absolute values. For each irreducible complex character chi of G, set R_S(chi,f)=det_C(lambda f acting by postcomposition on Hom_C[G](V_(dual chi),C tensor X)) and A_S(chi,f)=R_S(chi,f)/L_S^*(0,chi), where L_S^* is the first nonzero Laurent coefficient of the S-truncated Artin L-function and an empty determinant is1. Is A_S(chi,f) in Q(chi), with sigma(A_S(chi,f))=A_S(chi^sigma,f) for every sigma in Aut(C)?

    Why it matters

    A proof would give these analytically defined ratios precise algebraic locations and Galois symmetries. It would establish the rational leading-term relation without automatically proving integral refinements or constructing Stark units effectively.

    Status

    Open · subcases solved

    October 6, 2026: The October 4 manuscript Stark units for real quadratic fields and reciprocity laws provides claimed related progress in the real-quadratic setting. No full resolution of this rational leading-term conjecture for arbitrary finite Galois extensions and irreducible characters is established by the reviewed evidence. This target is distinct from the additional square-root refinement that the manuscript attributes to Tate and explicitly leaves open.

    Status reviewed

    Sources

    Research on ProofAtlas

    Related problems

    • #180 · Rank One Abelian Stark ConjectureRelated problemRank One Abelian Stark Conjecture

      Related questions with different scopes.

      Scope qualification: The rational Stark statement concerns leading terms and Galois equivariance using rational combinations of units. The rank-one statement also requires an actual unit and an abelian root extension. No implication between these exact formulations is established here; this does not assert logical independence.

    Problem family

    Explicit class field theory and Stark conjectures · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 147–197.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  153. Rank153Cramer Conjecture on the Limsup of Normalized Prime GapsAre the largest long-run gaps between consecutive primes on the scale of the square of the logarithm of the smaller prime, with the limiting upper ratio exactly one?Number theory & arithmetic geometryPrime gaps · 3 problems

    Exact statement

    Prove that lim sup_{n -> infinity} (p_{n+1} - p_n) / (log p_n)^2 = 1, where p_n is the n-th prime.

    Why it matters

    The assertion fixes an exact asymptotic constant for the largest normalized gaps, beyond a bound of the same order. It requires gaps approaching this normalized size arbitrarily far out while excluding any larger limiting upper value.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Prime gaps · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 144–200.

    Source-fit opponent count: 45.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  154. Rank154Smooth Four-Dimensional Schoenflies ProblemDoes every smoothly embedded three-sphere in the four-sphere cut it into two pieces whose closures are each smoothly equivalent to the standard four-dimensional ball?Geometry & topologySmooth four-dimensional spheres · 2 problems

    Exact statement

    Determine whether every smooth embedding of S^3 in S^4 is smoothly unknotted; equivalently, whether each of its two complementary closures is diffeomorphic to the standard 4-ball B^4.

    Why it matters

    A positive answer would rule out smooth knotting of three-spheres in the four-sphere. A negative answer would reveal an embedding whose complementary closures are not both smoothly equivalent to standard balls.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    Smooth four-dimensional spheres · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 139–207.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  155. Rank155Voevodsky's Smash-Nilpotence ConjectureIf an algebraic cycle with rational coefficients has zero intersection number against every complementary cycle, must multiplying copies of it on a product of sufficiently many copies of the variety make it zero modulo rational equivalence, for every smooth projective variety over any field?Number theory & arithmetic geometry

    Exact statement

    For every field k, every smooth projective k-variety X of dimension d, every integer 0<=r<=d, and every class alpha in CH^r(X)_Q (codimension-r cycles modulo rational equivalence with rational coefficients), assume deg(alpha.beta)=0 for every beta in CH^(d-r)(X)_Q. Voevodsky's smash-nilpotence assertion is that there exists an integer N>=1, depending on alpha, for which the N-fold EXTERNAL product alpha times ... times alpha is zero in CH^(Nr)(X^N)_Q, where X^N is the fibre product over k. No algebraic-closure, characteristic-zero or geometric-connectedness hypothesis is imposed; disconnected smooth projective schemes are interpreted on their pure-dimensional components. The reverse implication, smash-nilpotent implies numerically trivial, is known, so this is equality of numerical and smash equivalence for rational cycles. Neither intersection powers on X, composition powers of correspondences, integral coefficients, nor a uniform or computable exponent replaces this assertion.

    Why it matters

    A proof would characterize numerical triviality by actual vanishing after enough external products. The required exponent may depend on the cycle, and the products take place on powers of the variety rather than within the original variety.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Voevodsky's Smash-Nilpotence Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #155 · Voevodsky's Smash-Nilpotence Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 137–215.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  156. Rank156Uniform Boundedness Conjecture for Rational Points on CurvesFor each genus of at least two and each number field, is there one finite bound on rational points that works for every smooth curve of that genus over that field?Number theory & arithmetic geometry

    Exact statement

    For each integer g >= 2 and number field F, prove that there is a constant c(g,F) such that every smooth curve C of genus g over F has at most c(g,F) F-rational points.

    Why it matters

    A proof would make the number of rational points uniformly controllable while the curve varies. A disproof would produce arbitrarily large point counts even after both genus and number field are fixed.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Uniform Boundedness Conjecture for Rational Points on Curves · 1 problem in this edition. Family membership alone does not assert an implication.

    • #156 · Uniform Boundedness Conjecture for Rational Points on Curves (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 141–207.

    Source-fit opponent count: 39.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  157. Rank157Manins Conjecture on the Asymptotic Density of Rational Points on Fano VarietiesA point's height measures its arithmetic size. On every smooth Fano variety over a number field, does counting points up to anticanonical height B on a suitable open part give a leading term cB(log B)^(ρ−1), with ρ the Picard rank?Number theory & arithmetic geometry

    Exact statement

    Prove that for any smooth Fano variety X over a number field K with ample anticanonical bundle, the number of rational points of anticanonical height <= B on a Zariski open subset satisfies N(U, B) ~ c B (log B)^{rho(X)-1}.

    Why it matters

    The formula would connect the growth of rational-point counts to geometric invariants and the chosen height. It would predict a leading asymptotic on the selected open subset as the height cutoff increases.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Manins Conjecture on the Asymptotic Density of Rational Points on Fano Varieties · 1 problem in this edition. Family membership alone does not assert an implication.

    • #157 · Manins Conjecture on the Asymptotic Density of Rational Points on Fano Varieties (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 148–205.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  158. Rank158Dirichlet Divisor ProblemAfter subtracting the standard main term from the total number of divisors of the integers up to x, what is the optimal power of x that bounds the error as x grows, allowing an arbitrarily small extra exponent?Number theory & arithmetic geometryLattice point and divisor error bounds · 2 problemsOpen · disputed claim

    Exact statement

    Let d(n) count the positive divisors of n and Delta(x)=sum_{n<=x} d(n)-x log x-(2 gamma-1)x for real x>=2. Determine the optimal exponent alpha_2=inf{a: for every epsilon>0, Delta(x)=O_epsilon(x^(a+epsilon)) as x tends to infinity}.

    Why it matters

    This would identify the sharp power-law threshold for fluctuations in the cumulative divisor count. The target concerns the exponent, rather than finer logarithmic factors in the error.

    Status

    Open · disputed claim

    Open with unverified claims recorded. Mo (2011) and Jabbarov (May 2026 revision, Theorem 3.1) claim the quarter-exponent bound for the same unsmoothed error term. No independent validation is established here. Bordelles--Daval (January 2026) and Lamzouri (May 2026, after the newer claim) continue to treat that bound as conjectural. No proof claim is certified or adopted as a theorem.

    Status review date not recorded in this edition

    Sources

    Problem family

    Lattice point and divisor error bounds · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 146–215.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  159. Rank159Simplicity of the Nontrivial Zeros of the Riemann Zeta FunctionDoes every nontrivial zero of the Riemann zeta function have multiplicity exactly one, so that the function's derivative is nonzero at each such zero?Number theory & arithmetic geometry

    Exact statement

    Prove that zeta'(rho) is nonzero for every nontrivial zero rho of the Riemann zeta function; equivalently, every nontrivial zero of zeta(s) has multiplicity exactly one.

    Why it matters

    A proof would rule out repeated nontrivial zeros and fix their local multiplicity structure. This addresses how zeta vanishes at its zeros without asserting where those zeros lie.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Simplicity of the Nontrivial Zeros of the Riemann Zeta Function · 1 problem in this edition. Family membership alone does not assert an implication.

    • #159 · Simplicity of the Nontrivial Zeros of the Riemann Zeta Function (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 146–211.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  160. Rank160Capacity of the General Discrete Memoryless Relay ChannelWhat is the highest reliable communication rate through a memoryless channel when a relay can help using only what it has already observed? Find a computable exact answer for every finite-alphabet channel, with reliability measured by average error tending to zero.Cryptography, coding, information & optimizationOpen · disputed claim

    Exact statement

    Find a computable exact average-error Shannon-capacity characterization for every finite-alphabet discrete memoryless relay channel with law p(y,y_r|x,x_r), where the relay input at time i depends only on its observations before time i.

    Why it matters

    A computable characterization would determine the reliable communication limit for every channel in this model. A proof that no such characterization exists would instead establish a limit on computably describing those capacities.

    Status

    Open · disputed claim

    The general relay-channel capacity problem remains open. Ponniah’s proposed formula (first posted September 14, 2026) is challenged by Shiu’s September 16 counterexample: a strictly causal finite-alphabet channel achieves approximately 1.18872 bits/use, above the formula’s bound of approximately 1.04718. The inspected September 20 revision retains the formula and does not establish a resolution of that objection. Shiu credits ChatGPT-6 Astra with obtaining the counterexample. This refutes the proposed formula, not the possibility of a general capacity characterization.

    Status reviewed

    Sources

    Problem family

    Capacity of the General Discrete Memoryless Relay Channel · 1 problem in this edition. Family membership alone does not assert an implication.

    • #160 · Capacity of the General Discrete Memoryless Relay Channel (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 155–210.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  161. Rank161Weinstein Conjecture on periodic orbits of Reeb flowsA contact form determines a distinguished flow called its Reeb flow. On every compact contact manifold without boundary, must every such flow have a trajectory that returns to its starting point?Geometry & topology

    Exact statement

    Prove that every closed contact manifold admits a periodic orbit of the Reeb flow of any contact form, in all dimensions.

    Why it matters

    A proof would guarantee recurring motion for every flow in this geometric class. A counterexample would give a contact form whose Reeb dynamics avoid periodic trajectories entirely on a closed manifold.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Weinstein Conjecture on periodic orbits of Reeb flows · 1 problem in this edition. Family membership alone does not assert an implication.

    • #161 · Weinstein Conjecture on periodic orbits of Reeb flows (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 143–223.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  162. Rank162Conformal Invariance of the Critical Random-Cluster Model with q <= 4As a critical random-cluster model with cluster weight q≤4 is viewed at finer and finer lattice scales, does its separating interface approach the predicted Schramm–Loewner random curve, whose law is unchanged by conformal changes of coordinates?Probability, ergodic theory & dynamics

    Exact statement

    Prove that the exploration path of the critical random-cluster (FK) model with q <= 4 converges in distribution to a Schramm-Loewner evolution SLE_{kappa(q)} as the mesh size goes to zero, i.e., the scaling limit is conformally invariant.

    Why it matters

    A proof would identify the continuum distribution of the exploration path and establish its conformal invariance. It would connect the discrete lattice model to a precisely specified random-curve law in the scaling limit.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The release reports square-lattice Dobrushin-interface convergence to chordal SLE at the self-dual parameter for every fixed 1≤q≤4 on the stated Jordan-domain approximations, including q=4. For 0<q<1, the theorem assumes a bounded smooth Jordan domain and simple nearest-neighbor polygon approximations; slit and pinched boundaries are excluded. These regular-domain results do not establish an unrestricted domain-approximation or lattice-universality statement, so the broader target remains open. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Conformal Invariance of the Critical Random-Cluster Model with q <= 4 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #162 · Conformal Invariance of the Critical Random-Cluster Model with q <= 4 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 123–273.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  163. Rank163Separation of P_R from NP_R in the Blum–Shub–Smale ModelIn a computer model that performs exact real arithmetic at unit cost, is there a decision problem whose real-number certificates can be checked in polynomial time but which no deterministic polynomial-time machine can solve?Applied & computational mathematics

    Exact statement

    In the standard ordered-real Blum–Shub–Smale model, machines use finitely many built-in real constants, real-valued registers, exact field operations (+,-,multiplication,division where defined) and order/equality tests at unit cost. Inputs are finite real vectors and size is vector length. Let P_R be decision problems solved by deterministic machines in polynomially many steps, and NP_R those with polynomial-length real witnesses verifiable in P_R. The assertion is P_R != NP_R.

    Why it matters

    A separation would distinguish verification from computation when exact real operations have unit cost. The conclusion belongs to this real-number machine model and does not automatically establish the corresponding separation for binary-string computation.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Separation of P_R from NP_R in the Blum–Shub–Smale Model · 1 problem in this edition. Family membership alone does not assert an implication.

    • #163 · Separation of P_R from NP_R in the Blum–Shub–Smale Model (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 156–212.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  164. Rank164Omega(n log n) Boolean Circuit Lower Bound for Integer MultiplicationDoes multiplying two n-bit nonnegative integers and producing every bit of their product require at least a constant times n log n arbitrary two-input Boolean gates, however the circuit is arranged, for all sufficiently large n?Theoretical computer science

    Exact statement

    Let C_mult(n) be the least gate count of an acyclic Boolean circuit with 2n input bits, all 2n output bits of the product of the two nonnegative n-bit inputs, arbitrary binary Boolean gates, unrestricted fan-out and depth, and no uniformity requirement. There exist constants c>0 and N such that C_mult(n)>=c*n*log2(n) for every n>=N.

    Why it matters

    A proof would impose this size bound even with arbitrary two-input Boolean gates, unlimited depth and fan-out, and separate circuits for each input length. Refutation would defeat every proposed positive constant at arbitrarily large lengths.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Omega(n log n) Boolean Circuit Lower Bound for Integer Multiplication · 1 problem in this edition. Family membership alone does not assert an implication.

    • #164 · Omega(n log n) Boolean Circuit Lower Bound for Integer Multiplication (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 148–225.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  165. Rank165L-space ConjectureFor a closed oriented irreducible rational homology three-sphere, are these three conditions equivalent: its Heegaard Floer homology has more than minimal size; its nontrivial group of loops up to deformation admits a total ordering preserved by left multiplication; and it can be foliated by surfaces with a consistent transverse orientation, each leaf meeting a closed transverse curve?Geometry & topology

    Exact statement

    Prove or disprove: for a closed, oriented, irreducible rational homology 3-sphere Y, the following are equivalent: (1) Y is not a Heegaard Floer L-space; (2) pi_1(Y) is nontrivial and admits a left-invariant strict total order; (3) Y admits a coorientable taut foliation.

    Wording clarified after assessment

    Earlier comparisons used the wording below. A recorded review treats the published wording above as a clarification of the same intended problem. The wording review itself adds no comparison judgments.

    Earlier assessed wording

    Prove or disprove: for a closed, oriented, irreducible rational homology 3-sphere Y, the following are equivalent: (1) Y is not a Heegaard Floer L-space (its Heegaard Floer homology is not minimal); (2) pi_1(Y) is left-orderable; (3) Y admits a coorientable taut foliation.

    Wording review recorded .

    Why it matters

    A proof would identify the same class of manifolds through Heegaard Floer homology, an ordering of the fundamental group and a foliation condition. A counterexample would separate at least two of these descriptions.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    L-space Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #165 · L-space Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 136–235.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  166. Rank166Keating-Snaith Conjecture on Moments of the Riemann Zeta FunctionFor every real k>0, does the average of |ζ(1/2+it)|^(2k) over 0≤t≤T grow like (log T)^(k²) times the exact Keating–Snaith constant, giving the predicted statistics of zeta's size on the critical line?Number theory & arithmetic geometry

    Exact statement

    For every fixed real k>0, let I_k(T)=integral_0^T |zeta(1/2+it)|^(2k) dt. Prove that I_k(T) is asymptotic to a_k G(1+k)^2/G(1+2k) times T(log T)^(k^2), where G is the Barnes G-function and a_k=product_p (1-1/p)^(k^2) sum_(m>=0) (Gamma(m+k)/(Gamma(k)m!))^2 p^(-m).

    Why it matters

    The formula would determine both the leading growth and the exact leading constant for every positive moment on the critical line. These moments quantify different aspects of the zeta function's size distribution.

    Status

    Open · subcases solved

    October 6, 2026: De Faveri and Pandey (1 October) improve the sixth-moment upper bound to O(T^(37/30+epsilon)). This is not the conjectured asymptotic, even for the sixth moment, and does not resolve all fixed real moments. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Problem family

    Keating-Snaith Conjecture on Moments of the Riemann Zeta Function · 1 problem in this edition. Family membership alone does not assert an implication.

    • #166 · Keating-Snaith Conjecture on Moments of the Riemann Zeta Function (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 153–227.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  167. Rank167Grothendieck–Serre Conjecture on Principal Bundles over Regular Local RingsA principal bundle may become trivial after passing from a regular local ring to its fraction field. Must it already have been trivial over the ring itself, for every reductive group scheme over that ring?Algebra, representation & category theory

    Exact statement

    For every commutative Noetherian regular local ring R, with fraction field K=Frac(R), and every reductive R-group scheme G (smooth affine with connected reductive geometric fibers), every G-torsor over Spec(R) that becomes trivial over Spec(K) is already trivial over Spec(R). Equivalently, the natural map of nonabelian pointed sets H^1_et(R,G) -> H^1_et(K,G) has trivial kernel. Regular local rings here are domains. No field-containing, excellence, finite-type, unramified, quasi-split, isotropic or constant-group restriction is imposed.

    Why it matters

    A proof would make the generic fiber sufficient to detect triviality throughout this class of bundles. A counterexample would identify a bundle whose nontrivial structure disappears on passing to fractions.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Grothendieck–Serre Conjecture on Principal Bundles over Regular Local Rings · 1 problem in this edition. Family membership alone does not assert an implication.

    • #167 · Grothendieck–Serre Conjecture on Principal Bundles over Regular Local Rings (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 159–217.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  168. Rank168Algebraic Independence of e and piIs it impossible for any nonzero polynomial in two variables with rational coefficients to become zero when its variables are replaced by e and π?Number theory & arithmetic geometryExponential transcendence · 3 problems

    Exact statement

    Prove that no nonzero polynomial P in Q[X,Y] satisfies P(e,pi)=0.

    Why it matters

    A proof would exclude every polynomial relation between these two constants. It would also imply that their sum and product are transcendental, while making no blanket claim about arbitrary exponential expressions involving them.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Related problems

    Problem family

    Exponential transcendence · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 161–216.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  169. Rank169Kimura–O'Sullivan Finite-Dimensionality ConjectureDoes every smooth projective variety over any field have a rational Chow motive—the object encoding its algebraic-cycle information—that splits into an even part with a vanishing exterior power and an odd part with a vanishing symmetric power?Number theory & arithmetic geometry

    Exact statement

    For every field k and every smooth projective variety X over k, the Chow motive h(X) with rational coefficients in the ordinary symmetric, pseudo-abelian category CHM(k)_Q is Kimura–O'Sullivan finite-dimensional: there exist motives M+ and M− and positive integers n,m such that h(X) is isomorphic to M+ direct-sum M−, the exterior power Λ^n(M+) is zero, and the symmetric power Sym^m(M−) is zero. The summands may be zero. The field k is unrestricted; characteristic zero applies to the rational coefficient field, not to k. No effective construction, uniform n,m or classification is required.

    Why it matters

    A proof would impose this finite-dimensional structure on all such motives and make numerically trivial endomorphisms nilpotent under composition. It would not assert that their Chow groups have finite dimension as rational vector spaces.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Kimura–O'Sullivan Finite-Dimensionality Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #169 · Kimura–O'Sullivan Finite-Dimensionality Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 155–227.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  170. Rank170Transcendence of Apery's ConstantIs Apéry's constant, ζ(3) = 1 + 1/2³ + 1/3³ + ⋯, transcendental, meaning it satisfies no nonzero polynomial equation with rational coefficients?Number theory & arithmetic geometryOpen · disputed claim

    Exact statement

    Prove or disprove that the value zeta(3) is transcendental over the rational numbers.

    Why it matters

    A proof would rule out every algebraic equation for this infinite sum. A negative answer would place it among algebraic numbers and determine a different arithmetic nature for this particular odd zeta value.

    Status

    Open · disputed claim

    Status review date not recorded in this edition

    Sources

    Problem family

    Transcendence of Apery's Constant · 1 problem in this edition. Family membership alone does not assert an implication.

    • #170 · Transcendence of Apery's Constant (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 147–232.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  171. Rank171Beilinson-Soule Vanishing Conjecture for Rational Motivic CohomologyRational motivic cohomology packages information from algebraic cycles and K-theory. For every smooth scheme over a field, must these groups vanish in nonpositive cohomological degrees, except when both the degree and the weight are zero?Algebra, representation & category theory

    Exact statement

    Prove that for every smooth scheme X over a field k the rational motivic cohomology groups H^(n,i)(X), defined as the i-th graded piece for the gamma filtration on K_(2i-n)(X) tensored with the rationals, vanish whenever n <= 0 and the pair (n,i) is not (0,0).

    Why it matters

    A proof would impose a universal vanishing range on these graded pieces of algebraic K-theory. It would constrain which degrees and weights can carry rational motivic information for smooth schemes.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Beilinson-Soule Vanishing Conjecture for Rational Motivic Cohomology · 1 problem in this edition. Family membership alone does not assert an implication.

    • #171 · Beilinson-Soule Vanishing Conjecture for Rational Motivic Cohomology (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 159–235.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  172. Rank172Inhomogeneous Landau–Coulomb Global Regularity (HST All-Moment Data)In the spatially inhomogeneous Landau–Coulomb plasma model, does the locally constructed classical solution extend for all time from nonnegative Hölder-continuous data that decay faster than any power of velocity, uniformly in position, and have uniform positive-density patches at bounded velocities near every position?Analysis & PDE

    Exact statement

    Fix a positive physical constant a. On [0,infinity) times R^3_x times R^3_v consider the classical Landau equation (partial_t+v dot grad_x)f=Q(f,f), where Q(f,f)=div_v integral_{R^3} A(v-w)[f(t,x,w) grad_v f(t,x,v)-f(t,x,v) grad_w f(t,x,w)]dw and A(z)=a |z|^(-1)(I-z tensor z/|z|^2) for z!=0. For every nonnegative initial datum f_0 on R^6 with f_0 in the global Holder space C^alpha for some 0<alpha<1, sup_{x,v}<v>^k f_0(x,v)<infinity for every k>=0, and constants delta,r_0,R>0 such that for every x there exists |v_x|<R with f_0>=delta on the Euclidean R^6 ball B_{r_0}(x,v_x), does the classical solution branch constructed in HST Theorems 1.2–1.4 have maximal continuation time T_E=infinity? Here <v>=(1+|v|^2)^(1/2); the branch matches f_0 as t decreases to zero, is smooth for positive time, and has all polynomial velocity weights bounded on every finite time slab of its existence. The exact question is global extension of that branch, with the sufficiently decaying uniformly continuous kinetic weak uniqueness convention in HST Section 5, not uniqueness among all generalized weak solutions. The HST Coulomb continuation quantity is Psi(T)=esssup_{0<=t<=T,x} integral f(t,x,v)dv + esssup_{0<=t<=T,x,v} f(t,x,v). Finite Psi on every finite time slab is the associated continuation criterion, not an extra hypothesis presumed true. No smallness near equilibrium, finite total spatial mass or uniform bound across different initial data is imposed.

    Why it matters

    A proof would guarantee global continuation and positive-time smoothness for the entire stated data class, without assuming the data are close to equilibrium. Failure would concern this continuation class and need not rule out all generalized weak solutions.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Inhomogeneous Landau–Coulomb Global Regularity (HST All-Moment Data) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #172 · Inhomogeneous Landau–Coulomb Global Regularity (HST All-Moment Data) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 163–231.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  173. Rank173Furstenberg's x2 x3 conjecture on invariant measuresTake a probability distribution on the circle unchanged by doubling and tripling positions modulo one, which cannot be decomposed into a nontrivial mixture of other such invariant distributions. Must it be uniform or concentrated on finitely many points?Probability, ergodic theory & dynamics

    Exact statement

    Prove that every ergodic probability measure on the unit circle T that is invariant under both maps x -> 2x mod 1 and x -> 3x mod 1 is either the Lebesgue measure or supported on a finite set (equivalently, for coprime a,b >= 2, every ergodic x(a,b)-invariant measure is Lebesgue or finitely supported).

    Why it matters

    A proof would classify the probability distributions compatible with these simultaneous multiplicative symmetries. It would exclude distributions that are neither uniform nor concentrated on finitely many points under the stated ergodicity condition.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Furstenberg's x2 x3 conjecture on invariant measures · 1 problem in this edition. Family membership alone does not assert an implication.

    • #173 · Furstenberg's x2 x3 conjecture on invariant measures (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 162–234.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  174. Rank174Metric TSP Subtour LP Integrality Gap ConjectureFor a traveling-salesman problem with symmetric distances satisfying the triangle inequality, can the cheapest tour cost exceed the fractional Held–Karp lower bound by at most a factor of four-thirds, with that factor best possible?Cryptography, coding, information & optimization

    Exact statement

    Prove or disprove that the integrality gap of the subtour (Held-Karp) LP relaxation of the symmetric metric traveling salesman problem equals 4/3.

    Why it matters

    The answer would determine the worst possible discrepancy between actual tours and this fractional relaxation. A proof of four-thirds would supply a universal comparison between the relaxation's lower bound and the true optimum.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Metric TSP Subtour LP Integrality Gap Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #174 · Metric TSP Subtour LP Integrality Gap Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 139–257.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  175. Rank175Collatz conjectureStarting from any positive integer, must repeatedly halving even numbers and replacing odd numbers by three times the number plus one eventually reach one?Number theory & arithmetic geometry

    Exact statement

    Prove or disprove that, for every positive integer N, repeated application of N -> N/2 when N is even and N -> 3N+1 when N is odd eventually reaches 1.

    Why it matters

    A proof would control every trajectory generated by this single arithmetic rule. A disproof would reveal an orbit that never reaches one, through either a different cycle or an unbounded trajectory.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Collatz conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #175 · Collatz conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 150–259.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  176. Rank176Morton–Silverman Uniform Boundedness ConjectureUnder repeated application of an algebraic map on projective space, some rational points eventually repeat. Is their number uniformly bounded once the dimension, the map's degree and the degree of its number field are bounded? The map must be defined everywhere and have degree at least two.Number theory & arithmetic geometry

    Exact statement

    For all fixed integers N>=1, d>=2 and D>=1, there exists a finite constant C(N,d,D) such that for every number field K with [K:Q]<=D and every degree-d K-morphism phi:P^N_K->P^N_K, the set of K-rational preperiodic points has cardinality at most C(N,d,D). A point is preperiodic when its forward orbit under phi is finite. The morphism is everywhere defined, equivalently given by homogeneous degree-d forms with no common projective zero. The bound is independent of the field and map except for N,d,D; no effective formula or classification is required.

    Why it matters

    A proof would bound finite rational orbits uniformly across all maps and fields with those parameters. A counterexample would show unbounded orbit-point counts even while the three parameters remain fixed.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Morton–Silverman Uniform Boundedness Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #176 · Morton–Silverman Uniform Boundedness Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 165–239.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  177. Rank177Polynomial-Time Quantum Algorithm for Symmetric-Group Hidden SubgroupsSuppose an oracle labels permutations so two permutations get the same label exactly when they lie in the same coset of an unknown subgroup. Can a quantum computer recover that subgroup in time polynomial in the number of objects being permuted?Quantum information & computation

    Exact statement

    Give a polynomial-time quantum algorithm that recovers a subgroup hidden by an oracle over the symmetric group S_n.

    Why it matters

    An algorithm would efficiently recover generators for every hidden subgroup in this oracle setting. It would establish a uniform quantum method for extracting subgroup structure from the information supplied by the hiding oracle.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Polynomial-Time Quantum Algorithm for Symmetric-Group Hidden Subgroups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #177 · Polynomial-Time Quantum Algorithm for Symmetric-Group Hidden Subgroups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 147–255.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  178. Rank178Parity conjecture for elliptic curvesFor every elliptic curve over a number field, does its root-number sign tell whether the group of points over that field has even or odd rank: +1 for even and −1 for odd?Number theory & arithmetic geometryElliptic curve ranks and congruent numbers · 3 problems

    Exact statement

    For every number field K and every elliptic curve E/K, is w(E/K)=(-1)^rank_Z E(K), where rank_Z E(K) is the Mordell-Weil rank and w(E/K)=product_v w(E/K_v) is the global root number? The local factors are the usual Weil-Deligne root numbers at finite places, equal to1 at all but finitely many places, and -1 at every archimedean place; their definition does not assume analytic continuation of L(E/K,s).

    Why it matters

    A proof would determine whether the Mordell–Weil rank is even or odd using local root numbers. In particular, a negative product would guarantee a point of infinite order without necessarily constructing one or giving the exact rank.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. For E/Q with the specified low Selmer corank, equality of analytic and Mordell-Weil ranks gives the desired root-number parity through the functional equation over Q. The target includes all E over arbitrary number fields, including curves without the low-corank hypothesis; that universal assertion remains outside this result. Equality of analytic and actual ranks in the companion yields root-number parity over Q for the stated rational-two-torsion, low-2-Selmer-corank curves. The target quantifies over all curves over all number fields. The H10 undecidability assertion itself is not a general parity theorem. On the low-2-Selmer-corank E/Q subfamily, equality of actual and analytic ranks gives root-number parity via the functional equation. The target is universal over all curves and number fields; Goldfeld density/mean assertions are not a pointwise parity theorem for the remaining curves. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Related problems

    • #6 · Birch and Swinnerton-Dyer ConjectureRelated problemBirch and Swinnerton-Dyer Conjecture

      Related questions with different scopes.

      Scope qualification: The BSD entry concerns elliptic curves over the rationals, whereas the parity entry covers every number field. An implication between these exact scopes is not established here; this does not assert logical independence.

    Problem family

    Elliptic curve ranks and congruent numbers · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 167–232.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  179. Rank179Exact Consistency Strength of the Proper Forcing AxiomWhich large-cardinal assumption exactly matches the consistency strength of the Proper Forcing Axiom? The goal is matching consistency implications: assuming either theory is free of contradictions should establish the consistency of the other, over ordinary set theory.Logic, foundations & set theory

    Exact statement

    Determine the exact large-cardinal consistency strength of ZFC plus the Proper Forcing Axiom (PFA): identify its equiconsistency level by matching upper and lower consistency bounds.

    Why it matters

    A complete answer would locate PFA precisely within the large-cardinal consistency hierarchy. It would require matching consistency implications that establish both the necessary and sufficient strength.

    Status

    Open

    Open; the 2025 Oberwolfach report continues to identify the exact consistency strength of PFA as unresolved. Relative consistency from a supercompact cardinal is an upper bound, not an equiconsistency theorem for PFA.

    Status review date not recorded in this edition

    Sources

    Problem family

    Exact Consistency Strength of the Proper Forcing Axiom · 1 problem in this edition. Family membership alone does not assert an implication.

    • #179 · Exact Consistency Strength of the Proper Forcing Axiom (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 168–241.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  180. Rank180Rank One Abelian Stark ConjecturePartial zeta functions sort a number field's arithmetic by the symmetries of an abelian extension. In Stark's rank-one case, with a chosen place splitting completely, do their derivatives at zero come from logarithms of one unit's conjugates, with its e-th root giving an abelian extension, where e counts roots of unity in the larger field?Number theory & arithmetic geometryExplicit class field theory and Stark conjectures · 3 problems

    Exact statement

    Let K/F be an abelian extension of number fields with Galois group G, and let S be a finite set of places of F containing the archimedean and ramified places, with |S| >= 2 and at least one place v of S splitting completely in K. Prove that, for a fixed place w of K above v, there exists a unit u in U_{v,S} such that zeta'_{K/F,S}(sigma, 0) = -(1/e) log |u^sigma|_w for all sigma in G, where e is the number of roots of unity in K, and K(u^{1/e})/F is abelian.

    Why it matters

    A proof would turn these analytic derivatives into explicit identities involving a single arithmetic unit. The additional radical condition would connect that unit to the symmetry of a further number-field extension.

    Status

    Open · subcases solved

    October 6, 2026: The October 4 manuscript Stark units for real quadratic fields and reciprocity laws claims specialized real-quadratic progress. It explicitly omits the square-root refinement concerning abelianity of the root extension. Both the arbitrary-base-field scope and the root-extension requirement in the full statement here remain beyond the claimed result. This is claimed partial progress, not a resolution of this target. The omitted refinement is not needed for the manuscript’s separate SIC existence conclusion.

    Status reviewed

    Sources

    Related problems

    • #35 · Extension of Kroneckers Theorem on Abelian Fields to Any Algebraic Base FieldStronger assertionExtension of Kroneckers Theorem on Abelian Fields to Any Algebraic Base Field

      A proof of #35 would also establish this assertion.

    • #152 · Tate's rational leading-term Stark conjectureRelated problemTate's rational leading-term Stark conjecture

      Related questions with different scopes.

      Scope qualification: The rational Stark statement concerns leading terms and Galois equivariance using rational combinations of units. The rank-one statement also requires an actual unit and an abelian root extension. No implication between these exact formulations is established here; this does not assert logical independence.

    Problem family

    Explicit class field theory and Stark conjectures · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 162–249.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  181. Rank181Simpson's Motivicity Conjecture for Rigid Local SystemsDoes every irreducible rigid complex local system on a smooth connected quasi-projective variety, with finite-order determinant and quasi-unipotent boundary monodromy, come from the cohomology of a smooth projective family after restriction to a dense open set?Number theory & arithmetic geometry

    Exact statement

    Let X be a smooth connected complex quasi-projective variety with a smooth normal-crossings compactification, and let L be an irreducible complex local system of any finite positive rank. Assume its determinant has finite order, its monodromies around the boundary components are quasi-unipotent, and L is rigid: an isolated point in the moduli space with its determinant and boundary monodromy conjugacy classes fixed. Prove or disprove that L is of geometric origin: there exist a dense Zariski-open U in X, a smooth projective morphism f:Y to U, and an integer i>=0 such that L restricted to U is a direct summand of R^i f_* C_Y.

    Also known as

    Geometric origin of rigid local systems; Simpson's geometric-origin conjecture; Motivicity of rigid irreducible local systems

    Why it matters

    A positive answer would connect these rigid representations to algebraic geometry: each would occur as a direct summand of a family's cohomology. A counterexample would identify a rigid system outside this geometric origin.

    Status

    Open · subcases solved

    October 6, 2026: Simpson’s closedness theorem for strata of de Rham moduli is not the geometric-origin assertion for all rigid local systems required here. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Problem family

    Simpson's Motivicity Conjecture for Rigid Local Systems · 1 problem in this edition. Family membership alone does not assert an implication.

    • #181 · Simpson's Motivicity Conjecture for Rigid Local Systems (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 179–231.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  182. Rank182UCT Problem for Separable Nuclear Complex C*-AlgebrasCan KK-theory for every separable nuclear complex C*-algebra be computed from K-theory through the universal coefficient exact sequence? This must work with every separable partner algebra, whether or not either algebra has an identity.Algebra, representation & category theory

    Exact statement

    Decide whether EVERY separable nuclear complex C*-algebra A, unital or nonunital, satisfies the Rosenberg-Schochet UCT: for EVERY separable complex C*-algebra B, unital or nonunital, and each j in Z/2, the canonical UCT sequence 0 -> E_j(A,B) -> KK_j(A,B) -> H_j(A,B) -> 0 is exact. Here H_j(A,B)=direct_sum_(i=0,1) Hom_Z(K_i(A),K_(i+j)(B)) and E_j(A,B)=direct_sum_(i=0,1) Ext_Z^1(K_i(A),K_(i+j+1)(B)), with indices modulo2. The right map is the natural map on K-theory induced by KK classes; the left is the canonical inverse extension-class identification of the kernel, not an arbitrary chosen injection. No canonical or natural splitting is required. Equivalently, every such A belongs to the Rosenberg-Schochet bootstrap category of separable nuclear C*-algebras, or is KK-equivalent to a separable commutative complex C*-algebra.

    Why it matters

    A proof would make the prescribed K-theoretic computation of KK-groups available for every separable nuclear algebra. It would remove the UCT assumption from classification results wherever that is the remaining applicable hypothesis.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    UCT Problem for Separable Nuclear Complex C*-Algebras · 1 problem in this edition. Family membership alone does not assert an implication.

    • #182 · UCT Problem for Separable Nuclear Complex C*-Algebras (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 175–238.

    Source-fit opponent count: 38.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  183. Rank183Palis finite-attractor full-basin density conjectureCan every differentiable invertible map or flow on a compact connected manifold without boundary be perturbed so that almost every starting point is attracted to one of finitely many attractors, each with a dense orbit? This is asked at every finite differentiability order r≥1, with the perturbation arbitrarily small through r derivatives.Probability, ergodic theory & dynamics

    Exact statement

    For every compact connected boundaryless smooth manifold M and every finite integer r >= 1, prove the conjunction of the following two density assertions: (i) in Diff^r(M) with its C^r topology there is a dense subset D whose elements f have finitely many attractors whose basins of attraction together cover M up to a set of smooth-volume measure zero; (ii) the same holds in the space X^r(M) of C^r vector fields with its C^r topology, for the attractors and basins of the generated flows. Attractor follows Palis (2005) Sections 2.1 and 2.3: a compact invariant set A that is transitive (has a dense orbit) and attracts a set of positive volume, its basin being the set of points whose forward orbit (or forward flow trajectory) converges to A. The finite number of attractors may depend on the element of D; no uniform bound is required. Volume-zero is independent of the choice of smooth positive density on M. This is the first bullet of Palis's Global Conjecture only; physical measures, stochastic stability, metric stability under parameter perturbation, and one-dimensional-family statements are separate conjectures not included.

    Why it matters

    A proof would make systems with finitely many attracting destinations dense for both discrete and continuous time. Almost every initial condition in each approximating system would approach one of those attractors.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Palis finite-attractor full-basin density conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #183 · Palis finite-attractor full-basin density conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 159–258.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  184. Rank184Landau's Fourth Problem on Primes of the Form n^2+1Are there infinitely many positive integers n for which n² + 1 is prime, so that prime numbers occur indefinitely immediately after perfect squares?Number theory & arithmetic geometry

    Exact statement

    Prove or disprove that there are infinitely many primes of the form n^2+1 with n a positive integer.

    Why it matters

    A proof would guarantee endlessly many prime values of this particular quadratic polynomial. A negative answer would mean that every sufficiently large value of n² + 1 is composite.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Landau's Fourth Problem on Primes of the Form n^2+1 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #184 · Landau's Fourth Problem on Primes of the Form n^2+1 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 169–256.

    Source-fit opponent count: 32.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  185. Rank185Soliton Resolution for the Focusing Energy-Critical Wave Equation without Radial SymmetryIn dimensions three through five, must every solution of the focusing energy-critical wave equation that exists forever with bounded energy-space norm eventually separate into finitely many solitons—persistent localized waves—plus freely traveling radiation, with error tending to zero at all late times?Analysis & PDE

    Exact statement

    For d in {3,4,5}, consider the real focusing wave equation u_tt-Delta u=|u|^(4/(d-2))u on R^d, with energy data in dot-H1 x L2 and the usual strong energy solution on [0,infinity), uniformly bounded in that norm. Assertion: every such solution has, for all t->infinity, a decomposition of (u,u_t) into a finite-energy free wave, finitely many scaled and translated Lorentz transforms of fixed nonzero finite-energy stationary solutions, and a remainder tending to zero in dot-H1 x L2. More precisely, use the global CaseII decomposition(1.10) of DJKM1601.01871v3 with t replacing t_n: J>=0 is finite, each velocity ell_j has |ell_j|<1, lambda_j(t)>0 with lambda_j(t)/t->0, c_j(t)/t->ell_j, and for j!=k the sum lambda_j/lambda_k+lambda_k/lambda_j+|c_j-c_k|/lambda_j tends to infinity as in(1.9). No radiality, compact support, ground-state-only or distinct-velocity hypothesis is imposed.

    Why it matters

    A proof would describe the eventual structure of every solution in this class without requiring radial symmetry. The decomposition must hold throughout late-time evolution, which is stronger than holding only along selected sequences of times.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Soliton Resolution for the Focusing Energy-Critical Wave Equation without Radial Symmetry · 1 problem in this edition. Family membership alone does not assert an implication.

    • #185 · Soliton Resolution for the Focusing Energy-Critical Wave Equation without Radial Symmetry (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 172–250.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  186. Rank186Exact Asymptotic Waring Function G(k)For each exponent k at least three, what is the smallest number of positive kth powers whose sums represent every sufficiently large positive integer, allowing fewer summands when needed?Number theory & arithmetic geometry

    Exact statement

    For each integer k>=3, let G(k) be the least positive integer s for which every sufficiently large positive integer N is expressible as a sum of at most s positive integral k-th powers. Determine the entire function G(k) for all k>=3. Equivalently, zeros may be allowed and exactly s nonnegative integral k-th powers used; this is not the condition of exactly s positive summands.

    Why it matters

    Determining this function would give the exact eventual additive cost of representing integers by each kind of power. It would distinguish necessary numbers of summands from sufficient but excessive bounds.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Exact Asymptotic Waring Function G(k) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #186 · Exact Asymptotic Waring Function G(k) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 173–253.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  187. Rank187Local Smoothing Conjecture for Wave EquationsDoes averaging over time improve the Lᵖ regularity estimate for free waves by almost 1/p of a derivative, in every dimension n≥3 and for every p>2n/(n−1)?Analysis & PDEKakeya, restriction and smoothing estimates · 4 problems

    Exact statement

    For every n >= 3, p > 2n/(n-1), and epsilon > 0, prove ||exp(it sqrt(-Delta))f||_{L^p(R^n x [1,2])} <= C_{p,epsilon} ||f||_{W^{s_p-1/p+epsilon,p}(R^n)}, where s_p=(n-1)(1/2-1/p).

    Why it matters

    A proof would quantify how integrating a wave's behavior over time improves the regularity estimate. It would establish the specified gain across the entire dimension and exponent range, with every positive epsilon loss allowed.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. For n=3 the full wave target has p>3 and regularity 1-3/p+epsilon. The family states the three-dimensional Euclidean wave estimate throughout the full strict range, including every positive loss at the critical L^3 point, so it covers the n=3 part of the full assertion. It does not claim the estimates in all n>=4, and therefore cannot close the full all-dimensional target. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Sourced history

    First posed
    1991
    Attributed to
    Christopher Sogge
    First publication
    1991

    David Beltran, Jonathan Hickman, and Christopher D. Sogge, Local smoothing for Fourier integral operators, historical appendix

    Problem family

    Kakeya, restriction and smoothing estimates · 4 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 183–241.

    Source-fit opponent count: 32.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  188. Rank188Artin's primitive root conjectureFor each integer a that is neither a square nor −1, do the primes for which powers of a generate every nonzero residue class form the exact positive proportion predicted by Artin, including its correction factor?Number theory & arithmetic geometry

    Exact statement

    Prove or disprove that every integer a other than -1 that is not a perfect square is a primitive root modulo p for a positive natural density of primes p, with the density given by Artin's predicted correction factor.

    Why it matters

    A proof would quantify how often a fixed integer generates the multiplicative group modulo a prime. It would supply the predicted density for each permitted integer, beyond merely guaranteeing infinitely many such primes.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The source treats every admissible integer base but only proves at least c_a x/(log x)^2 primitive-root primes in each sufficiently large dyadic interval, hence infinitude. The full target requires positive natural density among primes with Artin's predicted correction factor; the stated lower bound permits density zero and gives no asymptotic constant. The simultaneous-prime-base companion is conditional on four inputs and must not upgrade this to the full density conjecture. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Artin's primitive root conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #188 · Artin's primitive root conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 174–256.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  189. Rank189Lehmer's Mahler Measure ProblemFor a monic integer polynomial, multiply the absolute values of its roots outside the unit circle to obtain its Mahler measure. Must every such measure greater than one stay at least a fixed positive distance above one?Number theory & arithmetic geometry

    Exact statement

    Prove that there exists an absolute epsilon > 0 such that every monic integer polynomial P with Mahler measure M(P) > 1 satisfies M(P) >= 1 + epsilon, or construct a sequence of such polynomials whose Mahler measures decrease to 1.

    Why it matters

    A gap would give a uniform restriction on how close these algebraic size measurements can come to their minimum above one. A sequence tending to one would show that no universal positive separation exists.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Lehmer's Mahler Measure Problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #189 · Lehmer's Mahler Measure Problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 171–256.

    Source-fit opponent count: 44.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  190. Rank190Bochner-Riesz conjectureBochner–Riesz means reconstruct functions from Fourier data with adjustable smoothing. In every dimension at least two, which smoothing strengths keep the reconstructed function uniformly controlled in each Lᵖ norm, for 1<p<∞?Analysis & PDEKakeya, restriction and smoothing estimates · 4 problems

    Exact statement

    Prove the sharp conjectured characterization of all orders delta greater than or equal to zero and exponents 1<p<infinity for which the Euclidean Bochner-Riesz means are uniformly bounded on L^p(R^d), in every dimension d at least 2.

    Why it matters

    The characterization would identify precisely when these smoothed Fourier reconstructions keep Lᵖ norms under uniform control. It would distinguish the allowable smoothing and integrability parameters from those where that control fails.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The abstract gives the full conjectured strict-order Bochner-Riesz range in dimension three, starting from L^3 boundedness for every positive order. The full target asks for a sharp characterization including delta=0 and all dimensions d>=2. No all-dimensional statement or complete boundary-order characterization is provided, so only the three-dimensional strict-range subcase is covered. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Kakeya, restriction and smoothing estimates · 4 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 135–313.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  191. Rank191Skolem Problem for Linear Recurrence SequencesIs there an algorithm that always decides whether a sequence defined by a linear recurrence over the integers or algebraic numbers ever contains a zero term?Logic, foundations & set theory

    Exact statement

    Decide algorithmically, given an integer (or algebraic) linear recurrence sequence, whether it has a zero term.

    Why it matters

    A positive resolution would replace an unbounded search through terms with a terminating decision procedure. A negative resolution would establish an inherent limit on algorithmic detection of zero occurrences.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Skolem Problem for Linear Recurrence Sequences · 1 problem in this edition. Family membership alone does not assert an implication.

    • #191 · Skolem Problem for Linear Recurrence Sequences (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 150–280.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  192. Rank192Polynomial Freiman-Ruzsa Conjecture for General Abelian GroupsIf adding a finite set to itself creates at most K times as many elements, must polynomially many translates of one convex coset progression—a higher-dimensional additive pattern combined with a subgroup—cover it? The progression must have dimension logarithmic in K and size at most a polynomial in K times the original set's size, in every abelian group.Combinatorics & discrete geometry

    Exact statement

    Prove that there is an absolute constant C such that for every finite subset A of an arbitrary abelian group with |A+A| <= K|A|, the set A can be covered by exp(C log K) translates of a centred convex coset progression of dimension at most C log K and size at most exp(C log K)|A|.

    Why it matters

    A proof would turn limited growth under addition into an efficient structural description. One absolute constant would control the covering count, progression size, and dimension across all abelian groups.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Polynomial Freiman-Ruzsa Conjecture for General Abelian Groups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #192 · Polynomial Freiman-Ruzsa Conjecture for General Abelian Groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 167–263.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  193. Rank193Brauer-Manin obstruction for number-field curves: Hasse-principle formFor a smooth projective curve of genus at least two over a number field, can failure to have a point over that field always be detected by the Brauer–Manin test, which checks the compatibility of solutions over all its local completions?Number theory & arithmetic geometry

    Exact statement

    For every number field k and every smooth projective geometrically connected curve C/k of genus at least two, if C(A_k)^Br is nonempty then C(k) is nonempty. Here Br(C)=H^2_et(C,G_m) is the full cohomological Brauer group, and C(A_k)^Br consists of adelic tuples (P_v) for which sum_v inv_v(alpha(P_v))=0 for every alpha in Br(C). Constants obey global reciprocity. Passing at archimedean places to connected components does not change nonemptiness. No rational base point, degree-one divisor class, Sha finiteness, rank bound or effective Jacobian presentation is assumed.

    Why it matters

    A proof would show that the full Brauer–Manin obstruction explains every failure of rational-point existence in this class. A counterexample would show that even all these compatibility tests can miss a global obstruction.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Brauer-Manin obstruction for number-field curves: Hasse-principle form · 1 problem in this edition. Family membership alone does not assert an implication.

    • #193 · Brauer-Manin obstruction for number-field curves: Hasse-principle form (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 182–269.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  194. Rank194Babai's diameter conjecture for nonabelian finite simple groupsIn a nonabelian finite simple group, can every element be written as a short product from any chosen generating set and its inverses, with length bounded by a fixed power of the logarithm of the group's size? The exponent must work for every group and generating set.Algebra, representation & category theory

    Exact statement

    Is there an absolute constant c>0 such that for every nonabelian finite simple group G and every generating set X of G, the undirected Cayley graph with steps X union X^{-1} has diameter at most (ln |G|)^c? Thus every element of G is a word of that length in X and its inverses, with the same c for all groups and generating sets.

    Why it matters

    A proof would uniformly bound word lengths regardless of how generators are chosen. It would guarantee short paths throughout every such Cayley graph using only a fixed power of the logarithm of group size.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Babai's diameter conjecture for nonabelian finite simple groups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #194 · Babai's diameter conjecture for nonabelian finite simple groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 186–264.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  195. Rank195Set Cover Conjecture (SeCoCo)For every fixed ε between 0 and 1, is there a fixed set-size bound k for which no bounded-error randomized algorithm solves k-Set Cover on an n-element universe in O*(2^((1−ε)n)) time? The bound k may depend on ε.Theoretical computer science

    Exact statement

    For every fixed epsilon in(0,1), does there exist a fixed positive integer k such that no classical bounded-error randomized algorithm solves k-Set Cover in worst-case time O*(2^((1-epsilon)n))? An instance consists of an n-element universe U, an explicitly listed family F of subsets of U each of cardinality at most k, and an integer t; the decision is whether at most t members of F cover U. The algorithm must succeed with probability at least2/3 on every instance. O* hides polynomial factors in the input length. Preserve the quantifier order: k may depend on epsilon; the assertion does not prohibit a k-dependent improvement for each separately fixed k.

    Also known as

    Set Cover Conjecture; Bounded-set-size Set Cover exponential-time conjecture

    Why it matters

    This asks whether the exponential running-time saving must shrink as the allowed set size grows. It leaves room for a different improvement for each separately fixed k.

    Status

    Open

    Status reviewed

    Sources

    Problem family

    Set Cover Conjecture (SeCoCo) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #195 · Set Cover Conjecture (SeCoCo) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 189–256.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  196. Rank196Polynomial Diameter Bound for PolytopesWhen moving between vertices of a polyhedron along its edges, is the number of steps always bounded by one polynomial in its dimension and number of flat bounding faces? The question includes unbounded polyhedra, provided they contain no complete line.Cryptography, coding, information & optimization

    Exact statement

    Prove or disprove that there is a polynomial p such that the vertex-edge graph of every pointed d-dimensional polyhedron, bounded or unbounded, with n facets has diameter at most p(n,d).

    Why it matters

    A proof would uniformly limit how far apart vertices can be in the edge graph. A counterexample family would show that this combinatorial travel distance can exceed every polynomial bound.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Polynomial Diameter Bound for Polytopes · 1 problem in this edition. Family membership alone does not assert an implication.

    • #196 · Polynomial Diameter Bound for Polytopes (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 190–269.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  197. Rank197Perfect-Matching Search in Deterministic NC (General Graphs)Can a deterministic parallel algorithm find a perfect matching—disjoint edges covering every vertex—or report that none exists in any finite simple undirected graph, using polynomially many processors and time polynomial in log n? The algorithm must have a logspace-uniform circuit description.Theoretical computer science

    Exact statement

    Assert the existence of one logspace-uniform family (C_n) of deterministic Boolean AND/OR/NOT circuits which, for every n>=1 and every finite simple undirected graph G on labelled vertices 1,...,n given by its n-by-n adjacency matrix, outputs a perfect matching of G if one exists and outputs a distinguished NO flag if none exists. Matching output is an n-by-n edge-indicator matrix, required to encode disjoint edges covering every vertex and belonging to G; its bits are ignored when NO is output. Circuit size is bounded by C n^c and depth by C (log(n+2))^k for constants C,c,k independent of n and G, and AND/OR fan-in is at most two. A deterministic logspace transducer on unary 1^n outputs the complete circuit description, including all designated outputs. This fixes ordinary logspace-uniform NC for the multi-output search relation. It does not conflate a search algorithm with an unweighted decision algorithm or restrict the input to bipartite or planar graphs.

    Why it matters

    A positive answer would give a uniform parallel algorithm that constructs the matching itself without randomness. A negative answer would establish a limitation on search in this precise circuit model.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Perfect-Matching Search in Deterministic NC (General Graphs) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #197 · Perfect-Matching Search in Deterministic NC (General Graphs) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 181–281.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  198. Rank198Virasoro Conjecture for Gromov–Witten InvariantsThe full Gromov–Witten generating function packages curve-counting invariants into one formal series. For every smooth projective complex variety, must its coefficients obey all the predicted Virasoro relations?Geometry & topology

    Exact statement

    For every smooth projective complex variety X, let Z_X be its total descendant Gromov–Witten potential in all genera and curve classes over the Novikov ring. For every integer k>=-1, the Eguchi–Hori–Xiong–Katz Virasoro operator L_k^X, with the conventions and formula of Getzler Section 1.4, annihilates Z_X: L_k^X Z_X=0, coefficient by coefficient, including the graded conventions for odd cohomology.

    Why it matters

    A proof would impose the stated family of identities across every genus and curve class, including odd-cohomology conventions. A counterexample would locate an operator and coefficient where the proposed identity fails.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The full target quantifies over every smooth projective complex variety. The complete-intersection theorem covers all genera, curve classes and cohomology insertions, including odd and primitive classes, for that class of varieties. The companion transfer theorem extends known cases to projectivizations of arbitrary rank-at-least-two bundles and their towers, but assumes the full constraints on the base. These classes and the conditional transfer do not cover every smooth projective target. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Virasoro Conjecture for Gromov–Witten Invariants · 1 problem in this edition. Family membership alone does not assert an implication.

    • #198 · Virasoro Conjecture for Gromov–Witten Invariants (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 188–273.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  199. Rank199Witten's Asymptotic Expansion Conjecture for SU(2) WRT InvariantsDo the SU(2) Witten–Reshetikhin–Turaev quantum invariants of every closed oriented three-manifold have the predicted expansion to all orders as the level grows, combining phases from flat SU(2) connections with fractional powers of the inverse level?Mathematical physics

    Exact statement

    For every closed oriented three-manifold Y, let WRT_k(Y) be the SU(2) Witten-Reshetikhin-Turaev invariant at level k-2 in the standard compact-gauge normalization WRT_k(S^3)=1. For every value S in CS(Y), the set of Chern-Simons actions modulo Z of flat SU(2) connections, there should exist a formal Puiseux Laurent series W_S(tau) in union_{n>=1} C((tau^(1/n))) such that WRT_k(Y) ~ sum_{S in CS(Y)} exp(2*pi*i*k*S) W_S(1/k) as integer k tends to infinity, as a full Poincare asymptotic expansion. The separately named growth-rate and explicit leading-coefficient conjectures are not additional requirements.

    Why it matters

    A proof would express the full asymptotic behavior through the specified phase-indexed formal series. The normalization fixes the sphere’s invariant at one, and the target leaves the series’ leading coefficients unspecified.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Witten's Asymptotic Expansion Conjecture for SU(2) WRT Invariants · 1 problem in this edition. Family membership alone does not assert an implication.

    • #199 · Witten's Asymptotic Expansion Conjecture for SU(2) WRT Invariants (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 198–261.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  200. Rank200Polynomial-Time Quantum Algorithm for the Dihedral Hidden Subgroup ProblemA function on a dihedral group hides a subgroup by taking the same value exactly on its cosets. Can a quantum algorithm recover that subgroup using time and memory polynomial in log N, where N is the number of rotations?Quantum information & computation

    Exact statement

    Find a polynomial-time quantum algorithm with poly(log N) operations and space for the hidden subgroup problem in dihedral groups D_N.

    Why it matters

    A construction would make subgroup recovery efficient relative to the logarithmic size parameter of the group. Both the operation count and memory usage must meet the bound; improving just one does not satisfy the target.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Polynomial-Time Quantum Algorithm for the Dihedral Hidden Subgroup Problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #200 · Polynomial-Time Quantum Algorithm for the Dihedral Hidden Subgroup Problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 170–301.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  201. Rank201Effective Roth TheoremFor any real irrational root α of an integer polynomial and any positive rational ε, can an algorithm compute a denominator cutoff beyond which no reduced fraction p/q satisfies |α−p/q|<1/q^(2+ε)?Number theory & arithmetic geometry

    Exact statement

    There exists one total computable algorithm which, given a real algebraic irrational alpha by its integer minimal polynomial and a rational isolating interval, and a positive rational epsilon, returns an integer Q>=1 such that every reduced rational p/q, with p an integer and q>=1 an integer, satisfying |alpha-p/q|<q^(-2-epsilon) has q<=Q. The algorithm must halt with this guarantee on every valid input. No time, space, practical-size or software-implementation requirement is imposed.

    Why it matters

    Such an algorithm would make the entire finite set of exceptionally close rational approximations computable from the inputs. Its existence would not promise practical runtime or a reasonably sized bound.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Effective Roth Theorem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #201 · Effective Roth Theorem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 183–283.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  202. Rank202Density of Hyperbolicity for Complex PolynomialsCan every complex polynomial of degree at least two be changed arbitrarily little, without changing its degree, so that each point where its derivative vanishes is drawn into an attracting periodic cycle under repeated application?Probability, ergodic theory & dynamicsDensity of hyperbolicity · 2 problems

    Exact statement

    For each complex polynomial T of degree at least two, determine whether T can be approximated by polynomials of the same degree all of whose critical points converge under iteration to attracting periodic cycles, that is, whether hyperbolic polynomials are dense in the space of complex polynomials of each degree.

    Why it matters

    A proof would guarantee that the specified attracting behavior can be obtained by arbitrarily small coefficient changes. A counterexample would exhibit a polynomial with an entire neighborhood avoiding that behavior.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Related problems

    Problem family

    Density of hyperbolicity · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 179–300.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  203. Rank203Hopf conjecture for positive sectional curvatureMust a closed even-dimensional manifold with positive curvature in every tangent two-plane have positive Euler characteristic—the alternating count of cells in a decomposition of the manifold?Geometry & topologyHopf curvature conjectures · 2 problems

    Exact statement

    Prove that every closed even-dimensional Riemannian manifold whose sectional curvatures are everywhere positive has positive Euler characteristic.

    Why it matters

    A proof would force the sign of a global topological invariant from a pointwise geometric condition. A counterexample would show that positive sectional curvature can coexist with a nonpositive Euler characteristic.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Hopf curvature conjectures · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 184–277.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  204. Rank204Zagier's dimension conjecture for classical multiple zeta valuesMultiple zeta values are nested infinite sums, grouped by the sum k of their exponents, called their weight. Is the number d_k of rationally independent values at each weight governed by d_k=d_(k−2)+d_(k−3), with d_0=1, d_1=0 and d_2=1?Number theory & arithmetic geometry

    Exact statement

    For positive integers k_1,...,k_r with k_1>=2, define the classical convergent real multiple zeta value zeta(k_1,...,k_r)=sum_(n_1>...>n_r>0) 1/(n_1^k_1 ... n_r^k_r), of weight k_1+...+k_r. For each k>=2 let Z_k be their Q-linear span at weight k, with Z_0=Q generated by the empty value1 and Z_1={0}. Define d_0=1,d_1=0,d_2=1 and d_k=d_(k-2)+d_(k-3) for k>=3. Is dim_Q Z_k=d_k for every k>=0?

    Why it matters

    A proof would count the independent multiple zeta values separately at each weight. It would determine exactly how many numbers are needed to express all values of that weight by rational linear combinations.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Zagier's dimension conjecture for classical multiple zeta values · 1 problem in this edition. Family membership alone does not assert an implication.

    • #204 · Zagier's dimension conjecture for classical multiple zeta values (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 201–269.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  205. Rank205Polynomial-Size Perfect Secret Sharing for General Access StructuresCan every access structure on n participants share a uniformly random one-bit secret with polynomially many bits per participant, with perfect reconstruction for authorized sets and perfect secrecy for all others? The polynomial must work for every access structure; computation need not be efficient.Cryptography, coding, information & optimization

    Exact statement

    Do there exist absolute constants C>0 and k>=0 such that, for every n>=1 and every nonempty upward-closed family A of subsets of [n] with the empty set not in A, a uniformly random one-bit secret S can be shared as finite-alphabet random variables X_1,...,X_n with max_i ceil(log_2 |supp(X_i)|)<=C n^k, such that every set in A reconstructs S with probability one and the joint shares of every set outside A are statistically independent of S? Schemes may be nonlinear and need not be efficiently computable. A proof of this uniform polynomial bound or its negation settles the target. The negation means that every proposed polynomial bound fails for some n and access structure; neither a full asymptotic formula nor an exponential lower bound is required.

    Also known as

    Polynomial share size for arbitrary perfect secret-sharing access structures; Superpolynomial lower-bound barrier for unrestricted one-bit secret sharing

    Why it matters

    This would bound the information that each participant must store for any prescribed pattern of authorized groups. A negative answer need only rule out every uniform polynomial bound; it need not prove an exponential lower bound.

    Status

    Open

    Status reviewed

    Sources

    Problem family

    Polynomial-Size Perfect Secret Sharing for General Access Structures · 1 problem in this edition. Family membership alone does not assert an implication.

    • #205 · Polynomial-Size Perfect Secret Sharing for General Access Structures (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 187–283.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  206. Rank206Berry random-wave conjecture: real negative-curvature BS formulationOn a closed connected negatively curved manifold of dimension at least two, do real high-frequency Laplace eigenfunctions look statistically like an isotropic Gaussian random wave when viewed near a uniformly chosen point at wavelength scale? This is asked for the full sequence in every real orthonormal eigenbasis.Mathematical physics

    Exact statement

    For every closed connected smooth Riemannian manifold (M,g) of dimension d>=2 with everywhere strictly negative sectional curvature, put mu=dvol_g/vol_g(M). For any real orthonormal eigenbasis of L2(M,mu) for the positive Laplace-Beltrami operator, order the eigenvalues nondecreasingly and omit the finite zero-eigenspace; write Delta_g phi_j=alpha_j^2 phi_j with alpha_j>0. If p is sampled with law mu, does the law of the pointed real-decorated manifold [M,alpha_j^2 g,p,phi_j] converge weakly, along the full positive-energy sequence, in Abert-Bergeron-Le Masson's local smooth topology on pointed isometry classes of real-decorated d-manifolds, to the law of [R^d,g_Euclidean,0,F]? Here F is the centered real stationary isotropic Gaussian field with covariance E[F(x)F(y)]=integral_{S^{d-1}} cos(<x-y,theta>) d sigma(theta), where sigma is probability surface measure on the unit sphere (unit variance and unit spectral radius, no extra 2*pi factor).

    Why it matters

    A proof would prescribe the local distribution of high-frequency eigenfunctions under these sampling and scaling rules. It would identify one Gaussian law for every allowed manifold and basis, including the full sequence of positive eigenvalues.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Berry random-wave conjecture: real negative-curvature BS formulation · 1 problem in this edition. Family membership alone does not assert an implication.

    • #206 · Berry random-wave conjecture: real negative-curvature BS formulation (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 197–275.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  207. Rank207Zauner's Conjecture: Unrestricted Complex SIC Existence in Every DimensionCan one choose d² equally separated complex lines in every dimension d≥2? Precisely, unit vectors along the lines must have the same squared overlap, |〈v,w〉|²=1/(d+1), for every distinct pair.Quantum information & computationResolution claim under review · historical rank

    Exact statement

    Decide whether for EVERY integer d>=2 there exist d^2 unit vectors v_1,...,v_(d^2) in the standard complex Hilbert space C^d with |<v_i,v_j>|^2=1/(d+1) whenever i!=j. Their complex lines are distinct. Equivalently, their rank-one projectors P_i=v_i v_i^* have sum_i P_i=d I_d and define SIC-POVM elements M_i=P_i/d; for i!=j the normalized trace is Tr(M_i M_j)=1/[d^2(d+1)]. No Weyl-Heisenberg covariance, order-three symmetry, explicit coordinates or uniform construction algorithm is imposed.

    Why it matters

    A proof would supply these symmetric quantum measurements in every finite dimension, without requiring any additional symmetry or explicit construction. A negative answer would identify a dimension where the exact equal-overlap geometry is impossible.

    Status

    Resolution claim under review

    Affirmative proof claim pending refereeing. A manuscript dated October 4, 2026 claims complex SIC-POVM existence in every dimension, the full target stated here, using reciprocity laws for real-quadratic Stark units and earlier SIC construction results of Appleby–Flammia–Kopp and Radchenko–Wheeler. The document discloses that almost all its arguments were generated by large language models and subsequently checked and edited by Radchenko and Wheeler; the models are not named. Preliminary mathematical reviews found no definite fatal error within their scope. The four supplied reviews used the same model family and do not constitute independent-family votes, outside referee reports or formal verification. Full independent validation remains unestablished in the evidence reviewed. No public formalization of this proof was identified as of October 6, 2026. The omitted square-root refinement of Stark’s conjecture is not required for the stated SIC implication.

    Status reviewed

    Sources

    Problem family

    Zauner's Conjecture: Unrestricted Complex SIC Existence in Every Dimension · 1 problem in this edition. Family membership alone does not assert an implication.

    • #207 · Zauner's Conjecture: Unrestricted Complex SIC Existence in Every Dimension (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 188–283.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  208. Rank208Parshin Conjecture on Higher K-Theory over Finite FieldsHigher algebraic K-theory records information about a variety beyond its vector bundles. For a smooth projective variety over a finite field, must every positive-degree K-theory class have finite order, becoming zero after multiplication by some positive integer?Number theory & arithmetic geometry

    Exact statement

    Prove that for every smooth projective variety X over a finite field and every integer n greater than zero, the higher algebraic K-group K_n(X) is torsion.

    Why it matters

    A proof would show that these higher K-groups contain only finite-order elements. Equivalently, their positive-degree information would disappear when rational coefficients are used.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Parshin Conjecture on Higher K-Theory over Finite Fields · 1 problem in this edition. Family membership alone does not assert an implication.

    • #208 · Parshin Conjecture on Higher K-Theory over Finite Fields (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 198–271.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  209. Rank209Shafarevich Freeness Conjecture over QabAfter adjoining every abelian extension of the rationals, is the remaining absolute Galois group a free profinite group on countably infinitely many generators? Here profinite freeness uses a generating basis converging to the identity.Number theory & arithmetic geometry

    Exact statement

    Let Qab be the maximal abelian extension of Q in an algebraic closure Qbar (equivalently the maximal cyclotomic extension). The absolute Galois group Gal(Qbar/Qab) is a free profinite group of countably infinite rank, in the standard profinite convention of a free basis converging to the identity.

    Why it matters

    A proof would determine the profinite structure of the Galois symmetries remaining after every abelian extension is included. It would identify that group with the stated universal free object.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Shafarevich Freeness Conjecture over Qab · 1 problem in this edition. Family membership alone does not assert an implication.

    • #209 · Shafarevich Freeness Conjecture over Qab (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 197–274.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  210. Rank210Sharp Erdos Distinct-Distances Conjecture in the PlaneMust every arrangement of n distinct points in the Euclidean plane determine at least a constant times n/√log n different pairwise distances, for sufficiently large n?Combinatorics & discrete geometry

    Exact statement

    Let D(n) be the minimum, over all sets of n distinct points in the Euclidean plane, of the number of distinct pairwise distances determined by the set. Prove D(n) = Omega(n/sqrt(log n)), matching up to a constant factor the square-lattice construction D(n) = O(n/sqrt(log n)).

    Why it matters

    A proof would determine the smallest possible number of planar distance values up to constant factors. The lower bound would match the distance count achieved by square-lattice arrangements.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Sharp Erdos Distinct-Distances Conjecture in the Plane · 1 problem in this edition. Family membership alone does not assert an implication.

    • #210 · Sharp Erdos Distinct-Distances Conjecture in the Plane (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 160–316.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  211. Rank211Polynomial-Time Distribution-Free PAC Learning of DNF FormulasAn unknown Boolean rule is an OR of s clauses, each an AND of bit conditions. Can it be learned from random labeled examples drawn from any unknown distribution, with small error and high confidence, in time polynomial in the number of bits, s, inverse error and log of inverse failure probability?Theoretical computer science

    Exact statement

    Given an unknown s-term DNF f on {0,1}^n and random labeled examples drawn from an arbitrary unknown distribution D, construct a possibly improper hypothesis in time polynomial in n, s, 1/epsilon, and log(1/delta) whose D-error is at most epsilon with probability at least 1-delta.

    Why it matters

    A proof would provide efficient learning of these Boolean rules using examples alone, without assumptions on how inputs are distributed. The output could use any hypothesis representation meeting the stated error guarantee.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Polynomial-Time Distribution-Free PAC Learning of DNF Formulas · 1 problem in this edition. Family membership alone does not assert an implication.

    • #211 · Polynomial-Time Distribution-Free PAC Learning of DNF Formulas (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 199–295.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  212. Rank212Shub-Smale τ-Conjecture for Integer ZerosA polynomial can be built from 1 and x using additions, subtractions and multiplications. Is the number of its distinct integer roots always bounded by a fixed polynomial in the length of its shortest such recipe, for every nonzero integer polynomial?Theoretical computer science

    Exact statement

    Determine whether universal constants a > 0 and c >= 1 exist such that every nonzero polynomial f in Z[x] has at most a(1 + tau(f))^c distinct integer zeros, where tau(f) is the shortest constant-free straight-line-program length using addition, subtraction, and multiplication, starting from 1 and x.

    Wording clarified after assessment

    Earlier comparisons used the wording below. A recorded review treats the published wording above as a clarification of the same intended problem. The wording review itself adds no comparison judgments.

    Earlier assessed wording

    Determine whether universal constants a and c exist such that every univariate integer polynomial f has at most a·τ(f)^c distinct integer zeros, where τ(f) is its shortest constant-free straight-line-program length using addition, subtraction, and multiplication.

    Wording review recorded .

    Why it matters

    A proof would bound the number of integer solutions by the length of a polynomial's arithmetic description. A counterexample family would have root counts exceeding every fixed polynomial bound in that length plus one.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Shub-Smale τ-Conjecture for Integer Zeros · 1 problem in this edition. Family membership alone does not assert an implication.

    • #212 · Shub-Smale τ-Conjecture for Integer Zeros (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 170–328.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  213. Rank213Chromatic Number of the Plane (Hadwiger--Nelson Problem)Allowing arbitrary colorings, with no requirement that color classes be measurable, what is the fewest colors needed for every point of the Euclidean plane so that points exactly one unit apart have different colors?Combinatorics & discrete geometry

    Exact statement

    Determine the chromatic number of the Euclidean plane: the minimum number of colours needed to colour every point so that points at distance exactly one have distinct colours, with ordinary arbitrary colourings in ZFC and no measurability requirement.

    Why it matters

    A resolution would pair a coloring of the entire plane with a proof that fewer colors cannot suffice. It would determine the exact coloring cost of this single geometric distance constraint.

    Status

    Open

    Earlier note (null): Open; primary sources give 5 <= chi(R^2) <= 7. No exact-value resolution was located in the review. October 7, 2026: OpenAI’s October 6 release reports partial progress. The source rules out arbitrary five-colorings of the plane and explicitly requires no regularity of color classes, matching the full ordinary-coloring model. Together with the seven-color upper bound it narrows the answer to six or seven. The exact chromatic number is not determined, so this is partial progress and must not remove the full problem. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Chromatic Number of the Plane (Hadwiger--Nelson Problem) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #213 · Chromatic Number of the Plane (Hadwiger--Nelson Problem) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 200–277.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  214. Rank214Capacity of the Binary Deletion ChannelFor every deletion probability between zero and one, how many information bits per input bit can be transmitted with average error tending to zero when bits independently disappear, the survivors keep their order, and neither sender nor receiver knows the deletion pattern?Cryptography, coding, information & optimization

    Exact statement

    Determine the exact average-error Shannon capacity C_del(d), in bits per input bit, of the binary deletion channel for every real d with 0<d<1. Each input bit is independently deleted with probability d; surviving bits retain their order, and neither terminal observes the deletion pattern.

    Why it matters

    An exact answer would identify the reliable communication limit for every nontrivial deletion probability. It would quantify the capacity loss from deletions without by itself providing efficient capacity-achieving codes.

    Status

    Open

    Open. The September 2026 primary preprint explicitly records that exact capacity is unknown and offers a uniform numerical approximation; this is partial progress, not an exact-function resolution.

    Status review date not recorded in this edition

    Sources

    Problem family

    Capacity of the Binary Deletion Channel · 1 problem in this edition. Family membership alone does not assert an implication.

    • #214 · Capacity of the Binary Deletion Channel (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 203–280.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  215. Rank215Kawamata's K-Equivalence Implies D-Equivalence ConjectureIf two smooth projective complex varieties have linearly equivalent canonical divisors after passage to a common smooth birational model, must their derived categories of coherent sheaves also be equivalent? The question links agreement of canonical geometry to agreement of these categories.Geometry & topology

    Exact statement

    Let X and Y be smooth projective complex varieties. If a common smooth variety Z has projective birational morphisms f:Z→X and g:Z→Y with f^*K_X linearly equivalent to g^*K_Y, prove an exact equivalence DbCoh(X)≃DbCoh(Y) of triangulated categories.

    Why it matters

    A proof would turn this agreement of canonical divisors into an equivalence of the varieties' derived categories. It would connect a birational geometric condition to the organization of coherent sheaves and their complexes.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Kawamata's K-Equivalence Implies D-Equivalence Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #215 · Kawamata's K-Equivalence Implies D-Equivalence Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 213–272.

    Source-fit opponent count: 37.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  216. Rank216Morrison-Kawamata Cone ConjectureFor a complex projective Q-factorial klt pair with numerically trivial log canonical divisor, do symmetries reduce both its effective nef cone and its effective movable cone to rational polyhedral fundamental domains? In other words, can finite-sided regions describe these divisor cones up to automorphisms and pseudo-automorphisms, respectively?Geometry & topology

    Exact statement

    For every projective Q-factorial klt pair (X, Delta) over the complex numbers with K_X + Delta numerically trivial, prove both that Aut(X, Delta) acting on the effective nef cone Nef^e(X) and that PsAut(X, Delta) acting on the effective movable cone Mov^e(X) have rational polyhedral fundamental domains.

    Why it matters

    A proof would reduce both potentially complicated divisor cones to finitely described rational polyhedral regions after accounting for their prescribed symmetries. A counterexample could reveal a failure in either cone.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Morrison-Kawamata Cone Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #216 · Morrison-Kawamata Cone Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 200–282.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  217. Rank217Global Existence of Classical Solutions to the Relativistic Vlasov-Maxwell SystemIn a three-dimensional relativistic plasma, particles and electromagnetic fields evolve together through the Vlasov–Maxwell equations. Does every sufficiently regular finite-energy initial state yield a classical solution for all future time?Analysis & PDE

    Exact statement

    Prove or disprove that the Cauchy problem for the three-dimensional relativistic Vlasov-Maxwell system has a global classical solution (f,E,B) for every sufficiently regular finite-energy initial datum.

    Why it matters

    A proof would guarantee that this kinetic description of particles and fields continues classically for the entire specified data class. A counterexample would show that regular finite-energy data can lose that form of evolution in finite time.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. For the three-dimensional one-species model without background charge, the manuscript reports global existence and uniqueness for arbitrarily large smooth nonnegative particle data compactly supported in position and momentum, with smooth bounded-derivative L2 electromagnetic fields satisfying the Gauss constraints. The full target permits sufficiently regular finite-energy data without those compact-support and derivative restrictions. That broader class remains unresolved by this result. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Global Existence of Classical Solutions to the Relativistic Vlasov-Maxwell System · 1 problem in this edition. Family membership alone does not assert an implication.

    • #217 · Global Existence of Classical Solutions to the Relativistic Vlasov-Maxwell System (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 192–297.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  218. Rank218Serre Uniformity QuestionFor elliptic curves over the rationals without complex multiplication, does the Galois action on points killed by multiplication by p become as large as possible for every sufficiently large prime p, with one threshold working for all such curves?Number theory & arithmetic geometry

    Exact statement

    Determine whether there exists a bound N > 0 such that for every non-CM elliptic curve E over Q and every prime p > N, the residual Galois representation rho_{E,p} on E[p] is surjective onto GL_2(F_p).

    Why it matters

    A proof would give a single cutoff valid for all such elliptic curves. A counterexample family would show that exceptional Galois actions can persist at arbitrarily large primes as the curve varies.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Serre Uniformity Question · 1 problem in this edition. Family membership alone does not assert an implication.

    • #218 · Serre Uniformity Question (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 205–300.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  219. Rank219Quantum versus Classical Communication Complexity of Total FunctionsTwo parties hold separate inputs and must compute a Boolean answer defined for every input pair. If they can do this with little quantum communication and small error, can a randomized classical protocol do it with only polynomially more communication and small error?Cryptography, coding, information & optimization

    Exact statement

    Prove that for every total Boolean function in the two-party communication model, its bounded-error randomized communication complexity is bounded by a polynomial in its bounded-error quantum communication complexity.

    Why it matters

    A proof would limit quantum communication savings to a polynomial relationship for total functions. A counterexample would demonstrate a larger separation while retaining an answer for every possible input.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Quantum versus Classical Communication Complexity of Total Functions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #219 · Quantum versus Classical Communication Complexity of Total Functions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 200–299.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  220. Rank22011/8-Conjecture on intersection forms of spin 4-manifoldsA four-manifold's intersection form has a rank and a signature, the difference between its positive and negative directions. For every smooth closed oriented spin four-manifold, must the rank be at least eleven-eighths of the absolute signature?Geometry & topology

    Exact statement

    Prove that every smooth closed oriented spin 4-manifold M satisfies b2(M) >= (11/8)|sigma(M)|.

    Why it matters

    A proof would impose a universal numerical restriction on the intersection forms of these manifolds. It would rule out candidate forms whose signature is too large relative to their rank.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    11/8-Conjecture on intersection forms of spin 4-manifolds · 1 problem in this edition. Family membership alone does not assert an implication.

    • #220 · 11/8-Conjecture on intersection forms of spin 4-manifolds (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 188–320.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  221. Rank221Existence of an odd perfect numberIs there an odd positive integer whose positive divisors, including the integer itself, sum to exactly twice that integer, or must every perfect number be even?Number theory & arithmetic geometry

    Exact statement

    Determine whether there exists an odd positive integer N satisfying sigma(N)=2N, where sigma is the sum-of-positive-divisors function; equivalently, prove that every perfect number is even or construct an odd perfect number.

    Why it matters

    A positive answer would establish a perfect number of odd parity. A negative answer would impose an absolute parity restriction on every integer satisfying this divisor-sum equation.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Existence of an odd perfect number · 1 problem in this edition. Family membership alone does not assert an implication.

    • #221 · Existence of an odd perfect number (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 203–292.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  222. Rank222MAJORITY is not in ACC0Can majority—whether at least half the input bits are one—ever be computed by polynomial-size Boolean circuits with a fixed number of layers, using AND, OR, NOT and gates testing divisibility of their input count by one fixed modulus? The AND, OR and divisibility gates may take arbitrarily many inputs.Theoretical computer science

    Exact statement

    Let MAJORITY_n(x)=1 exactly when sum_i x_i>=n/2. Define nonuniform ACC0 as Boolean families computed, for every input length n>=1, by circuits of polynomial size and some fixed constant depth using unbounded-fan-in AND, OR and MOD_m gates and fan-in-one NOT gates, with constants allowed and one fixed integer modulus m>=2 across all lengths; MOD_m outputs1 exactly when the input sum is divisible by m. Size counts input and internal gates; depth counts gate layers. The depth, modulus and polynomial size bound may depend on the family, not on n. Is the family (MAJORITY_n) outside ACC0?

    Why it matters

    A proof would establish a circuit-size limitation for this explicit counting task, across every fixed depth and modulus. A negative answer would show that a shallow modular circuit family can compute majority.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    MAJORITY is not in ACC0 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #222 · MAJORITY is not in ACC0 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 216–276.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  223. Rank223Breuil-Mezard conjecture: equality of the Galois and automorphic sidesA Galois representation modulo a prime encodes arithmetic symmetries using matrices. Do the Galois-side and locally algebraic automorphic-side constructions attached to it always produce the same representation?Number theory & arithmetic geometry

    Exact statement

    Conjecture (Breuil-Mezard): the equality sigma_Gal = sigma_Aut holds, where sigma_Gal and sigma_Aut are the Galois and automorphic (locally algebraic) representations attached to a mod p Galois representation.

    Why it matters

    A proof would identify the two representations obtained from these different constructions for every permitted input. A counterexample would pinpoint a mod p representation where the proposed agreement fails.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Breuil-Mezard conjecture: equality of the Galois and automorphic sides · 1 problem in this edition. Family membership alone does not assert an implication.

    • #223 · Breuil-Mezard conjecture: equality of the Galois and automorphic sides (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 186–314.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  224. Rank224Equidistribution of Galois Orbits of Special Points on Shimura VarietiesFor special points on a Shimura variety, do their finite sets of Galois conjugates become uniformly distributed according to the natural invariant probability measure, provided each fixed proper special subvariety contains only finitely many points of the sequence?Probability, ergodic theory & dynamics

    Exact statement

    For any sequence of special points on a Shimura variety that is eventually outside each proper special subvariety of the ambient variety, prove that its finite Galois orbits equidistribute with respect to normalized Haar measure. The qualification is pointwise in the subvariety: each fixed proper special subvariety contains only finitely many terms.

    Wording clarified after assessment

    Earlier comparisons used the wording below. A recorded review treats the published wording above as a clarification of the same intended problem. The wording review itself adds no comparison judgments.

    Earlier assessed wording

    For any sequence of special points on a Shimura variety eventually outside every special subvariety, prove that its finite Galois orbits equidistribute with respect to normalized Haar measure.

    Wording review recorded .

    Why it matters

    A proof would turn this precise avoidance condition into a distribution law for the points' arithmetic conjugates. A counterexample would reveal concentration or another deviation that persists despite that condition.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Equidistribution of Galois Orbits of Special Points on Shimura Varieties · 1 problem in this edition. Family membership alone does not assert an implication.

    • #224 · Equidistribution of Galois Orbits of Special Points on Shimura Varieties (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 176–339.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  225. Rank225Dynamical Mordell–Lang ConjectureRepeatedly apply an algebraic map to a point and record when its orbit lands in a fixed algebraic subvariety. In characteristic zero, are these return times always a finite union of arithmetic progressions and isolated times, for every everywhere-defined map of a quasiprojective variety?Number theory & arithmetic geometry

    Exact statement

    For every field K of characteristic zero, every quasiprojective variety X over K, every K-morphism f:X->X, every closed K-subvariety V of X and every point x in X(K), the set {n in Z_{≥0}: f^n(x) is in V(K)} is a finite union of arithmetic progressions {a+bm:m in Z_{≥0}} with a,b nonnegative integers. A zero step b=0 permits a singleton, and an empty union permits no returns. All iterates are defined because f is a morphism. No dominance, etaleness, smoothness, automorphism or degree restriction is imposed.

    Why it matters

    A proof would force these orbit-intersection times to follow finitely many recurring schedules with finitely many isolated exceptions. A counterexample would reveal a return-time pattern outside that arithmetic structure.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Dynamical Mordell–Lang Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #225 · Dynamical Mordell–Lang Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 226–281.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  226. Rank226The k-Server Conjecture for General Metric SpacesRequests arrive one at a time at locations in a metric space, and one of k servers must move to each requested location. Can a deterministic strategy always keep total travel within k times the best schedule that knew all requests in advance, plus a fixed additive constant?Theoretical computer science

    Exact statement

    For every integer k >= 1 and every metric space, prove that there is a deterministic online k-server algorithm whose movement cost on every request sequence is at most k times the offline optimum, up to an additive constant.

    Why it matters

    A proof would give a universal guarantee on the price of serving requests without knowing future requests. It would make the factor k attainable for every underlying metric space.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    The k-Server Conjecture for General Metric Spaces · 1 problem in this edition. Family membership alone does not assert an implication.

    • #226 · The k-Server Conjecture for General Metric Spaces (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 197–320.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  227. Rank227Serre's Conjecture II on Galois Cohomology of Simply Connected Semisimple GroupsA torsor is a space that becomes a copy of its symmetry group after extending the field. Over a perfect field of cohomological dimension at most two, must every torsor for a semisimple simply connected algebraic group already have a point over the original field?Algebra, representation & category theory

    Exact statement

    Let K be a perfect field of cohomological dimension at most 2. Prove that every principal homogeneous space under a semisimple simply connected linear algebraic K-group has a rational point, i.e. H^1(K,G) = 1 for every such group G.

    Why it matters

    A proof would eliminate all nontrivial first Galois-cohomology classes for these groups and fields. Every such torsor would admit a rational point, which identifies it with the group as a homogeneous space.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Serre's Conjecture II on Galois Cohomology of Simply Connected Semisimple Groups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #227 · Serre's Conjecture II on Galois Cohomology of Simply Connected Semisimple Groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 205–302.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  228. Rank228All-Pairs Shortest Paths in Truly Subcubic TimeCan we compute the exact shortest-path distances between every pair of vertices in a weighted graph in time with an exponent strictly below three? The saving in the exponent must stay fixed as the number of vertices grows.Cryptography, coding, information & optimizationResolution claim under review · historical rank

    Exact statement

    Determine whether exact all-pairs shortest paths on weighted n-vertex graphs can be computed in O(n^(3-epsilon)) time for some fixed epsilon greater than zero.

    Why it matters

    A positive answer would give an algorithm whose exponent beats three by a fixed amount. A negative answer would rule out that whole scale of exact running times in the applicable computation model.

    Status

    Resolution claim under review

    Claimed pending refereeing — integer case formally verified. Alman and Vassilevska Williams claim deterministic O(n^2.99942) time for polynomially bounded integer weights (Theorem 22), and Las Vegas O(n^2.998) expected time for real weights in the stated real RAM (Theorem 35), with no negative cycles. The integer theorem passed the archived Lean and Nanoda CI checks; the source coordinator also rebuilt it in Lean. Nanoda was not rerun locally. The formalization explicitly excludes the real-input runtime theorem, so the broader target remains pending and has no full-target Lean verification.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    All-Pairs Shortest Paths in Truly Subcubic Time · 1 problem in this edition. Family membership alone does not assert an implication.

    • #228 · All-Pairs Shortest Paths in Truly Subcubic Time (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 214–308.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  229. Rank229Anosov Diffeomorphism Classification ConjectureAn Anosov diffeomorphism has uniformly contracting and expanding directions. Must every such dynamical system become a hyperbolic algebraic automorphism of an infranilmanifold after a continuous change of coordinates?Probability, ergodic theory & dynamics

    Exact statement

    Prove that every Anosov diffeomorphism is topologically conjugate to a hyperbolic automorphism of an infranilmanifold.

    Why it matters

    A proof would classify all Anosov diffeomorphisms up to topological conjugacy using these algebraic models. A counterexample would establish an Anosov system whose dynamics cannot be represented by any such model.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Anosov Diffeomorphism Classification Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #229 · Anosov Diffeomorphism Classification Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 213–297.

    Source-fit opponent count: 39.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  230. Rank230Polynomial-Time Decidability of Semidefinite Programming FeasibilityCan an algorithm efficiently decide whether a rational symmetric matrix expression can be made positive semidefinite, meaning it has no negative eigenvalues? This asks for exact semidefinite-program feasibility in time polynomial in the number of bits needed to write the input.Cryptography, coding, information & optimization

    Exact statement

    Determine whether feasibility of a semidefinite program with rational symmetric input matrices can be decided in polynomial time in the Turing bit model.

    Why it matters

    A positive answer would provide an exact polynomial-time decision procedure, including instances where approximation alone does not settle feasibility. A negative result would delimit that algorithmic guarantee in its stated complexity model.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Polynomial-Time Decidability of Semidefinite Programming Feasibility · 1 problem in this edition. Family membership alone does not assert an implication.

    • #230 · Polynomial-Time Decidability of Semidefinite Programming Feasibility (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 176–351.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  231. Rank231Banach-Mazur Rotation ProblemSuppose a real Banach space has enough linear distance-preserving symmetries to move any unit vector to any other. If it is infinite-dimensional and separable, must its norm come from an inner product, making it a Hilbert space?Analysis & PDE

    Exact statement

    Let X be an infinite-dimensional separable real Banach space whose group of surjective linear isometries acts transitively on the unit sphere. Prove that X is linearly isometric to a Hilbert space, or construct a separable non-Hilbert counterexample.

    Why it matters

    A proof would show that complete symmetry of the unit sphere forces the norm to come from an inner product. A counterexample would separate this symmetry from Hilbert-space geometry.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Banach-Mazur Rotation Problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #231 · Banach-Mazur Rotation Problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 204–316.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  232. Rank232Vaught's conjectureA complete first-order theory describes a class of mathematical structures. In a countable language, can the number of distinct countable models, treating isomorphic models as the same, only be at most countable or as large as the set of real numbers?Logic, foundations & set theory

    Exact statement

    Prove or disprove that for every complete theory T in a countable first-order language, the number I(T,aleph_0) of pairwise nonisomorphic countable models of T is either at most countable or exactly 2^{aleph_0}.

    Why it matters

    A proof would exclude intermediate uncountable sizes from the spectrum of countable models. A counterexample would exhibit a theory whose nonisomorphic countable realizations occupy such an intermediate size.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Vaught's conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #232 · Vaught's conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 216–307.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  233. Rank233Constant-Query Locally Decodable Codes with Polynomial BlocklengthCan binary codes of length polynomial in message length recover any message bit using only a constant number of queries, even after a constant fraction of encoded positions is corrupted?Cryptography, coding, information & optimization

    Exact statement

    Construct binary codes of blocklength polynomial in the message length that permit recovery of any message bit with a constant number of queries from words corrupted in a constant fraction of positions, or prove that no such family exists.

    Why it matters

    A construction would combine polynomial storage expansion, constant local access and tolerance of widespread corruption. An impossibility proof would show that these three requirements cannot coexist in such a code family.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Constant-Query Locally Decodable Codes with Polynomial Blocklength · 1 problem in this edition. Family membership alone does not assert an implication.

    • #233 · Constant-Query Locally Decodable Codes with Polynomial Blocklength (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 223–306.

    Source-fit opponent count: 34.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  234. Rank234Infinitude of Mersenne PrimesAre there infinitely many prime exponents p for which the number 2^p−1 is itself prime, and hence infinitely many even perfect numbers?Number theory & arithmetic geometry

    Exact statement

    Prove or disprove that there are infinitely many primes p for which the Mersenne number 2^p - 1 is prime. Equivalently (by the Euclid-Euler theorem), prove or disprove that there are infinitely many even perfect numbers.

    Why it matters

    A resolution would decide whether these two linked families continue indefinitely or eventually stop. It would settle infinitude without requiring a formula for locating every member.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Infinitude of Mersenne Primes · 1 problem in this edition. Family membership alone does not assert an implication.

    • #234 · Infinitude of Mersenne Primes (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 194–378.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  235. Rank235Hopf conjecture for nonnegative sectional curvatureDoes nonnegative curvature in every tangent two-dimensional direction force a closed even-dimensional manifold to have nonnegative Euler characteristic, its alternating count of topological features?Geometry & topologyHopf curvature conjectures · 2 problems

    Exact statement

    Prove that every closed even-dimensional Riemannian manifold whose sectional curvatures are everywhere nonnegative has nonnegative Euler characteristic.

    Why it matters

    A proof would connect a local curvature inequality to a global topological invariant. It would prevent any closed even-dimensional manifold with negative Euler characteristic from carrying such a metric.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Hopf curvature conjectures · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 226–320.

    Source-fit opponent count: 34.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  236. Rank236Reliability Function of the Binary Symmetric Channel Below the Critical RateWhen each transmitted bit independently flips with probability 0<p<1/2 and there is no feedback, what is the best asymptotic exponent of average block-error decay? Determine it for fixed-blocklength codes at every positive rate below the channel's critical rate, optimizing over all codes with equally likely messages. Use the limsup of −log₂(error)/block length.Theoretical computer scienceChannel reliability · 1 problem

    Exact statement

    Determine the optimal fixed-blocklength reliability function of the memoryless binary symmetric channel without feedback for every crossover probability 0<p<1/2 and every rate 0<R<R_crit(p). With equiprobable messages, let P_e^*(n,R,p) be the minimum average block error under maximum-likelihood decoding among binary length-n codes of cardinality at least ceil(2^(nR)). Define E(R,p)=limsup_{n->infinity}[-log_2 P_e^*(n,R,p)]/n. Here H_2 is binary entropy and R_crit(p)=1-H_2(sqrt(p)/(sqrt(p)+sqrt(1-p))). Success requires matching achievable and converse exponents, or an equivalent evaluable exact characterization, for all these parameters. Restating the operational optimization or merely improving unmatched bounds is not a resolution.

    Why it matters

    An exact answer would identify the optimal tradeoff between information rate and error decay in this basic noisy channel. It would provide a performance benchmark; an efficient encoding or decoding algorithm would be a separate achievement.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Channel reliability · 1 problem in this edition. Family membership alone does not assert an implication.

    • #236 · Reliability Function of the Binary Symmetric Channel Below the Critical Rate (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 232–305.

    Source-fit opponent count: 34.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  237. Rank237Gauss Circle ProblemCount the integer grid points inside a disk centered at the origin and compare that count with the disk's area. What is the sharp size of the error as its radius r grows: is it bounded by a constant times r^(1/2+ε) for every ε>0?Number theory & arithmetic geometryLattice point and divisor error bounds · 2 problems

    Exact statement

    Determine the sharp order of the error term E(r) in N(r) = pi*r^2 + E(r), where N(r) counts the integer lattice points in the disk of radius r centered at the origin; in particular, prove or disprove that E(r) = O(r^(1/2+eps)) for every eps > 0.

    Why it matters

    A resolution would identify the precision with which area predicts this lattice-point count. The specified bound would constrain the fluctuations to barely above a square-root scale in the radius.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Lattice point and divisor error bounds · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 229–335.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  238. Rank238Equivalence or Strict Containment of QMA and QMA(2)Can two quantum proofs guaranteed to be unentangled certify more problems than one unrestricted quantum proof? In both cases the verifier must run in uniform polynomial time with bounded error, and the total proof length is polynomial in the input size.Quantum information & computationQuantum verification classes · 2 problems

    Exact statement

    Determine whether QMA(2)=QMA in the ordinary unrelativized, uniform polynomial-time quantum verification model. For a promise problem (Y,N), QMA(2) allows two polynomial-size quantum witnesses, required to be a product across the two prover registers: each YES input has some such witness accepted with probability at least 2/3, and each NO input is accepted with probability at most 1/3 for every such witness. QMA allows one unrestricted polynomial-size quantum witness with the same bounds. There is no requirement outside the promise. The verifier uses a fixed standard universal quantum computation model, with no oracle or advice. The two complete outcomes are equality and strict containment QMA properly contained in QMA(2).

    Why it matters

    Equality would show that the two-witness product-state promise adds no verification power. Strict containment would demonstrate that enforcing separation between the witnesses enables verification of additional promise problems.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Quantum verification classes · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 232–311.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  239. Rank239Classical Worst-Case Hardness Reduction for LWECan a classical algorithm establish the same worst-case lattice-hardness guarantee for Learning With Errors as Regev's quantum reduction, covering both the approximate shortest-vector and shortest-independent-vectors problems with the same parameters, polynomial modulus and no loss in lattice dimension?Cryptography, coding, information & optimization

    Exact statement

    Determine whether there is a classical (non-quantum) worst-case hardness reduction for the Learning With Errors problem that fully subsumes Regev's quantum reduction, i.e. reducing both GapSVP_gamma and SIVP_gamma to LWE for the same parameters, with polynomial modulus and no loss in lattice dimension.

    Why it matters

    A positive answer would connect the same worst-case lattice problems to Learning With Errors using classical computation throughout. It would preserve the specified parameter guarantees while removing the quantum step from that reduction.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Classical Worst-Case Hardness Reduction for LWE · 1 problem in this edition. Family membership alone does not assert an implication.

    • #239 · Classical Worst-Case Hardness Reduction for LWE (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 233–314.

    Source-fit opponent count: 51.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  240. Rank240Exact Unassisted Quantum Capacity of the Qubit Depolarizing ChannelA memoryless qubit depolarizing channel replaces its input by the maximally mixed state with probability p on each use. For every p from zero to one, how many qubits per use can it reliably transmit in the long run, using arbitrary block codes but no shared entanglement or auxiliary classical communication?Quantum information & computation

    Exact statement

    For every p in [0,1], determine the exact unassisted quantum capacity Q(D_p) of the memoryless qubit channel D_p(rho)=(1-p)rho+pI/2. Capacity is the supremum of asymptotically achievable rates in qubits per channel use for quantum-information transmission with vanishing error and arbitrary block coding, without preshared entanglement or auxiliary classical communication.

    Why it matters

    A resolution would identify the optimal qubits-per-use rate throughout the channel's noise range. It would specify exactly which transmission rates can be approached with arbitrarily small error using unassisted block codes.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Exact Unassisted Quantum Capacity of the Qubit Depolarizing Channel · 1 problem in this edition. Family membership alone does not assert an implication.

    • #240 · Exact Unassisted Quantum Capacity of the Qubit Depolarizing Channel (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 210–340.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  241. Rank241Finite Spectrum ProblemWhenever a first-order sentence has finite models of certain positive sizes, is the set of all remaining positive sizes also exactly the finite-model sizes of some first-order sentence?Logic, foundations & set theory

    Exact statement

    Determine whether the complement in the positive integers of every spectrum of a first-order sentence over finite structures is itself the spectrum of some first-order sentence.

    Why it matters

    A proof would make finite spectra closed under complement. A counterexample would distinguish describing the possible sizes of finite models from describing precisely the sizes excluded by another sentence.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Finite Spectrum Problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #241 · Finite Spectrum Problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 245–311.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  242. Rank242Zarankiewicz Problem for K_{t,t}-Free Bipartite GraphsIn a graph with n vertices split into two groups and edges only between groups, forbid t vertices on each side with every cross-edge present. For each fixed t≥2, what is the order of growth of the largest possible edge count as n increases?Combinatorics & discrete geometry

    Exact statement

    For every fixed integer t>=2, determine the asymptotic order as n tends to infinity of the maximum number of edges in an n-vertex finite simple bipartite graph containing no copy of K_{t,t}. Here n is the total number of vertices.

    Why it matters

    Matching asymptotic bounds would determine the density allowed by forbidding Kₜ,ₜ. The target asks for the order of growth as n increases, separately for each fixed t.

    Status

    Open · subcases solved

    Open with the order known for t=2 and t=3. The cited survey identifies tightness of the general Kovari--Sos--Turan upper bound for t>=4 as a central open problem.

    Status review date not recorded in this edition

    Sources

    Problem family

    Zarankiewicz Problem for K_{t,t}-Free Bipartite Graphs · 1 problem in this edition. Family membership alone does not assert an implication.

    • #242 · Zarankiewicz Problem for K_{t,t}-Free Bipartite Graphs (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 236–323.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  243. Rank243Bodirsky-Pinsker Tractability Conjecture for Infinite-Domain Constraint SatisfactionSome constraint problems allow infinitely many values but have a finite structural description. For finite-signature reducts of countable finitely bounded homogeneous structures, does excluding a primitive-positive encoding of graph three-coloring make every finite instance solvable in polynomial time?Theoretical computer science

    Exact statement

    For every finite-signature first-order reduct A of a countable finitely bounded homogeneous relational structure, determine whether absence of a primitive-positive construction of the three-vertex complete graph K3 implies that CSP(A), the finite-input homomorphism problem to the fixed template A, has a polynomial-time algorithm.

    Why it matters

    A proof would turn this structural exclusion into an efficient decision guarantee for finite inputs. A counterexample would show that the exclusion does not suffice to ensure tractability in this template class.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Bodirsky-Pinsker Tractability Conjecture for Infinite-Domain Constraint Satisfaction · 1 problem in this edition. Family membership alone does not assert an implication.

    • #243 · Bodirsky-Pinsker Tractability Conjecture for Infinite-Domain Constraint Satisfaction (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 232–342.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  244. Rank244Alperin–McKay ConjectureIn a finite group's representation theory, does each p-block have exactly as many height-zero irreducible characters as its Brauer-correspondent block in the normalizer of a defect group? This predicts that a count of representations of the whole group is determined by this local subgroup.Algebra, representation & category theory

    Exact statement

    For every finite group G, every prime p, every p-block B of G with defect group D, and its Brauer correspondent b in N_G(D), the numbers k_0(B) and k_0(b) of ordinary irreducible complex characters of height zero are equal. For a character chi in a p-block of a finite group H with defect group D, height zero means chi(1)_p=|H:D|_p, where a_p is the largest power of p dividing the positive integer a.

    Why it matters

    A proof would recover each global block's height-zero character count from its specified local normalizer block. It would establish equality of counts without requiring a canonical correspondence between individual characters.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Alperin–McKay Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #244 · Alperin–McKay Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 241–329.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  245. Rank245Planar homogeneous Mumford–Shah local-structure conjectureThe planar homogeneous Mumford–Shah model balances changes in a field against the length of its crack set. For a reduced absolute local energy minimizer, must every interior crack point look, after a C¹ coordinate change, like a line, a single tip, or three branches meeting at 120°?Analysis & PDEResolution claim under review · historical rank

    Exact statement

    Let Omega be any open subset of R^2. Consider every reduced absolute local minimizer (u,K) of E0(u,K;U)=integral_{U minus K}|grad u|^2 dx+H^1(K intersect U): K is relatively closed and countably 1-rectifiable, u is real and in W^{1,2}_loc(Omega minus K), E0 is finite on each relatively compact open U, and H^1(K intersect V)=0 implies K intersect V is empty for every open V contained in Omega. Absolute local minimality means E0(u,K;U)<=E0(v,J;U) for every relatively compact open U and every admissible locally finite-energy pair (v,J) for which (J symmetric-difference K) union the support of u-v is compactly contained in U; equality of functions off that compact set is understood almost everywhere on their common domain. There is no component-count or separation restriction on competitors. Must every x in K have a disk B_r(x) compactly contained in Omega and a C1 diffeomorphism Phi:B_r(x)->B_r(x), with Phi(x)=x and D Phi(x)=Id, such that Phi(K intersect B_r(x)) is one of: a diameter through x, one radius ending at x, or three radii ending at x with mutual120-degree angles?

    Why it matters

    A proof would classify the local structure at every interior point of the jump set into the three specified models. A counterexample would reveal another local geometry under the same unrestricted competitor rule.

    Status

    Resolution claim under review

    Earlier note (2026-10-06): Affirmative proof claim pending refereeing. Francesco Deangelis’s paper, first posted September 22, 2026 and revised October 5, claims the exact planar Mumford–Shah target, including the homogeneous case and unrestricted competitor topology. The author credits ChatGPT Astra with generating the initial proof strategy and manuscript draft, followed by author revision and responsibility. This is an AI-assisted proof claim; no full proof audit or public formalization was established in the October 6 review. October 7, 2026: Separately, OpenAI’s 6 October release reports C¹,α arcs, crack tips and 120-degree triple junctions for reduced absolute minimizers of its bounded-data fidelity functional on bounded Lipschitz planar domains. This is partial progress and does not establish the unrestricted homogeneous target. The new proof has not been independently audited, refereeing has not been independently confirmed, and no local Lean verification is claimed.

    Status reviewed

    Sources

    Problem family

    Planar homogeneous Mumford–Shah local-structure conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #245 · Planar homogeneous Mumford–Shah local-structure conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 256–318.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  246. Rank246Complex linearity of closed-surface mapping class groups by genusFor exactly which genera g≥3 can the orientation-preserving symmetries of a closed orientable surface, identified up to continuous deformation, be represented faithfully by finite-dimensional complex matrices? The matrix dimension may depend on g, the number of handles.Geometry & topology

    Exact statement

    For every integer g >= 3, let S_g be the closed connected orientable surface of genus g and Mod(S_g) its group of isotopy classes of orientation-preserving homeomorphisms. Determine exactly the set of such genera g for which there exist an integer n >= 1 and an injective homomorphism Mod(S_g) -> GL_n(C), with n allowed to depend on g. A complete resolution proves the answer for every g >= 3; one nonlinear genus, a bounded-dimensional obstruction, or an isolated family of representations without the remaining genus answers is partial progress.

    Why it matters

    A complete answer would identify every genus whose mapping classes admit a faithful finite matrix description. It would distinguish those cases from genera where every such representation loses group information.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Complex linearity of closed-surface mapping class groups by genus · 1 problem in this edition. Family membership alone does not assert an implication.

    • #246 · Complex linearity of closed-surface mapping class groups by genus (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 244–328.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  247. Rank247Michel Boundary Rigidity Conjecture for Simple ManifoldsCan distances through a manifold between boundary points determine its interior geometry, up to a smooth coordinate change fixing the boundary? The question concerns compact manifolds of dimension at least three with nonempty strictly convex boundary, unique geodesics between points and no conjugate points.Geometry & topology

    Exact statement

    Let M be a compact smooth manifold with nonempty boundary of dimension n>=3, and g,g' smooth simple Riemannian metrics: the boundary is strictly convex, every two points are joined by a unique geodesic, and there are no conjugate points. Assert that whenever d_g(x,y)=d_g'(x,y) for all boundary points x,y, there exists a diffeomorphism phi:M->M equal to the identity pointwise on the boundary with g'=phi^*g.

    Why it matters

    A proof would establish that these distance measurements determine the interior geometry within the stated simple-metric class. A counterexample would give indistinguishable boundary distances for interiors not related by an allowed change of coordinates.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Michel Boundary Rigidity Conjecture for Simple Manifolds · 1 problem in this edition. Family membership alone does not assert an implication.

    • #247 · Michel Boundary Rigidity Conjecture for Simple Manifolds (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 246–332.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  248. Rank248Polynomial-time explicit logarithmic Ramsey graphsCan one deterministic algorithm construct an N-vertex graph in time polynomial in N with no set larger than a fixed constant times log N whose vertices are either all pairwise joined or all pairwise unjoined?Theoretical computer science

    Exact statement

    Does there exist one deterministic algorithm A and constants C,c>0 such that, for every integer N>=2 given in unary, A outputs the adjacency matrix of a simple N-vertex graph in time at most O(N^c), with both its clique number and independence number at most C log_2 N? This is ordinary output-size polynomial explicitness, not poly(log N)-time adjacency queries.

    Why it matters

    A positive answer would make graphs with this logarithmic avoidance guarantee uniformly constructible in time polynomial in their size. It would produce the entire adjacency matrix, without requiring faster individual adjacency queries.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Polynomial-time explicit logarithmic Ramsey graphs · 1 problem in this edition. Family membership alone does not assert an implication.

    • #248 · Polynomial-time explicit logarithmic Ramsey graphs (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 249–343.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  249. Rank249Vassiliev Separation by Abelian-Valued Finite-Type InvariantsCan finite-type knot invariants distinguish every pair of different oriented knot types in the three-sphere? These invariants have finite order under the crossing-change relation, and may take values in any abelian group, with both the order and group chosen for the pair.Geometry & topology

    Exact statement

    For every pair of non-isotopic smooth closed oriented unframed knots K_1,K_2 in S^3, assert that there exist an abelian group A, a nonnegative integer d and an A-valued Vassiliev invariant v of order at most d with v(K_1) != v(K_2). Knots are taken up to ambient isotopy respecting their orientation, so a knot and its reverse are included when they are distinct oriented knot types. Extend v linearly from the free abelian group on oriented knot types to singular knots by the usual alternating positive/negative crossing-resolution relation; order at most d means vanishing on every singular knot with d+1 transverse double points. The coefficient group and finite order may depend on the pair. Rational-valued, integer-valued and Kontsevich-injectivity variants are related but are not asserted equivalent here.

    Why it matters

    A proof would make the full permitted finite-type invariant system detect oriented knot type. A counterexample would exhibit distinct oriented knots indistinguishable by every finite order and every allowed coefficient group.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Vassiliev Separation by Abelian-Valued Finite-Type Invariants · 1 problem in this edition. Family membership alone does not assert an implication.

    • #249 · Vassiliev Separation by Abelian-Valued Finite-Type Invariants (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 255–344.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  250. Rank250Grigorchuk Gap ConjectureCount the elements of a finitely generated group reachable in at most n generator steps. If this growth is strictly below exp(√n), when growth rates are compared up to constant rescaling of n and value, must it actually be bounded by a polynomial?Algebra, representation & category theory

    Exact statement

    For every finitely generated group G with finite generating set X, let gamma_G,X(n) count elements of word length at most n in X union X^(-1). For positive functions on positive integers, write f preceq g if some integer C>=1 satisfies f(n)<=C g(Cn) for every n>=1, and f prec g if f preceq g but not g preceq f. Prove that gamma_G,X prec exp(sqrt(n)) implies that G has polynomial growth, equivalently that G is virtually nilpotent. The growth comparison is independent of the finite generating set. It is strict comparison of growth classes, not merely an eventual numerical upper bound or non-strict domination.

    Why it matters

    A proof would exclude intermediate growth throughout this strict region and force such groups to be virtually nilpotent. A counterexample would place nonpolynomial growth below the proposed boundary.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Grigorchuk Gap Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #250 · Grigorchuk Gap Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 256–335.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  251. Rank251Deterministic polynomial-time exact bipartite matchingIn a graph with two groups of vertices and red or blue edges only between the groups, can a polynomial-time algorithm without randomness decide whether every vertex can be paired with exactly one other vertex using exactly k red edges?Theoretical computer science

    Exact statement

    Does there exist a deterministic algorithm which, for every explicitly encoded finite simple bipartite graph G=(U∪V,E), a red/blue colouring of E and an integer k encoded in binary, decides in time polynomial in the total input bit length whether G has a perfect matching with exactly k red edges? Unequal side sizes or k outside 0,...,|U| are immediate No instances.

    Why it matters

    A positive answer would give an exact-count matching decision procedure with time bounded polynomially in the encoded input length. It would remove randomness for this particular matching task.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Deterministic polynomial-time exact bipartite matching · 1 problem in this edition. Family membership alone does not assert an implication.

    • #251 · Deterministic polynomial-time exact bipartite matching (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 253–335.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  252. Rank252Margulis–Zimmer Commensurated-Subgroup Conjecture over Number FieldsIn an S-arithmetic group over a number field with total local rank at least two, must every commensurated subgroup be finite or commensurable with the arithmetic group obtained from a subset of the allowed places, possibly the full set? The question includes cocompact and mixed-place cases.Group theory

    Exact statement

    For every number field K, connected absolutely almost simple simply connected algebraic K-group G, finite set S of places containing all archimedean places, and subgroup Gamma of G(K) commensurable with G(O_S), suppose the sum over v in S of rank over K_v of G is at least two. For every subgroup N of Gamma such that N intersect gamma N gamma^(-1) has finite index in both N and gamma N gamma^(-1) for all gamma in Gamma, assert that N is finite or is commensurable with G(O_Sprime) for some set Sprime with all archimedean places contained in Sprime contained in S. Here O_S is the ring of S-integers and commensurability is inside G(K). Prove this universal assertion or give a counterexample. The target includes anisotropic/cocompact arithmetic lattices and mixed archimedean/nonarchimedean higher S-rank; it is not restricted to the non-uniform real-lattice cases.

    Also known as

    Margulis-Zimmer conjecture; Commensurated subgroup problem for higher-rank S-arithmetic groups

    Why it matters

    This would classify subgroups that differ from each conjugate by only finite index. It asks for that classification across the full stated higher-rank arithmetic setting, including cases beyond non-uniform real lattices.

    Status

    Open · subcases solved

    October 6, 2026: The Stuck-Zimmer paper states results about real Lie-group probability-preserving actions. It does not state the ranked commensurated-subgroup classification for all number-field S-arithmetic groups. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Problem family

    Margulis–Zimmer Commensurated-Subgroup Conjecture over Number Fields · 1 problem in this edition. Family membership alone does not assert an implication.

    • #252 · Margulis–Zimmer Commensurated-Subgroup Conjecture over Number Fields (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 256–339.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  253. Rank253Existence of NPT Bound EntanglementCan an entangled two-party quantum state have a negative eigenvalue after partially transposing one subsystem, yet yield no maximally entangled pairs at a positive asymptotic rate using local operations and classical communication?Quantum information & computation

    Exact statement

    Prove or disprove that there exists a bipartite entangled state with negative partial transpose that is nevertheless undistillable under LOCC.

    Why it matters

    Existence would show that negative partial transpose does not guarantee distillable entanglement. An impossibility proof would make this matrix criterion sufficient for distillation in every bipartite entangled state to which it applies.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Existence of NPT Bound Entanglement · 1 problem in this edition. Family membership alone does not assert an implication.

    • #253 · Existence of NPT Bound Entanglement (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 250–340.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  254. Rank254Zagier Conjecture on Dedekind Zeta ValuesPolylogarithms are higher relatives of the logarithm. Can every number field's Dedekind zeta value at an integer at least two be expressed using a determinant of single-valued polylogarithms, multiplied by a nonzero rational number and the predicted factors involving π and the field's discriminant?Number theory & arithmetic geometry

    Exact statement

    For every number field F of signature (r1,r2) and every integer k>=2, put n=r2 for even k and n=r1+r2 for odd k. There exist xi_1,...,xi_n in A_k(F) tensor Q, using the allowable-symbol group defined by Zagier--Gangl section2, such that zeta_F(k) belongs to Q^times times sqrt(abs(D_F)) pi^{k([F:Q]-n)} det(L_k(sigma_i(xi_j)))_{1<=i,j<=n}. Here L_k is the single-valued polylogarithm defined there by the Bernoulli-number combination of classical polylogarithms (imaginary part for even k, real part for odd k), extended Q-linearly to symbols. For even k take one embedding from each conjugate complex pair; for odd k also take all real embeddings. The empty determinant is1. This is existence of the determinant formula, not the stronger regulator-lattice or Bloch-group rank assertion.

    Why it matters

    A proof would express these zeta values through a finite determinant of polylogarithmic data over the same field. It would establish the stated formula without requiring the stronger regulator-lattice assertions.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Zagier Conjecture on Dedekind Zeta Values · 1 problem in this edition. Family membership alone does not assert an implication.

    • #254 · Zagier Conjecture on Dedekind Zeta Values (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 258–341.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  255. Rank255Positivity Problem for Linear Recurrence SequencesA sequence starts with finitely many rational values, then repeatedly forms a fixed rational linear combination of its preceding terms. Is there an algorithm that always finishes and decides whether every term is nonnegative?Logic, foundations & set theory

    Exact statement

    Select arbitrary finite order k>=1, rational constant coefficients a_1,...,a_k and rational initial values u_0,...,u_(k-1), each finitely encoded in binary, with u_(n+k)=sum_(i=1)^k a_i u_(n+k-i) for all n>=0. The assertion is that a total algorithm exists which, for EVERY such input, halts and correctly decides whether u_n>=0 for EVERY n>=0. There is no running-time bound. All coefficient values, including a_k=0, are permitted.

    Why it matters

    A positive answer would make an infinite sequence’s sign condition decidable from finite input, without imposing a speed requirement. A negative answer would establish undecidability for this exact recurrence model.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Positivity Problem for Linear Recurrence Sequences · 1 problem in this edition. Family membership alone does not assert an implication.

    • #255 · Positivity Problem for Linear Recurrence Sequences (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 264–330.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  256. Rank256Moonshine Uniqueness Conjecture (Strongly Rational Holomorphic VOA Form)Is the moonshine algebra uniquely determined by having central charge 24 and no weight-one states, within the class of strongly rational holomorphic vertex operator algebras? These are graded algebras with tightly controlled representation theory.Algebra, representation & category theory

    Exact statement

    Every strongly rational, holomorphic vertex operator algebra V of central charge 24 with V_1=0 is isomorphic to the moonshine vertex operator algebra V-natural. Here strongly rational means simple, of CFT type (V=sum_{n>=0} V_n and V_0=C1), self-contragredient, C_2-cofinite, and rational in the sense that every admissible V-module is completely reducible; holomorphic means that V is the only irreducible V-module up to isomorphism in this rational setting.

    Why it matters

    A proof would characterize the moonshine algebra uniquely using the stated structural and graded conditions. A counterexample would establish a different algebra sharing all those conditions.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Moonshine Uniqueness Conjecture (Strongly Rational Holomorphic VOA Form) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #256 · Moonshine Uniqueness Conjecture (Strongly Rational Holomorphic VOA Form) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 246–375.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  257. Rank257The Singer Conjecture on Vanishing of L2-Cohomology of Aspherical ManifoldsFor a closed manifold whose universal cover is contractible, must its L²-Betti numbers, which measure square-integrable topology on that cover, vanish outside the middle dimension? In odd dimensions, this predicts that they all vanish.Algebra, representation & category theory

    Exact statement

    Prove that for every closed aspherical n-manifold, all L2-Betti numbers vanish outside degree n/2; in particular, prove that all L2-Betti numbers vanish when n is odd.

    Why it matters

    A proof would constrain these analytic topological invariants to a single possible degree. A counterexample would show that asphericity alone allows L²-cohomological information away from the middle.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The full Singer target concerns every dimension and requires total L2-Betti vanishing in odd dimensions. The manuscript proves vanishing outside middle degree only in dimension four, even for the broader class of finite aspherical integral Poincare complexes and without orientability. It resolves the n=4 manifold subcase, but supplies no theorem for higher dimensions or the general odd-dimensional assertion, so the all-dimensional target remains open by these results. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    The Singer Conjecture on Vanishing of L2-Cohomology of Aspherical Manifolds · 1 problem in this edition. Family membership alone does not assert an implication.

    • #257 · The Singer Conjecture on Vanishing of L2-Cohomology of Aspherical Manifolds (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 255–360.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  258. Rank258Irrationality of Euler's ConstantIs Euler's constant, the limiting difference between the harmonic sum 1+1/2+⋯+1/n and the natural logarithm of n as n grows, an irrational number?Number theory & arithmetic geometry

    Exact statement

    Prove or disprove that Euler's constant gamma = lim_{n->infty} (1 + 1/2 + ... + 1/n - log n) = 0.5772156649... is irrational.

    Why it matters

    A resolution would decide whether this limit can equal a ratio of integers. It would settle that exact arithmetic distinction, independently of how accurately its decimal expansion is computed.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Irrationality of Euler's Constant · 1 problem in this edition. Family membership alone does not assert an implication.

    • #258 · Irrationality of Euler's Constant (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 238–387.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  259. Rank259Frey-Mazur ConjectureFor a prime p>17, suppose two elliptic curves over the rationals have the same Galois action on their points killed by multiplication by p. Must they be linked by an isogeny over the rationals, a nonconstant algebraic map preserving their group laws?Number theory & arithmetic geometry

    Exact statement

    For every prime p>17 and elliptic curves E,E′ over Q, prove that an isomorphism E[p]≅E′[p] as G_Q-modules implies that E and E′ are Q-isogenous.

    Why it matters

    A proof would make this finite torsion representation sufficient to determine the rational isogeny class. A counterexample would give distinct isogeny classes with indistinguishable p-torsion Galois modules at an allowed prime.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Frey-Mazur Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #259 · Frey-Mazur Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 244–381.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  260. Rank260Global Regularity of the 2D Inviscid Boussinesq EquationsIn a two-dimensional buoyancy-driven fluid without viscosity or thermal diffusion, can smooth finite-energy flow on the whole plane develop a singularity in finite time, or must it stay smooth forever? The governing equations are the inviscid Boussinesq equations.Analysis & PDE

    Exact statement

    Determine whether all smooth, finite-energy solutions of the two-dimensional inviscid Boussinesq equations (velocity coupled to a temperature or density by buoyancy, without dissipation) on R^2 remain smooth for all time, or whether finite-time singularity formation occurs for some C^infinity initial data; finite-time blow-up is known for some data on domains with corners and near half-space type domains.

    Why it matters

    A resolution would determine whether buoyancy coupled to an undamped fluid can cause finite-time loss of smoothness in this setting. It would distinguish global regularity from singularity formation for the stated planar model.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Global Regularity of the 2D Inviscid Boussinesq Equations · 1 problem in this edition. Family membership alone does not assert an implication.

    • #260 · Global Regularity of the 2D Inviscid Boussinesq Equations (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 244–438.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  261. Rank261Original Dubrovin conjecture for Fano manifoldsOn a smooth complex projective Fano manifold, can curve counting and the organization of algebraic sheaves reveal the same structure? Dubrovin predicts that generic semisimplicity of quantum cohomology occurs exactly when the derived category of sheaves has a full exceptional collection, with compatible numerical data where the quantum product converges and is semisimple.Geometry & topology

    Exact statement

    For every smooth projective complex Fano manifold F, prove the original Dubrovin conjecture in the form recalled in Galkin-Golyshev-Iritani Section 4.6: (1) the (big) quantum cohomology of F is generically semisimple if and only if the bounded derived category D^b_coh(F) admits a full exceptional collection; and, if the quantum cohomology is semisimple (quantum product convergent and semisimple at some point tau, with the monodromy data taken at tau and an admissible phase in that paper's conventions of Sections 2 and 4.5), there exists a full exceptional collection E_1,...,E_N such that (2) the Stokes matrix satisfies S_ij = chi(E_i,E_j) for all i,j, and (3) the central connection matrix factors as C = C'C'' where the columns of C'' are the components of the modified Chern characters Ch(E_j) and C' is an operator on cohomology commuting with cup product by c_1(F). The explicit Gamma-class formula for C' (Gamma Conjecture II) and the Hodge-Tate refined versions are not part of this target.

    Why it matters

    A proof would connect quantum cohomology to categorical generation, with Stokes entries given by Euler pairings and the central connection matrix factored through modified Chern characters. The other factor must commute with cup product by c₁; the stronger Gamma-class formula is separate.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Original Dubrovin conjecture for Fano manifolds · 1 problem in this edition. Family membership alone does not assert an implication.

    • #261 · Original Dubrovin conjecture for Fano manifolds (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 270–346.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  262. Rank262The Atiyah Conjecture on the Integrality of L2-Betti NumbersThe L²-Betti numbers are generalized counts of independent holes in a universal covering space. Must they be integers for every finite cell complex whose fundamental group has no nonidentity element of finite order?Algebra, representation & category theory

    Exact statement

    Prove that for every finite CW-complex X whose group G = π_1(X) is torsion-free, the L2-Betti numbers of the universal covering of X, i.e. the von Neumann dimensions of its L2-homology, are integers.

    Why it matters

    A proof would show that these von Neumann dimensions retain whole-number values under the torsion-free hypothesis. A counterexample would reveal nonintegral homological dimensions even when the group has no finite-order elements.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. For a finite connected aspherical integral Poincare complex of formal dimension four, the asserted vanishing leaves only b_2^(2). The L2 Euler characteristic identity then makes this remaining value equal to the ordinary integral Euler characteristic. Thus the result implies the integrality assertion for this special class (including closed aspherical four-manifolds). The Atiyah target quantifies over all finite CW-complexes with torsion-free fundamental group, without asphericity or dimension restrictions, so it is not resolved. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    The Atiyah Conjecture on the Integrality of L2-Betti Numbers · 1 problem in this edition. Family membership alone does not assert an implication.

    • #262 · The Atiyah Conjecture on the Integrality of L2-Betti Numbers (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 239–405.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  263. Rank263Erdos--Szekeres Convex Polygon ConjectureFor every n≥3, is 2^(n−2)+1 the smallest number of points in the plane that guarantees n vertices of a convex polygon when no three points lie on a line?Combinatorics & discrete geometry

    Exact statement

    Prove or disprove that ES(n)=2^(n-2)+1 for every integer n>=3, where ES(n) is the least N such that every set of N points in the Euclidean plane in general position (no three collinear) contains n points in convex position.

    Why it matters

    A proof would establish the proposed exact threshold for every polygon size. A counterexample would refute that formula for at least one size, without determining the whole threshold function.

    Status

    Open · subcases solved

    Open with solved small cases. The SoCG 2025 paper explicitly keeps the planar conjecture open; its counterexample concerns an ordered-hypergraph generalization.

    Status review date not recorded in this edition

    Sources

    Problem family

    Erdos--Szekeres Convex Polygon Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #263 · Erdos--Szekeres Convex Polygon Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 274–355.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  264. Rank264Erdős Girth ConjectureFor each fixed k≥2, can graphs on every sufficiently large number n of vertices have no cycles of length 2k or less, yet at least cₖn^(1+1/k) edges for some positive constant cₖ depending only on k?Combinatorics & discrete geometry

    Exact statement

    For every integer k>=2, there exist constants c_k>0 and N_k such that for every integer n>=N_k there is a finite simple undirected graph on n vertices with girth greater than 2k and at least c_k n^(1+1/k) edges. Forests have infinite girth. Equivalently ex(n,{C_3,...,C_(2k)})=Omega_k(n^(1+1/k)) as n tends to infinity. Constants and the threshold may depend on k; k is fixed before n grows.

    Why it matters

    A proof would guarantee dense graphs despite excluding all these short cycles. The positive constant and starting size may depend on k, but the guarantee must hold at every sufficiently large n.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Erdős Girth Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #264 · Erdős Girth Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 264–370.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  265. Rank265Montgomery's Large Value Conjecture for Dirichlet PolynomialsA Dirichlet polynomial is a short sum of oscillating terms n^(it). How often can one with N terms and coefficient magnitudes at most one exceed N^σ, for σ>1/2, at times at least one apart? Does Montgomery's bound, essentially N^(2−2σ) times a subpower factor in the time range, always hold for time ranges between N and a fixed power of N?Number theory & arithmetic geometryZeta zeros and large Dirichlet polynomial values · 2 problems

    Exact statement

    For every fixed sigma > 1/2 and N <= T <= N^{O(1)}, prove that if D(t) = sum_{N<n<=2N} b_n exp(it log n) has |b_n| <= 1 and W is a 1-separated subset of [0,T] on which |D(t)| > N^sigma, then |W| <= C(sigma) T^{o(1)} N^{2-2sigma}.

    Why it matters

    A proof would restrict how frequently these oscillating sums can become large. It would control exceptional peaks uniformly over all permitted coefficients and the specified polynomial range of observation times.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Zeta zeros and large Dirichlet polynomial values · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 242–413.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  266. Rank266Resolvent Degree of the General Degree-Seven Polynomial (Algebraic Form of Hilbert's Thirteenth Problem)How many variables are inherently needed in algebraic functions that solve the general degree-seven polynomial over the complex numbers, even when those functions may be composed in successive steps? Is the minimum exactly three, so two-variable algebraic functions cannot suffice?Algebra, representation & category theory

    Exact statement

    Prove that the resolvent degree rd_C(7) of the general degree-seven polynomial equals 3; equivalently, establish the lower bound rd_C(7) > 2.

    Why it matters

    A proof would establish the proposed lower bound on the number of variables needed in algebraic solution procedures for general septics. A negative answer would permit a reduction to at most two variables.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Resolvent Degree of the General Degree-Seven Polynomial (Algebraic Form of Hilbert's Thirteenth Problem) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #266 · Resolvent Degree of the General Degree-Seven Polynomial (Algebraic Form of Hilbert's Thirteenth Problem) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 256–389.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  267. Rank267Sample Compression Conjecture (Linear Size in VC Dimension)A concept class has VC dimension d, a measure of its ability to fit arbitrary labels. Can every labeled sample consistent with that class be compressed to only O(d) examples plus bounded side information, enough to reconstruct a rule agreeing with every label in the original sample?Theoretical computer science

    Exact statement

    Prove or disprove that every concept class of finite VC dimension d admits a sample compression scheme of size O(d): any realizable labeled sample can be summarized by O(d) examples plus bounded side information from which a hypothesis consistent with the full sample is reconstructed.

    Why it matters

    A proof would turn VC dimension into a linear bound on retained training examples sufficient for exact sample consistency. A counterexample would separate this capacity measure from the proposed compression guarantee.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Sample Compression Conjecture (Linear Size in VC Dimension) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #267 · Sample Compression Conjecture (Linear Size in VC Dimension) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 198–441.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  268. Rank268Tutte 5-Flow ConjectureA graph has no bridges if deleting any single edge never splits a connected component. Can every finite graph with this property carry nonzero integer flows of magnitude at most four on its oriented edges, with total inflow equal to total outflow at every vertex?Combinatorics & discrete geometry

    Exact statement

    Prove that every finite bridgeless graph admits an orientation and a flow f on its edges with integer values in {-4,-3,-2,-1,1,2,3,4}, such that the sum of outgoing values equals the sum of incoming values at every vertex.

    Wording clarified after assessment

    Earlier comparisons used the wording below. A recorded review treats the published wording above as a clarification of the same intended problem. The wording review itself adds no comparison judgments.

    Earlier assessed wording

    Prove that every bridgeless graph admits an orientation and a nonzero edge flow with values of magnitude at most 4 that satisfies flow conservation at every vertex.

    Wording review recorded .

    Why it matters

    A proof would make absence of bridges sufficient for a conserved integer flow using only the permitted eight nonzero values. A counterexample would show that some finite bridgeless graph cannot satisfy that bound.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Tutte 5-Flow Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #268 · Tutte 5-Flow Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 268–382.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  269. Rank269Tits Alternative for CAT(0) GroupsIf a discrete group acts properly and cocompactly by isometries on a proper CAT(0) geodesic space, must each finitely generated subgroup either have a solvable subgroup of finite index or contain a free group on two generators? Here properness of the space means that closed bounded sets are compact.Algebra, representation & category theory

    Exact statement

    Prove or disprove that every CAT(0) group satisfies the Tits alternative: if a discrete group G acts properly and cocompactly by isometries on a proper CAT(0) geodesic metric space, then every finitely generated subgroup H of G is either virtually solvable or contains a subgroup isomorphic to the free group F_2. Proper for the space means that closed bounded sets are compact; no geodesic-completeness, cube-complex or fixed-dimension hypothesis is imposed. Success is a proof for every such G and H, or a counterexample G,H violating both alternatives.

    Why it matters

    This asks whether nonpositive curvature forces a sharp algebraic dichotomy inside the group. A proof would extend the solvable-or-free description to every finitely generated subgroup of every group with such a geometric action.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Tits alternative for CAT(0) groups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #269 · Tits Alternative for CAT(0) Groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 278–362.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  270. Rank270C^r Closing LemmaA nonwandering point has neighborhoods that eventually return to meet themselves under iteration. Can such a point always be made exactly periodic by an arbitrarily small change to a diffeomorphism of a compact smooth manifold, keeping its derivatives through any chosen order r equally close?Probability, ergodic theory & dynamics

    Exact statement

    For every differentiability order r, compact smooth manifold M, C^r diffeomorphism f of M, nonwandering point p, and C^r neighborhood U of f, prove that some g in U has p as a periodic point.

    Why it matters

    A proof would turn recurrence near a point into exact periodicity at that same point under arbitrarily small permitted changes. A counterexample would reveal an obstruction at a specific differentiability level.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    C^r Closing Lemma · 1 problem in this edition. Family membership alone does not assert an implication.

    • #270 · C^r Closing Lemma (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 270–400.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  271. Rank271Marked Length Spectrum Rigidity ConjectureIf two negatively curved geometries on the same compact manifold without boundary give every closed-loop deformation class the same shortest length, must they agree after a smooth change of coordinates that can itself be deformed to the identity?Geometry & topology

    Exact statement

    Let g and h be negatively curved Riemannian metrics on the same closed manifold. Prove that equality of their marked length spectra implies that a diffeomorphism isotopic to the identity pulls h back to g.

    Why it matters

    A proof would reconstruct the metric, up to the allowed marking-preserving change of coordinates, from its closed-geodesic lengths. A counterexample would show that all these lengths can agree while the prescribed metric equivalence fails.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Marked Length Spectrum Rigidity Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #271 · Marked Length Spectrum Rigidity Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 285–358.

    Source-fit opponent count: 38.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  272. Rank272Exact Real Grothendieck ConstantHow much larger can a weighted sum of inner products become when we choose real unit vectors instead of signs +1 and −1? What is the smallest factor that bounds this advantage for every real matrix, regardless of its size or the vector dimension?Analysis & PDE

    Exact statement

    Determine the smallest real constant K_G such that, for every finite real m-by-n matrix (a_ij) and all unit vectors u_1,...,u_m,v_1,...,v_n in a real Hilbert space, abs(sum_ij a_ij <u_i,v_j>) <= K_G max_{epsilon_i,delta_j in {-1,1}} abs(sum_ij a_ij epsilon_i delta_j). The constant is uniform over both matrix sizes and Hilbert-space dimension.

    Why it matters

    The answer would give the exact worst-case loss in replacing vector choices by plus-or-minus signs. The same factor must work for every finite real matrix and every Hilbert-space dimension.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Exact Real Grothendieck Constant · 1 problem in this edition. Family membership alone does not assert an implication.

    • #272 · Exact Real Grothendieck Constant (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 279–380.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  273. Rank273Growth Rate of 3-Uniform Hypergraph Ramsey NumbersColor every triple from an N-element set red or blue. How large must N be to force t elements whose triples all have the same color? In particular, must this threshold be at least 2^(2^(ct)) for some fixed c>0 as t grows—a doubly exponential lower bound?Combinatorics & discrete geometry

    Exact statement

    Let r_3(t) be the least N such that every red-blue coloring of the triples of an N-element set has a monochromatic t-element set. Determine the asymptotic growth of r_3(t); in particular, prove or disprove that r_3(t) is at least 2^(2^(c t)) for some constant c>0 and all sufficiently large t.

    Why it matters

    A resolution would locate the size scale at which this higher-order coloring constraint forces uniformity. The proposed lower bound would require doubly exponential-sized examples to capture the worst-case obstruction.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Growth Rate of 3-Uniform Hypergraph Ramsey Numbers · 1 problem in this edition. Family membership alone does not assert an implication.

    • #273 · Growth Rate of 3-Uniform Hypergraph Ramsey Numbers (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 240–454.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  274. Rank274The Dynamic Optimality Conjecture for Splay TreesFor every sequence of searches, does a splay tree stay within a fixed constant factor of the best binary search tree strategy that knows all future requests, apart from initialization costs?Theoretical computer science

    Exact statement

    Prove that there is a universal constant C such that, for every access sequence, the cost of the splay-tree algorithm is at most C times the minimum cost of any offline dynamic binary search tree algorithm, up to the standard additive initialization term.

    Why it matters

    A proof would give splay trees a constant-factor guarantee against a tree algorithm that knows the entire sequence in advance. A counterexample family would show an unbounded gap from that benchmark.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    The Dynamic Optimality Conjecture for Splay Trees · 1 problem in this edition. Family membership alone does not assert an implication.

    • #274 · The Dynamic Optimality Conjecture for Splay Trees (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 267–379.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  275. Rank275Huisken's Generic Singularity Conjecture for Mean Curvature FlowAs a smooth closed hypersurface moves by mean curvature, must a generic choice of starting surface develop only spherical or cylindrical singularities, each with multiplicity one rather than overlapping sheets? The claim is for an open dense set of starting surfaces in every dimension.Analysis & PDE

    Exact statement

    Prove that, for a generic open dense set of smooth closed embedded initial hypersurfaces in R^(n+1), every mean-curvature-flow singularity is modeled by a multiplicity-one sphere or cylinder.

    Why it matters

    A proof would restrict singularities from this generic class of initial surfaces to the stated models. It would exclude other local shapes and higher multiplicities throughout that open dense set.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. For embedded compact surfaces in R^3 the source gives a unique multiplicity-one self-shrinker at each first-time singular point, without genericity. This supplies a restricted singularity-structure result related to Huisken. The full target demands an open dense set of initial hypersurfaces in arbitrary dimension with every singularity a sphere or cylinder. The abstract neither classifies the shrinker as spherical/cylindrical nor treats later singular times or general dimensions. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Huisken's Generic Singularity Conjecture for Mean Curvature Flow · 1 problem in this edition. Family membership alone does not assert an implication.

    • #275 · Huisken's Generic Singularity Conjecture for Mean Curvature Flow (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 268–398.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  276. Rank276Closed Geodesics Conjecture on Closed Riemannian ManifoldsDoes every closed connected Riemannian manifold of dimension at least two have infinitely many geometrically different prime closed geodesics?Geometry & topology

    Exact statement

    Does every closed connected smooth Riemannian manifold (M,g) of dimension at least two admit infinitely many geometrically distinct nonconstant prime closed geodesics?

    Also known as

    Infinitely many closed geodesics conjecture; Closed-geodesic multiplicity conjecture

    Why it matters

    Closed geodesics are periodic paths that locally follow the shortest route. The question asks for infinitely many different such paths on every manifold in this class; repeating one path does not count as a new geodesic.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The target ranges over every closed connected smooth Riemannian manifold of dimension at least two. The source covers standard spheres in every dimension at least two, manifolds with finite smooth spherical covers, and all closed three-manifolds, including degenerate metrics. Prime closed geodesics with pairwise distinct images satisfy the requested geometric distinctness. These are substantial subcases, but the abstracts do not cover arbitrary higher-dimensional manifolds outside the spherical-cover class, so they cannot close the unrestricted target. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Closed Geodesics Conjecture on Closed Riemannian Manifolds · 1 problem in this edition. Family membership alone does not assert an implication.

    • #276 · Closed Geodesics Conjecture on Closed Riemannian Manifolds (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 260–395.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  277. Rank277Congruence Subgroup Property for Mapping Class Groups of Surfaces of Genus at Least ThreeSurface symmetries up to deformation form the mapping class group. For a closed orientable surface with at least three handles, must every finite-index subgroup contain all symmetries acting trivially, up to conjugation, on some finite characteristic quotient of the surface's fundamental group?Geometry & topology

    Exact statement

    For every closed connected orientable surface S of genus g≥3 and every finite-index subgroup H of its orientation-preserving mapping class group Mod(S), there exists a finite-index characteristic subgroup N of π1(S) such that ker(Mod(S)→Out(π1(S)/N)) is contained in H.

    Why it matters

    A proof would show that these finite quotients detect enough of the mapping class group to refine every finite-index subgroup. A counterexample would exhibit a subgroup not containing any such congruence kernel.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Congruence Subgroup Property for Mapping Class Groups of Surfaces of Genus at Least Three · 1 problem in this edition. Family membership alone does not assert an implication.

    • #277 · Congruence Subgroup Property for Mapping Class Groups of Surfaces of Genus at Least Three (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 287–369.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  278. Rank278Woodin's HOD ConjectureHOD is the inner universe of hereditarily ordinal-definable sets. If an extendible cardinal exists, must arbitrarily large regular cardinals fail Woodin's omega-strong measurability condition relative to HOD—a technical test of the relationship between HOD and the surrounding universe?Logic, foundations & set theory

    Exact statement

    Prove or refute that ZFC plus the existence of an extendible cardinal implies there is a proper class of regular cardinals that are not omega-strongly measurable in HOD.

    Why it matters

    A proof would derive the stated restriction on HOD from the extendible-cardinal assumption. A refutation would show that the assumption can coexist with failure of this proper-class conclusion.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Woodin's HOD Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #278 · Woodin's HOD Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 278–389.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  279. Rank279The Online Boolean Matrix-Vector Multiplication ConjectureA Boolean matrix uses OR and AND in place of ordinary addition and multiplication. If n query vectors arrive one at a time and each product must be returned before the next arrives, is total time O(n^(3−ε)) impossible for every fixed ε>0, even with randomized algorithms allowed error at most one-third?Theoretical computer science

    Exact statement

    Prove or refute that for every epsilon > 0 there is no randomized algorithm which, given an n by n Boolean matrix and then n Boolean query vectors one at a time, outputs each Boolean-semiring product before receiving the next vector in total time O(n^(3-epsilon)) with error probability at most 1/3.

    Why it matters

    Each answer must be produced before the next vector arrives. A proof would establish the proposed online time barrier; a faster bounded-error algorithm would refute it.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    The Online Boolean Matrix-Vector Multiplication Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #279 · The Online Boolean Matrix-Vector Multiplication Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 264–406.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  280. Rank280Iwasawa mu = 0 Conjecture for Cyclotomic Z_p-ExtensionsAs a number field climbs its cyclotomic Zₚ tower, its ideal class groups can acquire more p-power torsion. For every prime p and every number field not abelian over the rationals, must the Iwasawa μ-invariant governing the fastest term in this growth be zero?Number theory & arithmetic geometryIwasawa invariants · 2 problems

    Exact statement

    For every number field F and every prime p, prove that the Iwasawa mu-invariant mu(X_infinity) of the unramified Iwasawa module for the cyclotomic Z_p-extension F_infinity/F is zero. The case in which F is abelian over Q is excluded as proved.

    Why it matters

    A proof would exclude a positive mu contribution throughout this remaining class of cyclotomic extensions. A counterexample would identify a field and prime where the unramified Iwasawa module has positive mu-invariant.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Iwasawa invariants · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 266–405.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  281. Rank281Arnold Chord Conjecture in Higher DimensionsA contact form defines a distinguished motion called its Reeb flow. In every closed co-oriented contact manifold of dimension at least five, must each closed Legendrian submanifold have a nonconstant trajectory starting and ending on it, for every compatible contact form?Geometry & topology

    Exact statement

    Prove that for every closed co-oriented contact manifold (Y, xi) of dimension 2n+1 with n >= 2, every contact form alpha for xi, and every closed Legendrian submanifold Lambda of Y, there exists a Reeb chord of Lambda: a non-constant trajectory of the Reeb vector field of alpha starting and ending on Lambda. Hutchings-Taubes proved the three-dimensional case (all Legendrian knots in all closed contact 3-manifolds); the conjecture is open in higher dimensions apart from special cases (subcritical Stein boundaries, conormal bundles, convex hypersurfaces).

    Why it matters

    A proof would guarantee a Reeb chord throughout the stated higher-dimensional setting. A counterexample would show that the contact and Legendrian conditions alone do not force such a returning trajectory.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Arnold Chord Conjecture in Higher Dimensions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #281 · Arnold Chord Conjecture in Higher Dimensions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 294–371.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  282. Rank282Legendre's Conjecture on Primes Between Consecutive SquaresFor every positive integer n, is there always at least one prime number strictly between the consecutive squares n² and (n+1)²?Number theory & arithmetic geometryPrime gaps · 3 problems

    Exact statement

    Prove that for every positive integer n there exists a prime p with n^2 < p < (n+1)^2.

    Why it matters

    A proof would guarantee that none of these increasingly long intervals is empty of primes. A counterexample would identify a specific gap covering an entire interval between consecutive squares.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Prime gaps · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 269–404.

    Source-fit opponent count: 32.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  283. Rank283Hartshorne Conjecture on complete intersections in codimension twoIn complex projective space of dimension at least seven, is every smooth subvariety two dimensions smaller than its ambient space defined by just two homogeneous polynomial equations, with the full scheme structure rather than only the same point set?Geometry & topology

    Exact statement

    Prove that for every integer N >= 7, every smooth codimension-two subvariety X of complex projective space P^N is the scheme-theoretic intersection of two hypersurfaces.

    Why it matters

    A proof would make the number of defining equations match the codimension throughout this range. A counterexample would exhibit smooth codimension-two geometry requiring more than two equations to define its full scheme structure.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Hartshorne Conjecture on complete intersections in codimension two · 1 problem in this edition. Family membership alone does not assert an implication.

    • #283 · Hartshorne Conjecture on complete intersections in codimension two (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 273–407.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  284. Rank284Gromov's Surface Subgroup Question for One-Ended Hyperbolic GroupsMust every one-ended word-hyperbolic group contain the fundamental group of a closed surface with negative Euler characteristic? Word-hyperbolic means its large-scale geometry has thin triangles; one-ended means its graph of generator steps has exactly one infinite component after any finite region is removed.Geometry & topology

    Exact statement

    Prove that every one-ended word-hyperbolic group contains a subgroup isomorphic to the fundamental group of a closed surface of negative Euler characteristic.

    Why it matters

    A proof would locate surface-group structure inside every group in this class. A counterexample would establish a one-ended hyperbolic group lacking all of the prescribed surface subgroups.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Gromov's Surface Subgroup Question for One-Ended Hyperbolic Groups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #284 · Gromov's Surface Subgroup Question for One-Ended Hyperbolic Groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 276–408.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  285. Rank285Shelah's Conjecture on NIP FieldsIf an infinite field is NIP—its definable families cannot encode arbitrary subset patterns—must it be separably closed, be real closed, or carry a nontrivial henselian valuation, a notion of size that supports lifting simple roots?Logic, foundations & set theoryNIP fields · 2 problems

    Exact statement

    Prove that every infinite field whose first-order theory (in the language of rings) has the independence property negated (is NIP, or dependent) is either separably closed, real closed, or admits a non-trivial henselian valuation; Johnson proved the conjecture for fields of finite dp-rank, and the general NIP case is open.

    Why it matters

    A proof would connect this logical tameness condition to one of three algebraic or valuation-theoretic alternatives. A counterexample would exhibit an infinite dependent field outside all three.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    NIP fields · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 291–389.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  286. Rank286Bass-Quillen conjectureDoes adding polynomial variables to a regular ring create any genuinely new algebraic vector bundles, or is every bundle on the resulting affine space obtained by extending one already defined over the original ring?Algebra, representation & category theory

    Exact statement

    Prove or disprove that for every regular ring R and every d >= 0, every vector bundle on affine d-space over R descends from R; equivalently, every finitely generated projective R[t_1,...,t_d]-module is extended from R.

    Why it matters

    A proof would show that adjoining polynomial coordinates introduces no new vector bundles beyond extensions of those on the base. A counterexample would identify a regular base where this invariance fails.

    Status

    Open · subcases solved

    October 6, 2026: Quillen’s cotangent-complex finiteness conjecture in characteristic two is different from the Bass-Quillen vector-bundle target. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Problem family

    Bass-Quillen conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #286 · Bass-Quillen conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 270–402.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  287. Rank287Koiran's Real Tau-Conjecture on Real Roots of Sums of Products of Sparse PolynomialsWrite a nonzero real polynomial as a sum of m products, each with k factors having at most t nonzero monomials. Is the number of its distinct real roots always bounded by A(mkt)^c for universal constants A and c, regardless of the degrees or the sizes of the real coefficients?Theoretical computer scienceAlgebraic computation lower bounds · 3 problems

    Exact statement

    Do there exist universal constants A>0 and c>=1 such that, for every positive integers m,k,t and every nonzero f in R[x] expressible as f=sum_(i=1)^m product_(j=1)^k f_ij with each f_ij in R[x] having at most t nonzero monomials, f has at most A(mkt)^c distinct real zeros? Exponents are ordinary nonnegative integers, coefficients arbitrary reals, and the bound is independent of degrees and coefficient magnitudes. Prove the universal bound or disprove existence of such constants.

    Why it matters

    A polynomial root bound in the size of this sparse description would connect real algebraic geometry to lower bounds for arithmetic computation. Its recorded consequences include VP ≠ VNP over the complex numbers.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Related problems

    • #31 · Valiants Algebraic P versus NP Conjecture (VP != VNP)Weaker assertionValiants Algebraic P versus NP Conjecture (VP != VNP)

      A proof of the uniform real-root bound would imply VP ≠ VNP over the complex numbers. A disproof of that root bound need not settle the algebraic-complexity separation.

      Condition: The stronger assertion is proved; a negative answer to it need not settle the weaker assertion.

      Scope qualification: The recorded consequence is over C, attributed to Tavenas by Briquel–Bürgisser. No converse or unrestricted all-fields consequence is asserted. Only the positive-answer implication is recorded. No converse or equivalence of resolution outcomes is established here.

    Problem family

    Algebraic computation lower bounds · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 291–389.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  288. Rank288Discontinuity of the q-State Potts Phase Transition on Z^d for d >= 3 and q >= 3In the nearest-neighbor Potts model, lattice spins can each take q states. With at least three states and at least three spatial dimensions, must the change between ordered and disordered phases always be abrupt rather than continuous?Mathematical physics

    Exact statement

    Prove that the order/disorder phase transition of the nearest-neighbor q-state Potts model on Z^d with d >= 3 is discontinuous (first-order) for every integer q >= 3.

    Why it matters

    A proof would establish first-order behavior across this entire parameter range. A counterexample would show that changing the number of states or dimension can produce a different transition type within the stated model.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Discontinuity of the q-State Potts Phase Transition on Z^d for d >= 3 and q >= 3 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #288 · Discontinuity of the q-State Potts Phase Transition on Z^d for d >= 3 and q >= 3 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 285–405.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  289. Rank289Néel Order for the Spin-1/2 Square-Lattice Heisenberg Antiferromagnet (Averaged Ground-State Order)In a square grid of interacting spin-1/2 particles, does the lowest-energy state retain checkerboard magnetic order as the grid grows? For the nearest-neighbor Heisenberg antiferromagnet on even periodic grids, the test is whether the mean square of alternating magnetization per site stays bounded away from zero.Mathematical physics

    Exact statement

    For each even L let H_L=sum_{<x,y>} S_x dot S_y be the nearest-neighbour isotropic spin-1/2 antiferromagnetic Heisenberg Hamiltonian on the periodic L by L square lattice, with coupling normalized to one and unique normalized ground state Phi_L. Put M_L=sum_x (-1)^(x_1+x_2) S_x^(3) and q_L=<Phi_L,(M_L/L^2)^2 Phi_L>. Assertion: liminf_{L->infinity, L even} q_L>0, equivalently there exist q>0 and L_0 such that q_L>=q for every even L>=L_0.

    Why it matters

    A proof would establish persistent averaged alternating magnetic order through a positive lower limit. Failure would allow this order measure to approach zero along some sequence of even lattice sizes.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Néel Order for the Spin-1/2 Square-Lattice Heisenberg Antiferromagnet (Averaged Ground-State Order) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #289 · Néel Order for the Spin-1/2 Square-Lattice Heisenberg Antiferromagnet (Averaged Ground-State Order) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 284–395.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  290. Rank290Mazur's Conjecture on the Real Topology of Rational PointsFor an algebraic variety defined over the rationals, take all its rational points and all their limits in the real topology. Must this set have only finitely many connected pieces?Number theory & arithmetic geometry

    Exact statement

    For every algebraic variety V over Q, prove or disprove that the closure of V(Q) in V(R), with the Euclidean topology, has finitely many connected components.

    Why it matters

    A proof would constrain how rational points and their real limits can split into disconnected pieces. A counterexample would exhibit infinitely many connected components in this closure for one variety.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Mazur's Conjecture on the Real Topology of Rational Points · 1 problem in this edition. Family membership alone does not assert an implication.

    • #290 · Mazur's Conjecture on the Real Topology of Rational Points (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 283–398.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  291. Rank291The Hadamard Matrix Existence ConjectureFor every positive integer k, does there exist a 4k×4k matrix of plus and minus ones whose distinct rows have dot product zero?Cryptography, coding, information & optimization

    Exact statement

    Prove that there exists a Hadamard matrix of order 4k for every positive integer k.

    Why it matters

    A proof would ensure the existence of perfectly orthogonal sign patterns in every order divisible by four. A counterexample would identify an order where that combination of size and orthogonality is impossible.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    The Hadamard Matrix Existence Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #291 · The Hadamard Matrix Existence Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 286–405.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  292. Rank292Nirenberg Problem on the Two-SphereWhich smooth Gaussian-curvature functions on the two-dimensional sphere can be produced by multiplying its standard round metric by a smooth positive factor?Geometry & topology

    Exact statement

    Characterize exactly the smooth functions K:S^2->R for which there is a smooth real function u on S^2 such that the metric e^(2u)g_round has Gaussian curvature K, where g_round is the standard unit round metric.

    Why it matters

    A complete characterization would identify exactly which Gaussian-curvature patterns are realizable within the round sphere's conformal class. It must account for both existence and every obstruction.

    Status

    Open · subcases solved

    Open with solved subclasses. Struwe 2023 records the general curvature-prescription challenge; Platt 2026 proves existence and symmetry information for particular prescribed curvatures. These results do not classify all smooth functions.

    Status review date not recorded in this edition

    Sources

    Problem family

    Nirenberg Problem on the Two-Sphere · 1 problem in this edition. Family membership alone does not assert an implication.

    • #292 · Nirenberg Problem on the Two-Sphere (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 297–388.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  293. Rank293Worst-case to average-case hardness reduction for the Learning Parity with Noise problemCan the typical difficulty of learning a hidden parity rule from examples corrupted at a fixed constant noise rate be established by a reduction from a worst-case decoding problem?Cryptography, coding, information & optimization

    Exact statement

    Decide whether average-case hardness of Learning Parity with Noise in the standard constant-noise regime can be established by a worst-case-to-average-case reduction from a worst-case decoding problem.

    Why it matters

    Such a reduction would connect hardness of typical noisy parity instances to hardness of decoding in the worst case. It would provide a foundation for that average-case hardness within the specified noise regime.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Worst-case to average-case hardness reduction for the Learning Parity with Noise problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #293 · Worst-case to average-case hardness reduction for the Learning Parity with Noise problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 280–417.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  294. Rank294Chowla Non-Vanishing Conjecture for Dirichlet L-Functions at the Central PointCan a primitive Dirichlet L-function ever vanish at its central point, s=1/2? The question includes every primitive character, real or complex, of either parity.Number theory & arithmetic geometry

    Exact statement

    For every integer q>=1 and every primitive Dirichlet character chi of conductor q, including real and nonreal characters and both even and odd parity, the analytically continued Dirichlet L-function satisfies L(1/2,chi) != 0. The conductor-one character is included with L(s,chi)=zeta(s), whose central value is already known to be nonzero; conductors with no primitive characters contribute no cases. No positivity, quantitative lower bound, effective constant or off-central zero-free region is part of the assertion.

    Why it matters

    A proof would exclude exact central zeros throughout this family. A counterexample would show that one occurs, but a zero at the central point would not by itself contradict the generalized Riemann hypothesis.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Chowla Non-Vanishing Conjecture for Dirichlet L-Functions at the Central Point · 1 problem in this edition. Family membership alone does not assert an implication.

    • #294 · Chowla Non-Vanishing Conjecture for Dirichlet L-Functions at the Central Point (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 303–390.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  295. Rank295Broues Abelian Defect Group ConjectureA representation block groups together related representations of a finite group. When its defect group is abelian, must the block's derived category be equivalent to that of its Brauer correspondent in the subgroup preserving the defect group?Algebra, representation & category theory

    Exact statement

    Prove that if B is a p-block of a finite group G with abelian defect group D and b is the Brauer correspondent block in N_G(D), then the bounded derived categories D^b(mod B) and D^b(mod b) are equivalent as triangulated categories.

    Why it matters

    A proof would connect the block’s module complexes and their homological relations to those of the corresponding block in a more local subgroup. Failure would limit this proposed categorical correspondence.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Broues Abelian Defect Group Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #295 · Broues Abelian Defect Group Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 285–400.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  296. Rank296Slice-Ribbon ConjectureIf a knot in the three-sphere bounds a smoothly embedded disc in the four-ball, must it also bound an immersed disc in the three-sphere with only ribbon singularities—double arcs where one sheet reaches the disc's boundary and the other stays in its interior?Geometry & topology

    Exact statement

    Is every smoothly slice knot in S^3 (one that bounds a smoothly embedded disk in the 4-ball) a ribbon knot (one bounding an immersed disk in S^3 with only ribbon singularities)?

    Why it matters

    A proof would represent every smoothly slice knot through this particular three-dimensional disc construction. A counterexample would separate smooth sliceness in the four-ball from the ribbon condition in the three-sphere.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Slice-Ribbon Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #296 · Slice-Ribbon Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 300–404.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  297. Rank297Jones Unknot ConjectureFor every knot in the three-sphere, does having Jones polynomial exactly equal to one guarantee that the knot can be untangled into a simple loop?Geometry & topology

    Exact statement

    Prove or disprove: a knot K in S^3 has Jones polynomial V_K = 1 if and only if K is the unknot (equivalently, every nontrivial knot has nontrivial Jones polynomial).

    Why it matters

    A proof would make the Jones polynomial a complete detector of the unknot. A counterexample would exhibit genuine knotting that this polynomial cannot distinguish from an unknotted loop.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Jones Unknot Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #297 · Jones Unknot Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 282–420.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  298. Rank298Graph reconstruction conjectureCan every finite simple undirected graph with at least three vertices be recovered, up to relabeling, from the complete collection of graphs obtained by deleting each vertex in turn, with repeated results counted?Combinatorics & discrete geometryGraph reconstruction · 2 problems

    Exact statement

    Prove or disprove that every finite simple undirected graph on at least three vertices is determined up to isomorphism by the multiset of its vertex-deleted subgraphs.

    Why it matters

    A proof would show that this collection of partial views determines the whole graph uniquely. A counterexample would give two different graphs that remain indistinguishable from all their vertex-deleted views.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Related problems

    • #483 · Edge Reconstruction Conjecture for GraphsVariantEdge Reconstruction Conjecture for Graphs

      A proof of this assertion would also establish #483.

      Scope qualification: The recorded direction concerns positive assertions only; no converse or whole binary-completion equivalence is asserted.

    Problem family

    Graph reconstruction · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 296–410.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  299. Rank299Dixmier ConjectureThe first Weyl algebra describes polynomial differentiation and multiplication in one variable. Over a characteristic-zero field, must every structure-preserving map from this algebra to itself be reversible?Algebra, representation & category theoryOpen · disputed claim

    Exact statement

    Let K be a field of characteristic zero and A_1(K) the first Weyl algebra. Prove that every K-algebra endomorphism of A_1(K) is an automorphism.

    Why it matters

    A proof would force all such structure-preserving self-maps to be automorphisms. A counterexample would identify an endomorphism of this algebra that fails to reach every element.

    Status

    Open · disputed claim

    Status review date not recorded in this edition

    Sources

    Problem family

    Dixmier Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #299 · Dixmier Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 291–411.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  300. Rank300Vinogradov Conjecture on the Least Quadratic Non-ResidueAs p ranges over odd primes, does the smallest positive integer that is not congruent to any square modulo p grow more slowly than every positive power of p, without relying on other unproved conjectures?Number theory & arithmetic geometry

    Exact statement

    Let n_p denote the least quadratic non-residue modulo an odd prime p. Prove unconditionally that for every eps > 0 one has n_p <<_eps p^eps.

    Why it matters

    A proof would guarantee a quadratic nonresidue below every fixed positive power scale, up to its permitted constant. Failure would show that one such scale cannot uniformly contain a nonresidue.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Vinogradov Conjecture on the Least Quadratic Non-Residue · 1 problem in this edition. Family membership alone does not assert an implication.

    • #300 · Vinogradov Conjecture on the Least Quadratic Non-Residue (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 311–393.

    Source-fit opponent count: 36.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  301. Rank301Polynomial-time unknot recognitionGiven a knot diagram, can an algorithm decide whether it represents an unknotted loop in time bounded by a polynomial in the input size?Geometry & topology

    Exact statement

    Determine whether there is a polynomial-time algorithm that, given a knot diagram or an equivalent finite encoding of a knot in three-space, decides whether the knot is the unknot.

    Why it matters

    A positive answer would make unknot recognition efficiently decidable in the worst case. A negative answer would show that every algorithm for this exact recognition task exceeds polynomial time on some inputs.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Polynomial-time unknot recognition · 1 problem in this edition. Family membership alone does not assert an implication.

    • #301 · Polynomial-time unknot recognition (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 269–452.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  302. Rank302Lions' Open Problem on Small-Time Global Exact Null Controllability of the Navier-Stokes Equations with Dirichlet Boundary ConditionsCan any admissible finite-energy incompressible fluid flow in a bounded domain be brought exactly to rest in any positive time by controlling velocity on just one nonempty open patch of its boundary? The flow follows Navier–Stokes, and velocity stays zero on the rest of the boundary.Applied & computational mathematics

    Exact statement

    Let Omega be a bounded domain, Gamma a nonempty open subset of its boundary, and T > 0. For every divergence-free initial velocity u_0 in the natural L^2 phase space, does there exist boundary data f supported on Gamma and a weak solution of the incompressible Navier-Stokes equations with u=f on Gamma, u=0 on the uncontrolled boundary, u(0)=u_0, and u(T)=0?

    Why it matters

    A proof would establish exact stopping through partial boundary control for arbitrary initial velocities in the stated phase space. It would guarantee a suitable weak solution and control even for an arbitrarily short positive time.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Lions' Open Problem on Small-Time Global Exact Null Controllability of the Navier-Stokes Equations with Dirichlet Boundary Conditions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #302 · Lions' Open Problem on Small-Time Global Exact Null Controllability of the Navier-Stokes Equations with Dirichlet Boundary Conditions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 287–427.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  303. Rank303Tutte's 3-Flow ConjectureCan every finite loopless multigraph remaining connected after deletion of up to three edges be oriented and assigned edge flows of one or two units, with inflow equaling outflow at every vertex?Combinatorics & discrete geometry

    Exact statement

    Every finite loopless multigraph G that is 4-edge-connected admits a nowhere-zero integer 3-flow: there exist an orientation D of E(G) and a function f:E(G)->{-2,-1,1,2} such that at each vertex the sum of f on incoming edges equals the sum on outgoing edges. Four-edge-connected means G remains connected after deletion of any set of at most3 edges. Parallel edges are allowed. Trivial one-vertex loopless graphs have the empty flow. This is the ordinary zero-boundary flow assertion, not the stronger claim of realizing every admissible prescribed boundary modulo3.

    Why it matters

    A proof would make four-edge-connectivity sufficient for a nowhere-zero integer 3-flow, even with parallel edges. A counterexample would identify a graph whose connectivity does not permit this bounded circulation.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Tutte's 3-Flow Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #303 · Tutte's 3-Flow Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 316–404.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  304. Rank304Existence of a Finitely Presented Group of Intermediate GrowthCan a group with finitely many generators and defining relations have intermediate growth: the number of distinct elements reachable in at most n generator steps has no polynomial upper bound, yet grows more slowly than every exponential rate?Algebra, representation & category theory

    Exact statement

    There exists a finitely presented group G with a finite generating set S whose word-ball growth gamma(n) is subexponential, meaning lim(n->infinity) log(gamma(n))/n=0, but is not bounded by any polynomial in n.

    Why it matters

    An example would show that a finite presentation permits intermediate growth. A nonexistence theorem would rule out every finitely presented group with this combination of subexponential growth and absence of a polynomial bound.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Existence of a Finitely Presented Group of Intermediate Growth · 1 problem in this edition. Family membership alone does not assert an implication.

    • #304 · Existence of a Finitely Presented Group of Intermediate Growth (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 310–405.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  305. Rank305Brauer's k(B) ConjectureFor a finite group and a prime p, can a representation block ever contain more distinct irreducible complex representations than there are elements in its defect group, the p-subgroup attached to the block?Algebra, representation & category theory

    Exact statement

    For every finite group G, every prime p, and every p-block B of G with defect group D, prove that k(B)=|Irr(B)|, the number of ordinary complex irreducible characters belonging to B, satisfies k(B)<=|D|.

    Why it matters

    A proof of the proposed bound would control this character count by the size of the associated p-group. A counterexample would show a block with more characters than that local bound permits.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Brauer's k(B) Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #305 · Brauer's k(B) Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 308–419.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  306. Rank306Bott Conjecture: nonnegatively curved simply connected manifolds are rationally ellipticIf a compact manifold has no noncontractible loops and admits nonnegative sectional curvature, must the Betti numbers of its based loop space—counts of independent homology classes—grow at most polynomially, over every coefficient field?Geometry & topology

    Exact statement

    Prove that every compact simply connected manifold M admitting a metric of nonnegative sectional curvature is rationally elliptic, i.e., the Betti numbers of its loop space grow at most polynomially for every field of coefficients.

    Why it matters

    A proof would constrain the topological complexity of loop spaces using the existence of a nonnegatively curved metric. A counterexample would show that such curvature permits faster growth under the stated coefficient condition.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. Under a sufficiently small scale-invariant trace-free Ricci L2 energy assumption, the nonnegatively curved simply connected four-manifold is one of S^4, CP^2, or S^2 x S^2. These classified spaces have the polynomial loop-homology growth required by the target, so the source covers this restricted subclass. Bott asks for every dimension and no Ricci-energy restriction; those quantifiers remain uncovered. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Bott Conjecture: nonnegatively curved simply connected manifolds are rationally elliptic · 1 problem in this edition. Family membership alone does not assert an implication.

    • #306 · Bott Conjecture: nonnegatively curved simply connected manifolds are rationally elliptic (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 300–428.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  307. Rank307Corona problem for the unit polydiscOn a product of two or more complex unit disks, suppose finitely many bounded holomorphic functions never all become arbitrarily small together. Can bounded holomorphic multipliers always make the sum of their products exactly one?Analysis & PDE

    Exact statement

    For every integer n>=2 and every finite integer m>=1, let f_1,...,f_m be scalar bounded holomorphic functions on the unit polydisc D^n={z in C^n: |z_k|<1 for all k}. If inf_{z in D^n} sum_{j=1}^m |f_j(z)|>0, then there exist scalar bounded holomorphic g_1,...,g_m on D^n such that sum_{j=1}^m f_j g_j=1 identically. No boundary extension, regularity, operator inequality or sharp solution-norm bound is required. Equivalently, for every n>=2, point evaluations from D^n are dense in the maximal ideal space of H-infinity(D^n).

    Why it matters

    A proof would establish scalar bounded holomorphic Bézout solvability on every such polydisc. It would guarantee existence of coefficients, without requiring boundary regularity or a sharp bound on their norms.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Corona problem for the unit polydisc · 1 problem in this edition. Family membership alone does not assert an implication.

    • #307 · Corona problem for the unit polydisc (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 311–411.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  308. Rank308Arithmeticity of Lattices in PU(n,1) for n at Least 4In complex dimension at least four, must every discrete symmetry group giving a finite-volume quotient of complex hyperbolic space be arithmetic, arising from number-theoretic constructions? Compact and noncompact quotients, including groups with torsion, are all included.Geometry & topology

    Exact statement

    For every integer n>=4, every lattice Γ in PU(n,1), the group of holomorphic isometries of complex hyperbolic n-space, is arithmetic. A lattice is a discrete subgroup of finite Haar covolume; both compact and noncompact finite-volume quotients and groups with torsion are included. Arithmeticity has its usual commensurability meaning via arithmetic subgroups of algebraic groups over Q projected through a compact-kernel real-group homomorphism, equivalently after the finite central lift to SU(n,1). No restriction to reflection lattices or one particular arithmetic construction is imposed. Here n is complex dimension, not real rank.

    Why it matters

    A proof would place all these finite-volume complex-hyperbolic lattices within arithmetic commensurability constructions. A single nonarithmetic example in an admissible dimension would refute the universal assertion.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Arithmeticity of Lattices in PU(n,1) for n at Least 4 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #308 · Arithmeticity of Lattices in PU(n,1) for n at Least 4 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 301–421.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  309. Rank309Sharp asymptotic order of minimal star discrepancy in each dimension d >= 3For N points in a unit cube of fixed dimension d≥3, how small can the worst counting error be over all boxes anchored at the origin? The error compares the number of points in a box with N times its volume; the goal is its best possible growth rate as N increases.Combinatorics & discrete geometry

    Exact statement

    For every fixed integer d >= 3, determine the asymptotic order, up to positive multiplicative constants depending only on d, of Delta_d(N)=inf_{P subset [0,1)^d, |P|=N} sup_{x in [0,1]^d} |#(P intersect product_j[0,x_j))-N product_j x_j| as N tends to infinity.

    Why it matters

    Matching asymptotic bounds would determine the best attainable counting uniformity as N grows. The error is measured in numbers of points, with comparison constants allowed to depend only on dimension.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Sharp asymptotic order of minimal star discrepancy in each dimension d >= 3 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #309 · Sharp asymptotic order of minimal star discrepancy in each dimension d >= 3 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 318–421.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  310. Rank310Existence of a p-Optimal Propositional Proof SystemIs there one efficiently checkable proof system for propositional tautologies into which proofs from every other such system can be translated in polynomial time?Theoretical computer scienceOptimal propositional proof systems · 2 problems

    Exact statement

    Determine whether there exists a Cook-Reckhow propositional proof system that polynomially simulates every other propositional proof system, including a polynomial-time translation of proofs.

    Why it matters

    A positive answer would provide a single proof framework that efficiently accommodates proofs from every competing system. A negative answer would rule out such a universally efficient proof translator.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Related problems

    • #125 · Existence of a Length-Optimal Propositional Proof SystemWeaker assertionExistence of a Length-Optimal Propositional Proof System

      A p-optimal proof system would also be length-optimal: polynomial-time proof translation gives a polynomial bound on the translated proof length.

      Scope qualification: No converse or proved strict separation is asserted. Length-optimality requires short proofs to exist, without requiring an efficient algorithm to find or translate them.

    Problem family

    Optimal propositional proof systems · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 309–421.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  311. Rank311Hilbert's Tenth Problem over C(t) with named tCan an algorithm always decide whether a polynomial equation has solutions that are complex rational functions of one formal variable t? The input coefficients are integer-coefficient polynomials in that same named t, and any finite number of unknowns is allowed.Number theory & arithmetic geometryHilbert’s tenth problem · 1 problem

    Exact statement

    Let t be a fixed transcendental indeterminate and C(t) the field of complex rational functions. Determine whether there exists an algorithm (Turing machine) which, given any n >= 1 and any polynomial f in Z[t][x_1,...,x_n], presented by its finite list of integer coefficients and monomial exponents in t and the x_i, halts and correctly decides whether f = 0 has a solution (x_1,...,x_n) in C(t)^n. Equivalently, determine whether the existential (positive existential) theory of the structure (C(t); 0, 1, +, -, *, t), with a constant symbol naming t and no constants for other complex numbers, is decidable. Inputs with arbitrary complex coefficients, and the ring language without a symbol for t, are not part of this target.

    Why it matters

    The answer would locate the decision boundary for these finitely encoded equations with the indeterminate t named. An algorithm must halt on every input; undecidability would rule out such a universal procedure for this exact coefficient language.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Hilbert’s tenth problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #311 · Hilbert's Tenth Problem over C(t) with named t (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 315–430.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  312. Rank312Igusa’s Standard p-Adic Monodromy ConjectureFor a nonconstant rational polynomial, must every pole s of its p-adic zeta integrals give the complex monodromy eigenvalue exp(2πi Re(s)) somewhere on its zero set? Only finitely many primes may be exceptional, depending on the polynomial alone, and every compactly supported locally constant weight is included.Number theory & arithmetic geometry

    Exact statement

    For every n≥1 and nonconstant f∈Q[x_1,...,x_n], there is a finite set S_f of primes such that for every p outside S_f, every complex-valued locally constant compactly supported function phi on Q_p^n, and every actual pole s_0 of the meromorphic continuation of Z_p(f,phi;s)=integral |f(x)|_p^s phi(x) dx, exp(2*pi*i*Re(s_0)) is an eigenvalue of the local complex monodromy of f at some point of {f=0}⊂C^n. Haar measure is normalized by vol(Z_p^n)=1; the nonzero normalization does not affect poles. This is the standard assertion of Veys Conjecture 2.12(1); S_f depends on f, not on phi. Strong Bernstein–Sato, origin-only, differential-form, topological, motivic and character variants are not silently identified with it.

    Why it matters

    A proof would connect the values exp(2πi Re(s₀)) from these poles to monodromy somewhere on the complex zero set. Refutation must defeat the allowance of finitely many exceptional primes.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Igusa’s Standard p-Adic Monodromy Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #312 · Igusa’s Standard p-Adic Monodromy Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 332–416.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  313. Rank313Iwaniec Conjecture for the Beurling-Ahlfors Transform NormFor complex-valued functions on the plane, is the sharp factor by which the Beurling–Ahlfors transform can enlarge their Lᵖ size exactly max(p,p/(p−1))−1 for every 1<p<∞? The Lᵖ norm measures size by integrating the pth power of the magnitude and taking the pth root.Analysis & PDE

    Exact statement

    For every 1<p<infinity, prove or disprove that the Beurling-Ahlfors transform B on complex-valued L^p(C) has operator norm ||B||_p=p^*-1, where p^*=max(p,p/(p-1)).

    Why it matters

    The equality would determine the transform’s sharp operator norm at every exponent. Failure would show that some functions experience greater amplification than this proposed universal formula allows.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Iwaniec Conjecture for the Beurling-Ahlfors Transform Norm · 1 problem in this edition. Family membership alone does not assert an implication.

    • #313 · Iwaniec Conjecture for the Beurling-Ahlfors Transform Norm (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 293–452.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  314. Rank314Partition Regularity of Pythagorean TriplesNo matter how the positive integers are colored with any finite number of colors, must some same-colored positive integers x, y, and z satisfy x²+y²=z²?Combinatorics & discrete geometry

    Exact statement

    Determine whether every finite coloring of the positive integers contains positive integers x, y, and z of one color satisfying x^2 + y^2 = z^2.

    Why it matters

    A proof would make a monochromatic Pythagorean triple unavoidable in every finite partition. A counterexample would provide a finite coloring that separates the entries of every positive Pythagorean triple.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Partition Regularity of Pythagorean Triples · 1 problem in this edition. Family membership alone does not assert an implication.

    • #314 · Partition Regularity of Pythagorean Triples (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 325–428.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  315. Rank315Bourgain's unbounded Sobolev orbit problem for cubic defocusing NLS on the square two-torusFor the defocusing cubic nonlinear Schrödinger equation on the standard square torus, can a single globally smooth wave develop arbitrarily large Hˢ norm—a measure of its fine-scale variation—at arbitrarily late times, for some s>1?Analysis & PDE

    Exact statement

    For the defocusing cubic nonlinear Schrödinger equation i partial_t u + Delta u = |u|^2 u on the standard square torus T^2=(R/2pi Z)^2, does there exist one global smooth complex-valued solution u and a real s>1 such that limsup_{t->+infinity} ||u(t)||_{H^s(T^2)} = infinity?

    Why it matters

    A positive answer would exhibit unbounded growth of a Sobolev norm along a single smooth evolution. A negative answer would bound each such orbit’s norm for every exponent s>1.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Bourgain's unbounded Sobolev orbit problem for cubic defocusing NLS on the square two-torus · 1 problem in this edition. Family membership alone does not assert an implication.

    • #315 · Bourgain's unbounded Sobolev orbit problem for cubic defocusing NLS on the square two-torus (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 335–422.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  316. Rank316Ivrii conjecture on periodic billiard orbitsFor billiards in any bounded smooth connected region of dimension at least two, do exactly periodic, nongrazing paths occupy zero measure among starting boundary positions and directions? Only trajectories whose reflections remain well-defined in both time directions are counted.Probability, ergodic theory & dynamics

    Exact statement

    For every integer n>=2 and every bounded connected open domain Ω in Euclidean R^n with C∞ boundary, prove the following assertion; neither convexity nor simple connectivity is assumed. Let M=B*∂Ω={(x,η) in T*∂Ω: |η|<1}, with the norm induced by the Euclidean metric and canonical symplectic (Liouville) volume μ=|ω^(n-1)|/(n-1)!, where ω is the canonical symplectic form. The billiard map β lifts (x,η) to the inward-pointing unit covector, follows the straight segment to its first subsequent boundary collision, and applies specular reflection, recording the tangential covector at that collision. Let M_reg be the set on which every forward and backward iterate is well-defined and nongrazing. For P=union over integers k>=2 of {z in M_reg: β^k(z)=z}, prove μ(P)=0. The integer k need not be the least period. Grazing collisions and any undefined singular continuation are excluded from M_reg; no rule continuing through them is imposed.

    Why it matters

    A proof would supply the dynamical hypothesis in Ivrii’s two-term asymptotic formula for Laplacian eigenvalue counts. A positive-measure periodic set would refute that hypothesis without itself disproving the spectral formula.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Ivrii conjecture on periodic billiard orbits · 1 problem in this edition. Family membership alone does not assert an implication.

    • #316 · Ivrii conjecture on periodic billiard orbits (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 332–423.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  317. Rank317Hopf Conjecture: no metric of positive sectional curvature on S^2 x S^2Can the four-dimensional product of two spheres, S²×S², be given a geometry with strictly positive sectional curvature—positive curvature in every tangent two-dimensional direction—everywhere?Geometry & topology

    Exact statement

    Determine whether the product S^2 x S^2 admits a Riemannian metric of positive sectional curvature; the Hopf conjecture asserts that no such metric exists.

    Why it matters

    Nonexistence would show that this topology prevents positive curvature in every tangent two-plane, regardless of the metric chosen. An example would demonstrate that such curvature is compatible with the product space.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The positive-sectional-curvature Einstein classification excludes S^2 x S^2 among closed positively curved Einstein four-manifolds. It therefore rules out a positively curved Einstein metric on that product. The full Hopf question allows every Riemannian metric, without an Einstein assumption, so the classification does not exclude all possible positively curved metrics on S^2 x S^2. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Hopf Conjecture: no metric of positive sectional curvature on S^2 x S^2 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #317 · Hopf Conjecture: no metric of positive sectional curvature on S^2 x S^2 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 303–453.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  318. Rank318The Word Problem for One-Relation MonoidsTake words built from finitely many symbols, with one rule allowing a substring to be replaced in either direction. Is there always an algorithm that finishes and decides whether two words can be made equal using that rule?Algebra, representation & category theory

    Exact statement

    Determine whether every monoid defined by a single defining relation has a solvable word problem: given a one-relation monoid presentation <a_1, ..., a_n | u = v> and two words over the generators, decide whether the two words represent the same element of the monoid; no undecidable one-relation monoid is known and no algorithm solving the general case is known.

    Why it matters

    A positive answer would make equality decidable throughout this presentation class. A negative answer would identify a one-relation monoid whose word equality cannot be decided by any total algorithm.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    The Word Problem for One-Relation Monoids · 1 problem in this edition. Family membership alone does not assert an implication.

    • #318 · The Word Problem for One-Relation Monoids (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 331–433.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  319. Rank319Kummer-Vandiver Conjecture on Cyclotomic Class NumbersFor the field generated by a primitive pth root of unity, take its largest real subfield. Must the prime p never divide the class number of that real field, the number of its ideal classes?Number theory & arithmetic geometry

    Exact statement

    Prove that the prime p does not divide the class number h+(Q(zeta_p)) of the maximal real subfield of the p-th cyclotomic field for all primes p.

    Why it matters

    A proof would exclude p-torsion from the ideal class group of each corresponding real cyclotomic field. A counterexample would identify a prime where this particular arithmetic obstruction survives.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Kummer-Vandiver Conjecture on Cyclotomic Class Numbers · 1 problem in this edition. Family membership alone does not assert an implication.

    • #319 · Kummer-Vandiver Conjecture on Cyclotomic Class Numbers (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 314–448.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  320. Rank320Brauer-Manin obstruction to the Hasse principle for K3 surfacesFor a K3 surface over a number field, suppose there is one choice of points over all completions that passes every Brauer–Manin test at once. Must the surface then have a point over the number field itself?Number theory & arithmetic geometry

    Exact statement

    For every number field k and every K3 surface X over k (smooth projective geometrically integral surface with trivial canonical bundle and H^1(X, O_X) = 0), prove that X(A_k)^{Br(X)} nonempty implies X(k) nonempty. Here Br(X) = H^2_et(X, G_m), and X(A_k)^{Br(X)} is the set of adelic points (x_v) such that for every alpha in Br(X) the sum over all places v of inv_v(alpha(x_v)) vanishes in Q/Z, i.e. simultaneous orthogonality of one adelic point to the entire Brauer group. No assertion about weak approximation, about density of X(k) in X(A_k)^{Br(X)}, or about the existence of a single obstructing class is included.

    Why it matters

    A proof would make the full Brauer–Manin obstruction sufficient for rational-point existence in this class. Failure would exhibit compatible local data passing every such condition while no rational point exists.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Brauer-Manin obstruction to the Hasse principle for K3 surfaces · 1 problem in this edition. Family membership alone does not assert an implication.

    • #320 · Brauer-Manin obstruction to the Hasse principle for K3 surfaces (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 334–438.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  321. Rank321Strong Compactness versus Supercompactness Equiconsistency ProblemStrongly compact and supercompact cardinals are two types of very large infinity. Are they equally strong as consistency assumptions over ZFC, or does supercompactness require stronger foundations? Here consistency means freedom from contradiction.Logic, foundations & set theory

    Exact statement

    Determine whether Con(ZFC plus the existence of a strongly compact cardinal) implies Con(ZFC plus the existence of a supercompact cardinal), or prove that the latter has strictly greater consistency strength.

    Why it matters

    The answer would determine whether these two large-cardinal hypotheses occupy the same consistency level. A separation would distinguish their foundational strength without merely comparing their definitions inside one model.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Strong Compactness versus Supercompactness Equiconsistency Problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #321 · Strong Compactness versus Supercompactness Equiconsistency Problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 323–450.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  322. Rank322Constant-Parameter Quantum Locally Testable CodesCan quantum error-correcting codes grow indefinitely while storing a constant fraction of their physical qubits as logical information, protecting against a constant fraction of errors, and admitting tests on only a bounded number of qubits whose rejection probability stays proportional to distance from the code?Quantum information & computation

    Exact statement

    Construct an infinite family of quantum locally testable codes with asymptotically constant rate, relative distance, soundness, and local-projector weight, or prove that no such family exists.

    Why it matters

    A construction would combine these four requirements without deterioration as the codes grow. An impossibility theorem would establish that at least one of them must weaken in any such family.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Constant-Parameter Quantum Locally Testable Codes · 1 problem in this edition. Family membership alone does not assert an implication.

    • #322 · Constant-Parameter Quantum Locally Testable Codes (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 269–507.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  323. Rank323Four Exponentials ConjectureIf each of two pairs of complex numbers is linearly independent over the rationals, must at least one exponential of their four cross-products be transcendental, satisfying no nonzero polynomial equation with integer coefficients?Number theory & arithmetic geometryExponential transcendence · 3 problems

    Exact statement

    Prove or disprove that, for complex numbers x1, x2 linearly independent over Q and y1, y2 linearly independent over Q, at least one of the four numbers exp(x_i y_j) (i, j in {1,2}) is transcendental.

    Why it matters

    A proof would force transcendence from these two independence conditions alone. A counterexample would produce two independent pairs for which all four exponentials satisfy polynomial equations with integer coefficients.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Related problems

    Problem family

    Exponential transcendence · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 320–462.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  324. Rank324Gromov-Lawson-Rosenberg Conjecture for Finite Fundamental GroupsFor a closed connected spin manifold of dimension at least five with finite fundamental group, is the real Rosenberg index the only obstruction to giving it everywhere positive scalar curvature? Does its vanishing guarantee that such a metric exists?Geometry & topology

    Exact statement

    For every finite group G and every closed connected smooth spin manifold M of dimension n>=5 with pi_1(M)=G, prove or disprove that M admits a positive-scalar-curvature metric if and only if its real Rosenberg index alpha(M) in KO_n(C*_r(G;R)) vanishes.

    Why it matters

    A proof would make this index a complete existence criterion for positive scalar curvature in the specified class. A counterexample would separate the index condition from that geometric possibility.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Gromov-Lawson-Rosenberg Conjecture for Finite Fundamental Groups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #324 · Gromov-Lawson-Rosenberg Conjecture for Finite Fundamental Groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 332–456.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  325. Rank325Lieb-Thirring Conjecture on the Sharp Constant for Riesz Means of Order One in Three DimensionsFor a three-dimensional Schrödinger operator, is the total magnitude of its negative bound-state energies always at most 1/(15π²) times the integral of |V|^(5/2) over the region where the potential V is negative, with this constant best possible?Mathematical physics

    Exact statement

    Prove that the optimal constant L_{1,3} in Tr(-Delta+V)_- <= L_{1,3} integral_{R^3} V_-(x)^(5/2) dx equals the semiclassical constant L_{1,3}^{cl}=1/(15*pi^2).

    Why it matters

    A proof would make the semiclassical value the sharp constant in this energy bound. A potential requiring a larger constant would show that the proposed value is insufficient.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Lieb-Thirring Conjecture on the Sharp Constant for Riesz Means of Order One in Three Dimensions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #325 · Lieb-Thirring Conjecture on the Sharp Constant for Riesz Means of Order One in Three Dimensions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 338–444.

    Source-fit opponent count: 34.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  326. Rank326Černý Conjecture on Shortest Reset Words of Synchronizing AutomataA finite-state machine can be reset if one input sequence sends every starting state to the same final state. For every such deterministic machine with n states, is there always a reset sequence of at most (n−1)² symbols?Theoretical computer science

    Exact statement

    Prove or refute that every synchronizing deterministic finite automaton with n states admits a reset (synchronizing) word of length at most (n-1)^2, i.e., that the reset threshold of every n-state synchronizing automaton does not exceed (n-1)^2.

    Why it matters

    A proof would give a universal quadratic limit on the length of a reset sequence. A counterexample would show that some synchronizing machine needs a longer sequence despite having only n states.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Černý Conjecture on Shortest Reset Words of Synchronizing Automata · 1 problem in this edition. Family membership alone does not assert an implication.

    • #326 · Černý Conjecture on Shortest Reset Words of Synchronizing Automata (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 296–486.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  327. Rank327Asymptotic Turan Numbers of Even CyclesFor each fixed even cycle length of at least four, what is the order of growth of the largest number of edges in an n-vertex simple graph avoiding that cycle as n increases?Combinatorics & discrete geometry

    Exact statement

    For every fixed integer k >= 2, determine the order of magnitude, as n tends to infinity, of ex(n,C_{2k}), the maximum number of edges of an n-vertex finite simple graph containing no cycle of length 2k.

    Why it matters

    Matching bounds would determine the exact order of magnitude of the maximum edge count. The question forbids one specified cycle length, so its answer measures that particular restriction on graph density.

    Status

    Open · subcases solved

    Open with the order known for k=2,3,5; the September 2026 primary source still reports no matching general lower bound. Exact leading constants are a different target.

    Status review date not recorded in this edition

    Sources

    Problem family

    Asymptotic Turan Numbers of Even Cycles · 1 problem in this edition. Family membership alone does not assert an implication.

    • #327 · Asymptotic Turan Numbers of Even Cycles (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 345–444.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  328. Rank328Beal's Conjecture (Tijdeman-Zagier Conjecture)Whenever positive integers satisfy Aˣ+Bʸ=Cᶻ with all three exponents greater than two, must a single prime divide all three bases A, B, and C?Number theory & arithmetic geometryEquations in perfect powers · 2 problems

    Exact statement

    Prove that if A^x + B^y = C^z with A, B, C, x, y, z positive integers and x, y, z all greater than 2, then A, B, and C have a common prime factor.

    Why it matters

    A proof would rule out every such equation whose three bases have no common prime factor. A counterexample would exhibit an admissible power-sum identity without that shared divisibility.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Equations in perfect powers · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 319–463.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  329. Rank329Positive Combinatorial Rule for Kronecker CoefficientsCan every Kronecker coefficient be obtained by counting a concrete family of combinatorial objects, without subtraction? These coefficients count how often irreducible representations of a symmetric group appear in tensor products.Algebra, representation & category theory

    Exact statement

    Find a positive combinatorial interpretation or rule for the Kronecker coefficients g_{lambda,mu,nu}, equivalently the multiplicities of irreducible symmetric-group representations indexed by nu in tensor products of those indexed by lambda and mu, for all partitions lambda, mu, and nu of n.

    Why it matters

    Such a rule would interpret every Kronecker coefficient as a count of specified objects. It would give a uniform combinatorial account of these tensor-product multiplicities for all triples of partitions of the same size.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Positive Combinatorial Rule for Kronecker Coefficients · 1 problem in this edition. Family membership alone does not assert an implication.

    • #329 · Positive Combinatorial Rule for Kronecker Coefficients (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 291–522.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  330. Rank330NP-Hardness of Constant-Factor Approximation for Densest k-SubgraphIn a given graph, choose exactly k vertices to capture as many edges as possible. Is there a fixed approximation factor greater than one for which this task is NP-hard under deterministic polynomial-time reductions?Theoretical computer science

    Exact statement

    Let OPT(G,k)=max_{S subseteq V(G), |S|=k}|E(G[S])| for a finite simple undirected input graph G. Prove or refute the assertion that there exists an absolute constant epsilon>0 for which the promise problem of distinguishing OPT(G,k)>=t from OPT(G,k)<t/(1+epsilon), with input integers 1<=k<=|V(G)| and t>=1, is NP-hard under deterministic polynomial-time gap reductions. The target is any fixed nontrivial constant approximation gap for exactly k vertices, based on NP-hardness rather than a stronger average-case or exponential-time assumption.

    Why it matters

    A proof would establish constant-gap hardness for selecting exactly k vertices, with no stronger complexity assumption in the reduction. It would rule out polynomial-time solutions to this promise problem unless P equals NP.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    NP-Hardness of Constant-Factor Approximation for Densest k-Subgraph · 1 problem in this edition. Family membership alone does not assert an implication.

    • #330 · NP-Hardness of Constant-Factor Approximation for Densest k-Subgraph (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 352–433.

    Source-fit opponent count: 40.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  331. Rank331Shelah's Categoricity Conjecture for L_omega1,omegaIn the infinitary logic Lω₁,ω, formulas allow countable “and” and “or” combinations. If a sentence has exactly one model up to isomorphism at one cardinality at or above the specified Hanf threshold, must it have exactly one at every cardinality at or above that threshold?Logic, foundations & set theory

    Exact statement

    Prove or refute the strong categoricity transfer for L_omega1,omega: if an L_omega1,omega sentence is categorical in some cardinal at least the Hanf threshold delta_omega1, then it is categorical in every cardinal at least that threshold.

    Why it matters

    A proof would transfer uniqueness of models across all cardinalities in the stated range. A counterexample would show that uniqueness at one sufficiently large size need not propagate throughout that range.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The release reports an eventual categoricity threshold for abstract elementary classes of bounded Löwenheim–Skolem number, without prescribing its numerical value. A separate result under CH supplies a counterexample to a specified AEC threshold principle. Neither the required named threshold nor definability by one L_omega1,omega sentence is established for that counterexample. These results therefore neither prove nor refute the exact prescribed-threshold sentence formulation here; the CH conclusion is conditional nonprovability for a different AEC principle, not full independence of this target. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Shelah's Categoricity Conjecture for L_omega1,omega · 1 problem in this edition. Family membership alone does not assert an implication.

    • #331 · Shelah's Categoricity Conjecture for L_omega1,omega (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 320–466.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  332. Rank332Computability of the Shannon Capacity of Finite GraphsA graph's Shannon capacity measures how many messages can be distinguished without error when the graph describes which symbols can be confused. Can an algorithm approximate it to any requested accuracy for every nonempty finite simple graph, with guaranteed termination but no speed requirement?Cryptography, coding, information & optimization

    Exact statement

    Determine whether an algorithm exists which, given the exact adjacency matrix of any nonempty finite simple graph G and an integer t>=1, halts and returns a rational a with |a-Theta(G)|<=2^(-t), where Theta(G)=sup_{m>=1} alpha(G^{boxtimes m})^(1/m), alpha is independence number and boxtimes is the strong graph product. No running-time bound is imposed.

    Why it matters

    A positive answer would make the supremum over all strong graph powers computably accessible to arbitrary precision. A negative answer would rule out every total approximation algorithm, regardless of running time.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Computability of the Shannon Capacity of Finite Graphs · 1 problem in this edition. Family membership alone does not assert an implication.

    • #332 · Computability of the Shannon Capacity of Finite Graphs (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 337–468.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  333. Rank333Margulis Bounded Diagonal-Orbit ConjectureConsider unit-volume lattices in three dimensions, stretched independently along the coordinate directions while preserving volume. If a lattice's full orbit under these diagonal transformations is bounded, must it contain all its accumulation points?Probability, ergodic theory & dynamics

    Exact statement

    Let A be the diagonal subgroup of SL(3,R). Prove or disprove that every bounded A-orbit in the homogeneous space SL(3,R)/SL(3,Z) is closed.

    Why it matters

    A proof would turn boundedness into a strong orbit-closure restriction in this space of lattices. A counterexample would exhibit bounded diagonal motion whose orbit omits some of its limiting points.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Margulis Bounded Diagonal-Orbit Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #333 · Margulis Bounded Diagonal-Orbit Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 302–499.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  334. Rank334The Cherlin-Zilber Algebraicity Conjecture for Simple Groups of Finite Morley RankMust every infinite simple group with finite Morley rank—a model-theoretic notion of dimension—be isomorphic to the point group of a linear algebraic group over an algebraically closed field? Such algebraic groups are described by polynomial equations in matrix entries.Logic, foundations & set theory

    Exact statement

    Prove that every infinite simple group of finite Morley rank is isomorphic to the group of rational points of a linear algebraic group over an algebraically closed field.

    Why it matters

    A proof would identify all groups in this logical classification with algebraic groups of the stated kind. A counterexample would exhibit an infinite simple group whose finite Morley rank does not force that algebraic form.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    The Cherlin-Zilber Algebraicity Conjecture for Simple Groups of Finite Morley Rank · 1 problem in this edition. Family membership alone does not assert an implication.

    • #334 · The Cherlin-Zilber Algebraicity Conjecture for Simple Groups of Finite Morley Rank (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 331–472.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  335. Rank335Pugh–Shub Stable Ergodicity (C^r Density in Partial Hyperbolicity)On a compact connected manifold without boundary, can every volume-preserving partially hyperbolic map be approximated in every Cʳ topology, r≥2 including ∞, by one that remains ergodic under small C¹ perturbations through C² volume-preserving maps? Ergodic means there is no invariant region of intermediate volume.Probability, ergodic theory & dynamics

    Exact statement

    For every compact connected boundaryless smooth manifold M with a smooth probability volume m, and every r in [2,infinity] (including the C^infinity topology), is the following true? In the space PH_m^r(M) of volume-preserving C^r partially hyperbolic diffeomorphisms, maps that are C^1-stably ergodic among C^2 volume-preserving diffeomorphisms are C^r-dense. Thus every f in PH_m^r(M) and every relative C^r neighborhood U of f contain g such that some C^1 neighborhood V of g has every C^2 volume-preserving h in V ergodic for m. Use the usual C^r topology, with the Holder convention for nonintegral finite r. Partial hyperbolicity means a continuous Df-invariant dominated splitting TM=E^u+E^c+E^s: E^u and E^s are nontrivial, E^c may be zero, and for some common integer N>=1 the unstable unit vectors expand by >2, stable unit vectors contract by <1/2, and earlier nonzero bundles dominate later ones by a factor >2 under Df^N, pointwise at the same base point in that order. This is the ACW stability convention with the r>=2 range documented by RHRHU, not density among all conservative diffeomorphisms or only C^1-density.

    Why it matters

    A proof would guarantee robust ergodic behavior arbitrarily close in each stated smooth topology. A disproof would identify a regularity and a neighborhood within the partially hyperbolic class where that approximation fails.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Pugh–Shub Stable Ergodicity (C^r Density in Partial Hyperbolicity) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #335 · Pugh–Shub Stable Ergodicity (C^r Density in Partial Hyperbolicity) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 351–451.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  336. Rank336Hilbert's Third Problem for Spherical and Hyperbolic Scissors CongruenceIn spherical and hyperbolic three-dimensional geometry, can any two polyhedra with the same volume and Dehn invariant be cut into finitely many matching pieces? The Dehn invariant records a combination of edge lengths and dihedral angles.Geometry & topology

    Exact statement

    Prove or disprove Dehn invariant sufficiency in spherical and hyperbolic three-dimensional geometry: for X equal to H^3 and S^3, show whether volume is injective on the kernel of the Dehn invariant, equivalently whether volume together with Dehn invariant classifies polyhedra up to scissors congruence in those geometries.

    Why it matters

    A positive answer would make these two invariants sufficient to classify scissors congruence in the stated geometries. Failure would reveal distinctions between polyhedra that neither invariant detects.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Hilbert's Third Problem for Spherical and Hyperbolic Scissors Congruence · 1 problem in this edition. Family membership alone does not assert an implication.

    • #336 · Hilbert's Third Problem for Spherical and Hyperbolic Scissors Congruence (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 335–463.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  337. Rank337Gromov's Filling Area ConjectureMust every compact orientable Riemannian surface filling a circle of circumference 2π have area at least 2π if no route through the surface shortens any boundary distance?Geometry & topology

    Exact statement

    For every compact orientable Riemannian surface M whose boundary is isometric to a circle C of circumference 2π and satisfies d_M(x,y)=d_C(x,y) for all x,y in C, prove Area(M) is at least 2π.

    Why it matters

    A proof would impose a minimum area on fillings that preserve all distances around the boundary circle. A counterexample would achieve the same boundary distances using less area.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Gromov's Filling Area Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #337 · Gromov's Filling Area Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 342–452.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  338. Rank338Corona Problem for the Unit Ball in Several Complex VariablesOn a complex unit ball of dimension at least two, suppose finitely many bounded holomorphic functions never all become arbitrarily small together. Can bounded holomorphic multipliers always make the sum of their products exactly one?Analysis & PDE

    Exact statement

    For every n ≥ 2, if f_1,...,f_m are bounded holomorphic functions on the unit ball B^n in C^n and inf_{z∈B^n} sum_j |f_j(z)| > 0, prove or disprove that there exist bounded holomorphic g_1,...,g_m on B^n with sum_j f_j g_j = 1.

    Why it matters

    A proof would guarantee bounded holomorphic solutions to this multiplication-and-addition equation whenever the tuple stays uniformly away from simultaneous zero. A counterexample would expose an obstruction despite that lower bound.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Corona Problem for the Unit Ball in Several Complex Variables · 1 problem in this edition. Family membership alone does not assert an implication.

    • #338 · Corona Problem for the Unit Ball in Several Complex Variables (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 327–472.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  339. Rank339Conforti–Cornuéjols Conjecture: Packing Property Implies Max-Flow Min-CutDoes every finite clutter with the packing property also have the max-flow min-cut property? Both properties use the stated finite-optimum convention, so an infeasible covering program or an unbounded dual is not a counterexample.Combinatorics & discrete geometry

    Exact statement

    For every finite clutter H, prove or disprove that its packing property implies its max-flow min-cut property. Use the following precise linear-programming convention, with M the edge-by-vertex0/1 incidence matrix: the packing property says that min{w^T x:Mx>=1,x>=0}=max{1^T y:M^T y<=w,y>=0} has integral optimal solutions on both sides for every w in {0,1,+infinity}^V for which the optimum is finite. An infinite cost forces that primal coordinate to0 and omits the corresponding dual upper-bound constraint. The MFMC property requires integral optimal solutions on both sides for every finite nonnegative integral w. Equivalently, packing means the covering/edge-packing min-max equality holds for all minors with the usual feasible-case convention. The reverse implication MFMC implies packing is already known.

    Scope qualification. Apply the stated finite-optimum/feasible-case convention to both packing and MFMC. An infeasible covering LP or unbounded dual is not a failure of the required integral optima. In particular, a clutter containing an empty edge does not constitute a counterexample to this canonical conjecture.

    Also known as

    Conforti–Cornuéjols packing conjecture; Packing property implies MFMC for clutters

    Why it matters

    This would promote the packing condition to integral primal and dual optima for every finite nonnegative integral weight vector in the feasible, finite-optimum cases. The reverse implication is already known.

    Status

    Open · subcases solved

    October 6, 2026: The generalized packing-covering result concerns generalized Hamming weights and covering radii of linear codes, not the Conforti-Cornuéjols clutter MFMC conjecture. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Problem family

    Conforti–Cornuéjols Conjecture: Packing Property Implies Max-Flow Min-Cut · 1 problem in this edition. Family membership alone does not assert an implication.

    • #339 · Conforti–Cornuéjols Conjecture: Packing Property Implies Max-Flow Min-Cut (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 346–451.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  340. Rank340Yau's First Eigenvalue Conjecture for Minimal HypersurfacesA minimal hypersurface is stationary for area. For every closed, connected, embedded n-dimensional example in the unit sphere, is its lowest nonzero Laplace eigenvalue—the value governing its first nonconstant vibration mode—always n?Geometry & topologyOpen · disputed claim

    Exact statement

    Prove that every connected closed embedded minimal hypersurface M^n in the unit sphere S^(n+1) has first nonzero Laplace-Beltrami eigenvalue equal to n.

    Why it matters

    A proof would make this first spectral value depend only on dimension across the stated class. A counterexample would reveal an embedded minimal hypersurface with a different first eigenvalue.

    Status

    Open · disputed claim

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Yau's First Eigenvalue Conjecture for Minimal Hypersurfaces · 1 problem in this edition. Family membership alone does not assert an implication.

    • #340 · Yau's First Eigenvalue Conjecture for Minimal Hypersurfaces (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 338–470.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  341. Rank341Coleman-Oort conjecture: eventual special-subvariety exclusionIn sufficiently large genus, are there no positive-dimensional special families of abelian varieties whose general member is the Jacobian of a smooth complex curve? Here the special families are irreducible Shimura subvarieties; the claim concerns generic containment, not isolated points.Number theory & arithmetic geometry

    Exact statement

    There exists an integer g0 such that for every integer g>=g0 the complex Siegel moduli space A_g of principally polarized abelian varieties contains no positive-dimensional irreducible special (Shimura) subvariety S generically contained in the smooth-Jacobian Torelli locus T_g^o. Precisely, if T_g is the Zariski closure of the image of the moduli of smooth genus-g complex curves under the Torelli map, the prohibited S satisfies S subset T_g and S intersection T_g^o nonempty, hence open dense in S. No fixed, optimal or effective threshold is prescribed. The statement excludes neither isolated CM points nor special subvarieties wholly in the decomposable boundary.

    Why it matters

    A proof would exclude these special families in all sufficiently large genera without requiring an explicit threshold. A disproof would require such families in unbounded genera; isolated CM points and families wholly in the decomposable boundary are outside this exclusion.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Coleman-Oort conjecture: eventual special-subvariety exclusion · 1 problem in this edition. Family membership alone does not assert an implication.

    • #341 · Coleman-Oort conjecture: eventual special-subvariety exclusion (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 351–457.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  342. Rank342Blackadar–Kirchberg Quasidiagonality Problem for Stably Finite Nuclear C*-AlgebrasIs every separable nuclear stably finite complex C*-algebra quasidiagonal, including nonunital algebras and without adding simplicity, UCT or a faithful-trace hypothesis?Algebra, representation & category theory

    Exact statement

    Prove or disprove that every separable nuclear stably finite complex C*-algebra A, unital or nonunital, is quasidiagonal. Here quasidiagonality means that there are completely positive contractive maps phi_n:A to M_{k(n)}(C) with ||phi_n(ab)-phi_n(a)phi_n(b)|| tending to zero and ||phi_n(a)|| tending to ||a|| for every a,b in A. No simplicity, UCT, faithful-trace or regularity hypothesis is added. Stable finiteness has its usual C*-algebra meaning; existence of a faithful trace is not part of the target.

    Scope qualification. The Blackadar-Kirchberg target asserts a sufficient condition: every separable nuclear stably finite complex C*-algebra is quasidiagonal. Nuclearity and stable finiteness remain hypotheses on A; the target does not assert that quasidiagonality implies nuclearity or gives a biconditional characterization over all separable C*-algebras. Both unital and nonunital algebras are included, with no added simplicity, UCT or faithful-trace assumption.

    Also known as

    Blackadar–Kirchberg quasidiagonality conjecture; Nuclear stable finiteness implies quasidiagonality; Blackadar–Kirchberg NF problem

    Why it matters

    This would provide the specified finite-matrix approximations throughout that class of algebras. It asserts a sufficient condition for quasidiagonality, not a converse or a characterization of all separable C*-algebras.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Blackadar–Kirchberg Quasidiagonality Problem for Stably Finite Nuclear C*-Algebras · 1 problem in this edition. Family membership alone does not assert an implication.

    • #342 · Blackadar–Kirchberg Quasidiagonality Problem for Stably Finite Nuclear C*-Algebras (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 350–457.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  343. Rank343Lebesgue measure of the boundary of the Mandelbrot setDoes the boundary of the Mandelbrot set, defined by bounded orbits of zero under z↦z²+c, occupy zero area or positive area in the complex plane?Probability, ergodic theory & dynamicsMandelbrot set boundary · 2 problems

    Exact statement

    Let M={c in C: the forward orbit of 0 under z maps to z2+c is bounded}. Does its boundary have two-dimensional Lebesgue measure zero, or positive two-dimensional Lebesgue measure?

    Why it matters

    The answer would determine whether the transition boundary between bounded and unbounded critical orbits carries positive planar area. A zero-area result would place that boundary in a planar null set.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Mandelbrot set boundary · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 337–485.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  344. Rank344Global Regularity of the 3D Incompressible Hall-MHD Equations for Large DataFor the three-dimensional incompressible Hall magnetohydrodynamics equations, which couple conducting-fluid motion to a magnetic field with the Hall effect, can smooth solutions always continue smoothly for all time, even when the initial data are large?Analysis & PDE

    Exact statement

    Prove or disprove that smooth solutions to the three-dimensional incompressible Hall-magnetohydrodynamics system exist globally in time for large (non-small) initial data. Known results provide local well-posedness for large data in critical spaces together with continuation criteria, and global well-posedness only under smallness assumptions on the initial data.

    Why it matters

    A proof would exclude finite-time breakdown throughout the specified large-data regime. A negative answer would reveal admissible initial data for which the asserted global smooth evolution fails.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Global Regularity of the 3D Incompressible Hall-MHD Equations for Large Data · 1 problem in this edition. Family membership alone does not assert an implication.

    • #344 · Global Regularity of the 3D Incompressible Hall-MHD Equations for Large Data (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 303–536.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  345. Rank345Arnold-Givental Lagrangian intersection conjectureTake the nonempty real part L of a compact symplectic manifold. Whenever a Hamiltonian motion makes L meet its image transversely, must there be at least as many intersection points as the sum of L's mod-two Betti numbers, which count its independent homological features?Geometry & topology

    Exact statement

    For a compact real symplectic manifold (M, omega, tau) with compact nonempty real part L = Fix(tau), and for every Hamiltonian diffeomorphism phi such that L and phi(L) meet transversely, prove that the number of intersection points is at least the total Z/2 Betti number of L.

    Why it matters

    A proof would force intersection points from the topology of the real part. A counterexample would show that the proposed topological count can exceed the intersections surviving Hamiltonian motion.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Arnold-Givental Lagrangian intersection conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #345 · Arnold-Givental Lagrangian intersection conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 333–476.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  346. Rank346Green's conjecture on syzygies of canonical curvesCan one invariant of a curve, its Clifford index, predict exactly how far the linear relations in its canonical equations continue? Green's conjecture proposes a sharp cutoff for every smooth non-hyperelliptic projective curve.Algebra, representation & category theory

    Exact statement

    For every smooth non-hyperelliptic projective curve C, prove Green's canonical syzygy conjecture: the vanishing pattern of K_{p,1}(C,K_C) is determined by the Clifford index of C, equivalently K_{l,1}(C,K_C)=0 for all l>=p if and only if Cliff(C)>g-p-2.

    Why it matters

    A proof would link the curve’s special linear systems to the linear part of its canonical syzygies. A counterexample would exhibit a curve whose algebraic relations depart from that prediction.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Green's conjecture on syzygies of canonical curves · 1 problem in this edition. Family membership alone does not assert an implication.

    • #346 · Green's conjecture on syzygies of canonical curves (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 329–487.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  347. Rank347Thomson problemFor every positive integer N, which configurations of N identical point charges constrained to the unit sphere minimize their total Coulomb repulsion energy?Mathematical physics

    Exact statement

    For each positive integer N, determine the configuration of N equal point charges on the unit sphere that minimizes total Coulomb energy.

    Why it matters

    A complete answer would identify globally optimal charge arrangements for every N and prove their optimality. Finding stable or locally minimizing arrangements alone would not establish the required energy minimum.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Thomson problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #347 · Thomson problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 356–470.

    Source-fit opponent count: 36.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  348. Rank348Brown-Erdos-Sos Conjecture on Sparse Triple SystemsTake distinct triples from n points, with their number at least a fixed positive multiple of n². For every positive choice of that multiple and every fixed k≥3, must a sufficiently large collection contain k triples that together use at most k+3 points?Combinatorics & discrete geometrySparse triple systems · 2 problems

    Exact statement

    For every fixed integer k>=3 and every epsilon>0, does there exist N=N(k,epsilon) such that every 3-uniform hypergraph on n>=N vertices with at least epsilon n^2 edges contains k distinct edges whose union has at most k+3 vertices? Equivalently, for each fixed k, the maximum size of an n-vertex 3-graph avoiding such a configuration is o(n^2). The target is the full (k+3,k) conjecture, not the separate (k+2,k) asymptotic-constant problem.

    Why it matters

    A proof would force this small configuration whenever the edge count retains any positive quadratic fraction. Equivalently, avoiding it would require the edge count divided by n² to tend to zero.

    Status

    Open · subcases solved

    October 6, 2026: This paper studies the (k+2,k) quadratic-density extremal problem. The ranked Brown-Erdos-Sos target is the separate (k+3,k) o(n^2) problem. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Problem family

    Sparse triple systems · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 365–460.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  349. Rank349Li-Li Undirected Multiple-Unicast Network Coding ConjectureIn a finite undirected network carrying independent source-to-destination messages, can encoding and mixing messages ever improve the rates achieved by splitting traffic among routes? The conjecture says no for zero-error transmission, when an edge's capacity is shared between its two directions.Cryptography, coding, information & optimization

    Exact statement

    For every finite undirected edge-capacitated network and every finite collection of independent source-destination unicast sessions, prove or disprove that the zero-error network-coding capacity region equals the fractional multicommodity-routing capacity region. Each undirected edge has a shared capacity bounding the sum of transmission rates in its two directions; intermediate nodes may causally encode received messages in the coding model. Fractional routing may split each session over paths. The target uses the standard zero-error formulation, not multicast or directed networks.

    Why it matters

    A proof would show that causal message encoding provides no rate advantage over splitting traffic among paths in this model. A counterexample would exhibit a zero-error rate vector attainable by coding but not routing.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Li-Li Undirected Multiple-Unicast Network Coding Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #349 · Li-Li Undirected Multiple-Unicast Network Coding Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 364–462.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  350. Rank350Matroid Secretary ConjectureItems arrive in random order, revealing their nonnegative weights only as they arrive; each acceptance is final. Under any known matroid's rules for allowable sets, can one algorithm obtain, in expectation, a fixed positive fraction of the best total weight possible with full advance knowledge?Theoretical computer science

    Exact statement

    Determine whether there exist an absolute constant c>0 and an online algorithm A such that, for every finite known matroid M=(E,I) and every adversarially fixed nonnegative weight function w, A receives the elements in uniformly random order, learns an element's weight only when it arrives, makes irrevocable acceptance decisions, always maintains an independent accepted set ALG, and satisfies E[w(ALG)] >= c max_{S in I} w(S). The expectation is over the arrival order and any internal randomness of A; A may use the known matroid but not unseen weights.

    Why it matters

    A proof would guarantee a constant share of the best total weight despite decisions preceding knowledge of later weights. The constant must work uniformly across all finite known matroids and all permitted weight assignments.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Matroid Secretary Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #350 · Matroid Secretary Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 355–474.

    Source-fit opponent count: 34.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  351. Rank351Fermat-Catalan Conjecture on Finiteness of Coprime Power TriplesAre there only finitely many pairwise coprime positive-integer triples (x,y,z) satisfying x^p+y^q=z^r for some integers p,q,r greater than one with 1/p+1/q+1/r less than one?Number theory & arithmetic geometryEquations in perfect powers · 2 problems

    Exact statement

    Prove that there are only finitely many pairwise coprime positive-integer triples (x,y,z) for which x^p + y^q = z^r for integers p,q,r > 1 satisfying 1/p + 1/q + 1/r < 1.

    Why it matters

    A proof would establish finiteness across all permitted exponent choices together. It would show that allowing the exponents to vary cannot produce infinitely many distinct primitive base triples.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Equations in perfect powers · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 350–482.

    Source-fit opponent count: 35.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  352. Rank352Bartnik's cosmological splitting conjectureMust every connected, time-oriented smooth spacetime of dimension at least two, with no boundary, be a static product (R × S, −dt² + h) if it is globally hyperbolic, has a compact Cauchy hypersurface, is timelike geodesically complete in both time directions, and has nonnegative Ricci curvature on timelike vectors? The spatial factor S would be compact.Mathematical physicsCosmological splitting · 1 problem

    Exact statement

    For every connected, time-oriented smooth Lorentzian manifold (M,g) of dimension n >= 2, without boundary, assume global hyperbolicity with a compact Cauchy hypersurface, timelike geodesic completeness in both time directions, and Ric_g(v,v) >= 0 for every timelike tangent vector v. Must (M,g) be globally isometric to (R x S, -dt^2 + h), where (S,h) is a compact Riemannian (n-1)-manifold? Use signature (-,+,...,+); the timelike convergence condition is equivalently nonnegative on all causal vectors by continuity.

    Why it matters

    A proof would force an unchanging spatial geometry from these global causal, completeness and curvature assumptions. A counterexample would exhibit a nonstatic spacetime satisfying all of them.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Cosmological splitting · 1 problem in this edition. Family membership alone does not assert an implication.

    • #352 · Bartnik's cosmological splitting conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 373–471.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  353. Rank353Chen-Yang Volume Conjecture for Turaev-Viro InvariantsDoes the exponential growth of Turaev–Viro invariants recover the volume of every finite-volume hyperbolic three-manifold in the closed, cusped, or totally geodesic boundary classes? Use the Chen–Yang normalization, at odd levels r and root exp(2πi/r).Geometry & topology

    Exact statement

    For every finite-volume hyperbolic three-manifold M in the scope of Chen-Yang Conjecture 1.1—closed, cusped, or with totally geodesic boundary—prove or disprove that lim_{r -> infinity, r odd} (2*pi/r)*log(TV_r(M;exp(2*pi*i/r))) = Vol(M). TV is the scalar Turaev-Viro state-sum invariant in their convention, using the specified extension for nonempty boundary and a compact core for cusps. The different standard root exp(pi*i/r), and the separate complex-valued Reshetikhin-Turaev/Chern-Simons refinement, are not part of this target.

    Why it matters

    A proof would recover hyperbolic volume from the exponential growth of this scalar state-sum invariant. It would apply with the specified boundary extension and cusp compact-core conventions.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Chen-Yang Volume Conjecture for Turaev-Viro Invariants · 1 problem in this edition. Family membership alone does not assert an implication.

    • #353 · Chen-Yang Volume Conjecture for Turaev-Viro Invariants (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 379–465.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  354. Rank354Smallest number of variables for undecidable Hilbert's Tenth Problem over ZWhat is the smallest number n of variables for which no algorithm can decide whether an arbitrary integer-coefficient polynomial in at most n variables has an integer solution?Logic, foundations & set theory

    Exact statement

    Determine the least integer n for which no algorithm decides whether an arbitrary integer-coefficient polynomial in at most n variables has an integer zero.

    Why it matters

    A complete answer would locate the precise variable-count boundary for undecidability. It requires an impossibility proof at that count together with decision procedures for all smaller counts.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Smallest number of variables for undecidable Hilbert's Tenth Problem over Z · 1 problem in this edition. Family membership alone does not assert an implication.

    • #354 · Smallest number of variables for undecidable Hilbert's Tenth Problem over Z (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 325–511.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  355. Rank355Whitehead Asphericity ConjectureA space is aspherical if every sphere of dimension two or higher mapped into it can be continuously contracted. If a two-dimensional cell complex is aspherical, must every connected subcomplex be aspherical too?Geometry & topologyOpen · disputed claim

    Exact statement

    Decide whether every connected subcomplex of an aspherical two-dimensional CW-complex is aspherical, equivalently whether every subpresentation of an aspherical presentation is aspherical.

    Why it matters

    A proof would show that asphericity survives passage to connected subcomplexes in dimension two. A counterexample would reveal higher homotopy in a subcomplex despite its absence from the surrounding complex.

    Status

    Open · disputed claim

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Whitehead Asphericity Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #355 · Whitehead Asphericity Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 342–500.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  356. Rank356Clemens Conjecture for Rational Curves on a General Quintic ThreefoldFor a general smooth degree-five hypersurface in complex projective four-space, are there only finitely many smooth rational curves of each fixed degree, with each unable to move even infinitesimally within the hypersurface?Geometry & topology

    Exact statement

    Prove that a general smooth quintic threefold in complex projective four-space contains only finitely many smooth rational curves of each fixed positive degree, with the expected normal-bundle rigidity.

    Why it matters

    A proof would make these curves finite and rigid degree by degree. Failure would identify either an infinite collection at a fixed degree or a curve violating the specified rigidity requirement.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Clemens Conjecture for Rational Curves on a General Quintic Threefold · 1 problem in this edition. Family membership alone does not assert an implication.

    • #356 · Clemens Conjecture for Rational Curves on a General Quintic Threefold (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 372–483.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  357. Rank357Gauss's Real Quadratic Class Number One ConjectureAre there infinitely many real quadratic fields Q(√d), with squarefree d>1, whose ordinary ideal class number is one?Number theory & arithmetic geometry

    Exact statement

    Resolve whether infinitely many squarefree integers d>1 have ordinary ideal class number h(Q(sqrt(d)))=|Cl(O_d)|=1, where O_d is the full ring of integers: Z[(1+sqrt(d))/2] for d=1 mod 4 and Z[sqrt(d)] for d=2 or 3 mod 4, and Cl(O_d) is the group of nonzero fractional ideals modulo all nonzero principal fractional ideals.

    Also known as

    Real quadratic class-number-one infinitude; Infinitely many real quadratic number fields of class number one; Gauss real quadratic class-number-one conjecture

    Why it matters

    Class number one means every nonzero ideal class is principal in the full ring of integers. The conjecture asks whether this arithmetic simplicity occurs for infinitely many real quadratic fields.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Gauss's Real Quadratic Class Number One Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #357 · Gauss's Real Quadratic Class Number One Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 341–525.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  358. Rank358Minkowski's conjecture on products of non-homogeneous linear formsIn every dimension n, does every translated lattice with unit-volume fundamental cells contain a point whose absolute coordinates have product at most 2⁻ⁿ?Number theory & arithmetic geometry

    Exact statement

    For every dimension n>=1, every real nonsingular n by n matrix A with |det A|=1 and every translation x in R^n, there exists an integer vector m in Z^n such that the product over i=1,...,n of |x_i+(Am)_i| is at most 2^(-n). The quantifier requires an ACTUAL integer witness at the closed bound, not merely an infimum no greater than the bound. Equivalently, scaling a general real nonsingular matrix B and its translation by |det B|^(-1/n) gives the same actual-witness assertion with bound |det B|/2^n; all coefficients and translations are real. No equality characterization of extremal lattices, well-rounded Euclidean covering assertion, positive-characteristic analogue or constructive algorithm is appended. Preserve all dimensions, not merely a selected high-dimensional open remainder.

    Why it matters

    The result would guarantee a point attaining the closed product bound for every such translated lattice. Merely approaching the bound without a qualifying point would not supply the required witness.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Minkowski's conjecture on products of non-homogeneous linear forms · 1 problem in this edition. Family membership alone does not assert an implication.

    • #358 · Minkowski's conjecture on products of non-homogeneous linear forms (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 387–477.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  359. Rank359Zilber's Exponential-Algebraic Closedness ConjectureTreat complex variables and their exponentials as independent coordinates, producing an algebraic variety. If it is irreducible, free and rotund—conditions on rational linear and multiplicative constraints and on dimensions after quotients—must it contain a genuine point (z, exp(z))?Logic, foundations & set theoryThe complex exponential field · 2 problems

    Exact statement

    For every positive integer n and irreducible complex algebraic subvariety V of C^n x (C*)^n which is free and rotund, decide whether V contains (z,exp(z)) for some z in C^n. Free means that neither projection is contained in a translate of a proper Q-defined linear subspace or a proper algebraic subtorus, respectively. Rotund means dim(pi_L(V)) >= n-dim(L) for every Q-defined complex linear subspace L of C^n, where pi_L is the quotient to C^n/L x (C*)^n/exp(L). This is ordinary exponential-algebraic closedness, without an extra generic-transcendence-degree requirement.

    Why it matters

    The result would decide whether these algebraic freedom and dimension conditions force a solution to the corresponding polynomial-exponential equations. The target asks for existence without an additional generic-transcendence-degree requirement.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    The complex exponential field · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 383–477.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  360. Rank360Polynomial-Time Solvability of Condon's Simple Stochastic GamesIn a two-player game with turns and random moves, one player tries to reach a target while the other tries to prevent it. Can the optimal probability of reaching the target be computed in polynomial time for Condon's finite simple stochastic games, with binary choices and rational probabilities?Theoretical computer sciencePolynomial-time algorithms for finite games · 3 problems

    Exact statement

    Determine whether Condon's simple stochastic games can be solved in polynomial time: given a two-player zero-sum turn-based stochastic game with binary decisions, rational transition probabilities, and a reachability objective, compute the game value, equivalently deciding whether the start-vertex value reaches a given rational threshold.

    Why it matters

    An algorithm would efficiently compute the optimal reachability value and answer rational-threshold questions for every game in scope. An impossibility proof would place that general task beyond polynomial time in the stated model.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. A reachability game can use an absorbing success state of reward one and reward zero elsewhere, making pathwise liminf average reward the indicator of eventual success. Shifting by the rational comparison threshold and clearing denominators reduces the full threshold decision to the publisher’s release's nonnegative expected liminf problem with binary signed integer rewards and rational chance probabilities. The resulting guarantee is quasipolynomial, not the full polynomial-time bound; it does not close the target. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Related problems

    • #72 · Polynomial-Time Solvability of Mean-Payoff GamesConditional implicationPolynomial-Time Solvability of Mean-Payoff Games

      A proof of this assertion would also establish #72.

      Condition: An affirmative simple-stochastic-games-in-P theorem implies deterministic-mean-payoff-in-P. Its contrapositive transfers a negative mean-payoff answer upward; a negative simple-stochastic-games answer need not resolve deterministic mean payoff.

      Scope qualification: An affirmative simple-stochastic-games-in-P theorem implies deterministic-mean-payoff-in-P. Its contrapositive transfers a negative mean-payoff answer upward; a negative simple-stochastic-games answer need not resolve deterministic mean payoff.

    Problem family

    Polynomial-time algorithms for finite games · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 333–530.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  361. Rank361GNRS ConjectureFor graph families closed under deleting vertices or edges and contracting edges, and excluding at least one graph, do network cuts estimate simultaneous routing capacity within a fixed factor? The factor may depend on the family, but not on network size or traffic demands.Theoretical computer science

    Exact statement

    Prove or disprove that every nontrivial minor-closed family of finite undirected graphs has uniformly bounded multi-commodity flow-cut gap, equivalently uniformly bounded L1 distortion for its shortest-path metrics, with the constant depending only on the excluded minor family.

    Why it matters

    A positive answer would give uniform control of both flow-cut gaps and distortion when these graph distances are represented in L1. A counterexample would show that excluding a fixed minor does not ensure such control.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The full GNRS target quantifies over every nontrivial minor-closed graph family. The two manuscripts cover the planar family and each family of bounded treewidth, for finite connected graphs with arbitrary positive edge lengths, with distortion constants uniform in graph size and dependent only on the permitted family parameter. These are genuine GNRS subcases; arbitrary excluded-minor families are not covered merely by planar and bounded-treewidth cases. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    GNRS Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #361 · GNRS Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 360–510.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  362. Rank362Coding Theory and Combinatorics Problem CC-6: Existence of Finite Projective Planes of Non-Prime-Power OrdersIn a finite projective plane, any two points determine a line and any two lines meet. Can one exist with 13 points on each line, meaning order 12, or at another composite non-prime-power order not ruled out by the Bruck–Ryser–Chowla theorem?Cryptography, coding, information & optimizationFinite projective planes · 2 problems

    Exact statement

    Does there exist any finite projective plane of order 12, or of any other composite non-prime-power order not ruled out by the Bruck-Ryser-Chowla theorem?

    Why it matters

    One construction at an admissible order would settle the existence question affirmatively. A negative resolution would need to rule out every order in the stated class, not only order 12.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Finite projective planes · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 381–491.

    Source-fit opponent count: 34.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  363. Rank363Boundedness of Compact Hyperkahler Manifolds: Finiteness of Deformation Types in Each DimensionIn each fixed even complex dimension, are there only finitely many deformation types of compact irreducible holomorphic symplectic manifolds? Deformations may pass through connected smooth proper holomorphic families; no projectivity, polarization or fibration assumption is imposed.Geometry & topologyHyperkähler fibrations and deformation types · 1 problem

    Exact statement

    For each integer n>=1, are there only finitely many deformation-equivalence classes of compact irreducible holomorphic symplectic complex manifolds of complex dimension 2n? Such a manifold is smooth, simply connected and compact Kähler, with H^0(X,Omega_X^2) spanned by a holomorphic two-form nondegenerate everywhere. Deformation equivalence is the equivalence generated by occurrence as fibers in smooth proper holomorphic families of such manifolds over connected bases. No projectivity, polarization, lattice, second-Betti-number or fibration restriction is imposed. Resolve by proving finiteness in every fixed dimension or showing infinitely many deformation classes in one fixed dimension.

    Why it matters

    Finiteness would place a basic limit on the variety of these manifolds in any one dimension. It would establish that only finitely many deformation classes occur, without requiring an explicit classification of all of them.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Related problems

    • #366 · Hyperkähler SYZ (Lagrangian Fibration) ConjectureSource-reported proved · historical rankRelated problemHyperkähler SYZ (Lagrangian Fibration) Conjecture

      Hyperkähler SYZ, together with the recorded fibration theorem, would give deformation finiteness in the scope where the second Betti number is at least five.

      Scope qualification: This conditional consequence covers only part of the finiteness target. No unconditional implication between the two complete conjectures is established here.

    Problem family

    Hyperkähler fibrations and deformation types · 1 problem in this edition. Family membership alone does not assert an implication.

    • #363 · Boundedness of Compact Hyperkahler Manifolds: Finiteness of Deformation Types in Each Dimension (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 354–503.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  364. Rank364Pólya's Conjecture for Dirichlet EigenvaluesA region's Dirichlet eigenvalues describe its vibration modes with the boundary held fixed. For every bounded region in dimension d≥2, must its kth eigenvalue be at least (2π)²(k/(ω_dV))^(2/d), where V is the region's volume and ω_d is the unit-ball volume?Analysis & PDE

    Exact statement

    For every bounded open domain Ω⊂R^d of finite volume, d≥2, prove that every Dirichlet-Laplacian eigenvalue satisfies λ_k(Ω)≥(2π)^2(k/(ω_d|Ω|))^{2/d} for all k≥1, where ω_d is the volume of the unit d-ball.

    Why it matters

    A proof would turn this volume-based spectral scale into a bound for every eigenvalue, including the lowest ones. A counterexample would show that some domain and index fall below it.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Pólya's Conjecture for Dirichlet Eigenvalues · 1 problem in this edition. Family membership alone does not assert an implication.

    • #364 · Pólya's Conjecture for Dirichlet Eigenvalues (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 358–506.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  365. Rank365Finiteness of central configurations for the n-body problem, n >= 4For n≥4 gravitating bodies with any fixed positive masses, are there only finitely many central-configuration shapes—arrangements in which every acceleration is proportional to position relative to the center of mass, with the same proportionality factor?Mathematical physics

    Exact statement

    For the n-body problem, n >= 4, show that the number of central configurations is finite for all choices of masses m_i > 0, or find a counterexample.

    Why it matters

    A proof would establish finiteness for each number of bodies n≥4 and each fixed choice of positive masses. A counterexample would identify masses for which infinitely many central configurations occur.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Finiteness of central configurations for the n-body problem, n >= 4 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #365 · Finiteness of central configurations for the n-body problem, n >= 4 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 376–489.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  366. Rank366Turan Conjecture on the Turan Density of K_4^{(3)}Is 5/9 the limiting maximum proportion of triples that can be chosen from a large vertex set without choosing all four triples on any four vertices?Combinatorics & discrete geometry

    Exact statement

    Prove or refute that the Turan density of the complete 3-uniform hypergraph K_4^(3) on four vertices is 5/9.

    Why it matters

    A proof would determine the asymptotic edge-density limit for this forbidden three-uniform hypergraph. Refutation would show that the proposed threshold does not describe the largest possible density.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Turan Conjecture on the Turan Density of K_4^{(3)} · 1 problem in this edition. Family membership alone does not assert an implication.

    • #366 · Turan Conjecture on the Turan Density of K_4^{(3)} (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 350–537.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  367. Rank367Fourier Entropy-Influence ConjectureFor a function of binary inputs with a binary output, does its average sensitivity to flipping input bits control how widely its Fourier spectrum is spread? The conjecture asks for one universal bound on spectral entropy as a multiple of total influence.Theoretical computer science

    Exact statement

    Prove or refute the Friedgut-Kalai conjecture that there exists a universal constant C such that for every Boolean function f: {-1,1}^n -> {-1,1}, the spectral entropy of f (the Shannon entropy of the distribution given by its squared Fourier coefficients) is at most C times its total influence I(f).

    Why it matters

    A proof would bound spectral spread by sensitivity to changing input bits. A counterexample family would separate these quantities by making their ratio arbitrarily large as the functions vary.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Fourier Entropy-Influence Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #367 · Fourier Entropy-Influence Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 373–501.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  368. Rank368Realizable Limit Shapes in Two-Dimensional First-Passage PercolationOn the square lattice, assign independent edge-crossing times with a common nonnegative Borel probability law. Assume zero-time edges have probability below the square lattice's percolation threshold and the minimum of four independent times has finite second moment. Exactly which compact convex sets arise as deterministic large-scale growth shapes as this law varies?Probability, ergodic theory & dynamics

    Exact statement

    In nearest-neighbour first-passage percolation on Z^2, let edge weights be i.i.d. with Borel probability law nu on [0,infinity), nu({0}) < p_c(2), and E[min(t_1,t_2,t_3,t_4)^2] < infinity for four independent weights of law nu. Characterize exactly the compact convex sets B_nu that occur as deterministic limit shapes as nu varies over this class.

    Why it matters

    A necessary-and-sufficient characterization would identify the large-scale growth geometries obtainable from independent, identically distributed passage times under these assumptions. It would distinguish realizable shapes from those the model cannot produce.

    Status

    Open

    Open. The cited survey distinguishes the unresolved i.i.d. shape classification from the solved stationary-ergodic analogue; no exact-target closure was located in the recency review.

    Status review date not recorded in this edition

    Sources

    Problem family

    Realizable Limit Shapes in Two-Dimensional First-Passage Percolation · 1 problem in this edition. Family membership alone does not assert an implication.

    • #368 · Realizable Limit Shapes in Two-Dimensional First-Passage Percolation (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 393–488.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  369. Rank369Martin's Conjecture for Definable Turing-Invariant FunctionsAssuming the Axiom of Determinacy, must every Turing-invariant transformation, above some baseline computational degree, either have constant computational strength or preserve or increase the input's strength? Are the increasing transformations then ordered by iterations of the Turing jump?Logic, foundations & set theory

    Exact statement

    Under the Axiom of Determinacy, classify Turing-invariant functions on a cone by proving that each is constant or increasing and that the increasing functions are ordered by iterates of the Turing jump, or refute one of these conclusions.

    Why it matters

    A proof would classify these transformations of computational degree using constants and the jump hierarchy. A counterexample would identify cone behavior outside one of the two proposed classification clauses.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Martin's Conjecture for Definable Turing-Invariant Functions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #369 · Martin's Conjecture for Definable Turing-Invariant Functions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 362–513.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  370. Rank370Frankl Union-Closed Sets ConjectureIn every finite family containing a nonempty set and closed under taking unions, must some element belong to at least half of the sets?Combinatorics & discrete geometry

    Exact statement

    Prove that in every finite union-closed family of sets other than the empty set, there exists an element that belongs to at least half of the members of the family.

    Why it matters

    A proof would force one element to occur frequently solely from closure under unions. A counterexample would show that every element can remain below half despite that closure property.

    Status

    Open

    October 6, 2026: The claimed union-closed frequency lower bound 0.3828852549667978 improves prior bounds but remains below the conjectured 1/2. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Frankl Union-Closed Sets Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #370 · Frankl Union-Closed Sets Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 371–518.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  371. Rank371Polynomial-Time P-Matrix Linear Complementarity ProblemA P-matrix guarantees a unique solution to a linear complementarity problem, which combines linear equations, nonnegativity, and conditions preventing paired quantities from both being positive. Can that solution always be found in polynomial time for rational inputs?Theoretical computer science

    Exact statement

    Determine whether the unique solution of every linear complementarity problem with a rational P-matrix and rational right-hand side can be computed in time polynomial in the bit length of the input.

    Why it matters

    A positive answer would turn guaranteed uniqueness into an efficient exact solution method for this full input class. A negative answer would show that uniqueness does not ensure polynomial-time solvability here.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Polynomial-Time P-Matrix Linear Complementarity Problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #371 · Polynomial-Time P-Matrix Linear Complementarity Problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 379–501.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  372. Rank372Balanced L4 boundedness of the trilinear Hilbert transformThe trilinear Hilbert transform combines three functions sampled at x+t, x+2t, and x+3t through a singular integral. Is its output's L⁴ᐟ³ norm always bounded by one constant times the product of the three L⁴ input norms, for smooth rapidly decaying inputs?Analysis & PDE

    Exact statement

    For Schwartz functions f,g,h:R->C, let T(f,g,h)(x)=p.v. integral_R f(x+t)g(x+2t)h(x+3t) dt/t. Is there a finite constant C such that ||T(f,g,h)||_{L^(4/3)(R)} <= C ||f||_{L^4(R)}||g||_{L^4(R)}||h||_{L^4(R)} for every such triple?

    Why it matters

    A proof would control this singular integral uniformly over all triples, for its fixed shifts x+t, x+2t, and x+3t. Refutation would give triples whose normalized output size exceeds every constant.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The manuscript reports L3 × L3 × L3 → L1 boundedness for complex Schwartz inputs. The target here instead asks for L4 × L4 × L4 → L(4/3). A further localized four-linear estimate is stated, but the published theorem statements do not give the representation or interpolation result transferring it to the exact principal-value operator at the required exponents. The L3 estimate is partial progress; the full L4 target remains open. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Balanced L4 boundedness of the trilinear Hilbert transform · 1 problem in this edition. Family membership alone does not assert an implication.

    • #372 · Balanced L4 boundedness of the trilinear Hilbert transform (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 382–498.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  373. Rank373Rationality Implies C2-Cofiniteness for Vertex Operator AlgebrasMust a complex vertex operator algebra of CFT type have finite-dimensional V/C2(V) whenever every admissible module is a direct sum of simple admissible modules? No unitarity or self-contragredience assumption is added.Representation theory

    Exact statement

    For every complex vertex operator algebra V of CFT type, V=direct_sum_(n>=0) V_n with finite-dimensional graded pieces and V_0=C1, assume rationality in the sense that every admissible (N-gradable weak) V-module is a direct sum of simple admissible modules. Must dim_C(V/C2(V)) be finite, where C2(V)=span_C{u_(-2)v:u,v in V}? Prove the implication for all such V, or construct a rational CFT-type VOA with infinite-dimensional C2 quotient. No self-contragredient, unitarity or pre-existing C2-cofiniteness hypothesis is imposed. Semisimplicity restricted to ordinary modules alone is not the rationality hypothesis.

    Also known as

    Zhu rationality implies C2-cofiniteness conjecture

    Why it matters

    This asks whether the stated representation-theoretic rationality forces the separate finiteness condition C2-cofiniteness. Semisimplicity of ordinary modules alone is a weaker hypothesis and is not the question here.

    Status

    Open

    Status reviewed

    Sources

    Problem family

    Rationality Implies C2-Cofiniteness for Vertex Operator Algebras · 1 problem in this edition. Family membership alone does not assert an implication.

    • #373 · Rationality Implies C2-Cofiniteness for Vertex Operator Algebras (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 392–507.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  374. Rank374Colmezs Conjecture on Faltings Heights of CM Abelian VarietiesDoes Colmez's formula give the arithmetic height of every abelian variety with complex multiplication? It predicts the stable Faltings height, a measure of arithmetic complexity, from logarithmic derivatives of L-functions determined by the variety's complex multiplication.Number theory & arithmetic geometry

    Exact statement

    Prove for every CM abelian variety that its stable Faltings height is the Colmez linear combination of logarithmic derivatives at s=0 of the Artin L-functions determined by its CM type.

    Why it matters

    A proof would express this arithmetic height through analytic values determined by the CM type. A counterexample would expose a failure of that predicted identity for a particular type.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Colmezs Conjecture on Faltings Heights of CM Abelian Varieties · 1 problem in this edition. Family membership alone does not assert an implication.

    • #374 · Colmezs Conjecture on Faltings Heights of CM Abelian Varieties (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 387–505.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  375. Rank375Greenberg's Conjecture on Vanishing of Iwasawa Invariants of Totally Real FieldsClass numbers measure the failure of unique factorization in number fields. For every totally real field and prime p, does the power of p dividing these class numbers stay bounded along the cyclotomic ℤₚ tower of fields?Number theory & arithmetic geometryIwasawa invariants · 2 problems

    Exact statement

    Prove that for every totally real number field F and every prime p, the Iwasawa invariants lambda_p(F) and mu_p(F) of the cyclotomic Z_p-extension vanish; equivalently, the p-parts of class numbers in the tower remain bounded.

    Why it matters

    A proof would force both associated Iwasawa invariants, λ and μ, to vanish. A counterexample would reveal unbounded p-primary class-number growth for some totally real field and prime.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Iwasawa invariants · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 384–522.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  376. Rank376Uniqueness of Tangent Cones for Area-Minimizing Integral Currents in Higher CodimensionIf we keep zooming in on an area-minimizing generalized surface, must its limiting cone be independent of the sequence of magnifications? Ask this at every interior point of an integral current of dimension at least three in Euclidean space of dimension at least two higher.Analysis & PDE

    Exact statement

    Let T be an m-dimensional area-minimizing integral current in R^(m+n), with m>=3 and n>=2. Prove that at every interior point p all weak limits, as r decreases to 0, of the rescaled currents induced by x -> (x-p)/r coincide.

    Why it matters

    A unique cone would make the local limiting shape independent of the chosen sequence of magnifications. Distinct limits at one point would demonstrate an ambiguity that area minimization alone cannot remove.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Uniqueness of Tangent Cones for Area-Minimizing Integral Currents in Higher Codimension · 1 problem in this edition. Family membership alone does not assert an implication.

    • #376 · Uniqueness of Tangent Cones for Area-Minimizing Integral Currents in Higher Codimension (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 371–532.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  377. Rank377Von Neumann's Problem for II_1 FactorsMust every nonamenable II₁ factor, a type of infinite-dimensional operator algebra, contain a copy of the standard von Neumann algebra built from the free group on two generators?Analysis & PDE

    Exact statement

    Determine whether every nonamenable II_1 factor contains a von Neumann subalgebra isomorphic to the free group factor L(F_2), or construct a nonamenable II_1 factor with no such subfactor.

    Why it matters

    The answer would determine whether nonamenability always forces this particular free-group structure inside a II_1 factor. A counterexample would separate nonamenability from containment of L(F_2).

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Von Neumann's Problem for II_1 Factors · 1 problem in this edition. Family membership alone does not assert an implication.

    • #377 · Von Neumann's Problem for II_1 Factors (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 382–527.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  378. Rank378Exact value of the site-percolation threshold on the square latticeWhen each site of the square lattice is independently occupied with the same probability, what is the exact critical probability for an infinite nearest-neighbor occupied cluster?Probability, ergodic theory & dynamics

    Exact statement

    Determine the exact critical probability p_c for Bernoulli site percolation on the nearest-neighbor square lattice Z^2. The corresponding bond-percolation threshold is p_c = 1/2, whereas the site threshold is known only numerically, p_c approximately 0.592746, and no exact value is known.

    Why it matters

    An exact value would locate the transition between the two connectivity regimes of this site model. It would give a precise threshold for occupied vertices, preserving the distinction from percolation defined by occupied edges.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Exact value of the site-percolation threshold on the square lattice · 1 problem in this edition. Family membership alone does not assert an implication.

    • #378 · Exact value of the site-percolation threshold on the square lattice (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 370–586.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  379. Rank379Sums of Three Cubes Representation ConjectureApart from integers leaving remainder 4 or 5 when divided by 9, can every integer be written as a sum of three integer cubes? More strongly, does each allowed integer have infinitely many such representations?Number theory & arithmetic geometry

    Exact statement

    Prove that every integer k not congruent to 4 or 5 modulo 9 admits a representation k = x^3 + y^3 + z^3 in integers, with Heath-Brown's refinement asking for infinitely many such representations for each eligible k.

    Why it matters

    Existence would show that these congruence exclusions account for every failure of representation. The stronger infinitude statement would ensure that each eligible integer has endlessly many representations, a separate requirement from finding one.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Sums of Three Cubes Representation Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #379 · Sums of Three Cubes Representation Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 379–533.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  380. Rank380Uniqueness of the KPZ Fixed Point Among Scaling-Covariant Local FieldsThe KPZ fixed point describes random fluctuations across space and time. Is it the only nontrivial field obeying the conjecture's locality condition and scaling-covariance laws, which relate the field's behavior at different scales?Probability, ergodic theory & dynamics

    Exact statement

    Prove that the KPZ fixed point is the unique non-trivial space-time field satisfying the locality condition and the specified scaling-covariance properties.

    Why it matters

    A proof would let the stated locality and rescaling laws identify a candidate scaling-limit field as the KPZ fixed point. Another nontrivial field satisfying those laws would show that the conditions do not uniquely identify it.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Uniqueness of the KPZ Fixed Point Among Scaling-Covariant Local Fields · 1 problem in this edition. Family membership alone does not assert an implication.

    • #380 · Uniqueness of the KPZ Fixed Point Among Scaling-Covariant Local Fields (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 369–566.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  381. Rank381Piltz Divisor Problem for k-Fold Divisor Error TermsFor each fixed k≥3, what is the optimal power-law exponent for the error when the number of ordered k-tuples of positive integers with product at most x is approximated by its standard main term, allowing an arbitrarily small extra power of x?Number theory & arithmetic geometry

    Exact statement

    For each integer k>=3, let d_k(n) count the ordered k-tuples of positive integers with product n, and let Delta_k(x)=sum_{n<=x} d_k(n)-Res_{s=1}(zeta(s)^k x^s/s). Determine alpha_k=inf{a: for every epsilon>0, Delta_k(x)=O_{k,epsilon}(x^(a+epsilon)) as x tends to infinity}.

    Why it matters

    A full answer would give the sharp error-exponent threshold for every higher divisor sum. The target asks for these optimal exponents without assuming a particular conjectured value or a finer logarithmic error bound.

    Status

    Open

    Open. Bellotti--Yang and the 2025/2026 number-field paper provide bounds, and Baluyot--Castillo provide sign-change results. None identifies the complete pointwise exponent sequence.

    Status review date not recorded in this edition

    Sources

    Problem family

    Piltz Divisor Problem for k-Fold Divisor Error Terms · 1 problem in this edition. Family membership alone does not assert an implication.

    • #381 · Piltz Divisor Problem for k-Fold Divisor Error Terms (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 421–511.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  382. Rank382Erdős Matching ConjectureChoose k-element subsets of n objects, forbidding s mutually disjoint choices, with n≥ks. Is the maximum size always the larger of two constructions: all k-subsets inside ks−1 objects, or all k-subsets meeting a fixed set of s−1 objects?Combinatorics & discrete geometry

    Exact statement

    For all positive integers k,s,n with n>=ks, let F be a family of distinct k-element subsets of [n]={1,...,n} containing no s pairwise disjoint members. The conjecture asserts |F|<=max{binom(ks-1,k), binom(n,k)-binom(n-s+1,k)}. Here binom(a,k)=0 for integers 0<=a<k. For s=1 the family is empty, and k=1 is included as a trivial case. The two terms are attained by all k-subsets of a fixed (ks-1)-set and by all k-subsets meeting a fixed (s-1)-set, respectively. Their role is sharpness of the bound, not an additional classification of all equality families.

    Why it matters

    A negative answer would determine the exact largest size of every such family through the larger of the two constructions. A larger family would refute the formula without requiring a classification of all equality cases.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Erdős Matching Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #382 · Erdős Matching Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 422–508.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  383. Rank383Lovász Conjecture on Hamiltonian Paths in Vertex-Transitive GraphsDoes every finite connected graph whose symmetries can move any vertex to any other contain a path that visits every vertex exactly once?Combinatorics & discrete geometryOpen · disputed claim

    Exact statement

    Prove or disprove that every finite connected vertex-transitive graph contains a Hamiltonian path.

    Why it matters

    A proof would turn vertex symmetry and connectedness into a universal guarantee of a spanning path. A counterexample would show that these conditions still permit an obstruction to visiting every vertex without repetition.

    Status

    Open · disputed claim

    October 6, 2026: A claimed n^(1-o(1)) long-path bound for vertex-transitive graphs does not guarantee a Hamiltonian path. Preserve the existing disputed-claim qualification on the full target. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Lovász Conjecture on Hamiltonian Paths in Vertex-Transitive Graphs · 1 problem in this edition. Family membership alone does not assert an implication.

    • #383 · Lovász Conjecture on Hamiltonian Paths in Vertex-Transitive Graphs (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 378–558.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  384. Rank384Deterministic Linear-Time Minimum Spanning Tree ProblemCan a deterministic algorithm always find the cheapest tree connecting all vertices in time proportional to the number of edges? The graph is connected and undirected with distinct real edge weights, and the algorithm must use the pointer-machine model, comparing weights only two at a time.Theoretical computer science

    Exact statement

    Determine whether there exists a uniform deterministic comparison-based algorithm that, for every connected undirected graph with n vertices and m edges carrying distinct real-valued weights, computes a minimum spanning tree in worst-case O(m) time on a pointer machine, with edge weights accessed only by binary comparisons, in the model of Pettie and Ramachandran (2002).

    Why it matters

    A positive answer would establish a linear worst-case time guarantee in the specified computational model. An impossibility proof would show that no uniform deterministic algorithm in that model can achieve the guarantee.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Deterministic Linear-Time Minimum Spanning Tree Problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #384 · Deterministic Linear-Time Minimum Spanning Tree Problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 419–513.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  385. Rank385Exact critical exponents of oriented percolation in 1+1 dimensionsAt the critical threshold for random connections directed forward in time on a one-dimensional lattice, what are the exact power-law exponents for survival, expected cluster size, and spatial spread? The model uses nearest-neighbor connections.Probability, ergodic theory & dynamics

    Exact statement

    For nearest-neighbor oriented percolation in 1+1 space-time dimensions at criticality, determine the exact values of rho, eta, and nu governing survival probability, the expected number of vertices connected from the origin at time t, and the gyration radius, respectively.

    Why it matters

    These values would specify how survival, cluster mass, and spatial spread scale at criticality. Failure of one defining power-law limit would instead show that the corresponding scaling description needs revision.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Exact critical exponents of oriented percolation in 1+1 dimensions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #385 · Exact critical exponents of oriented percolation in 1+1 dimensions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 364–582.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  386. Rank386Decidability of Finite Identity Bases for Finite SemigroupsIs there an algorithm that always halts and decides, from a finite semigroup's multiplication table, whether all its identities follow from a finite set of identities in the ordinary semigroup language?Algebra & logic

    Exact statement

    Does a single algorithm exist which, on input the multiplication table of any finite semigroup S, halts and correctly decides whether all semigroup identities valid in S follow in equational logic from some finite set of identities? Identities are equalities of nonempty words in the one-associative-binary-operation semigroup signature; no identity constant, inverse or involution is added. Resolve by a total decision algorithm with a correctness proof, or by a proof that no such algorithm exists. The target is finite basedness itself, not inherent non-finite basedness or finite elementary-inclusion bases.

    Also known as

    Tarski finite basis problem for finite semigroups; Tarski-Perkins semigroup decision problem

    Why it matters

    This would decide whether a finite algebra's entire identity theory has a finite equational basis. A negative resolution would prove that no single total decision algorithm handles all finite semigroups.

    Status

    Open

    Status reviewed

    Sources

    Problem family

    Decidability of Finite Identity Bases for Finite Semigroups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #386 · Decidability of Finite Identity Bases for Finite Semigroups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 411–525.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  387. Rank387Maximal Number of Mutually Unbiased Bases in Composite DimensionsTwo orthonormal bases are mutually unbiased if every vector from one overlaps equally in magnitude with every vector from the other. What is the maximum number of such bases in each dimension that is composite but not a prime power—in particular, can dimension six support seven?Quantum information & computation

    Exact statement

    Determine the maximal number K of mutually unbiased bases in a D-dimensional Hilbert space for non-prime-power composite dimensions D (in particular, decide whether K = 7 MUBs exist in dimension D = 6).

    Why it matters

    The answer would determine how many bases can coexist with the prescribed equal-overlap relationships. In dimension six, it would distinguish the possibility of seven such bases from a strictly smaller maximum.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s 6 October 2026 release reports N(6)=3, resolving the dimension-six subcase and excluding seven mutually unbiased bases there. The exclusion of four arbitrary complex bases depends on complete execution of the documented computer-assisted pipeline under the stated binary64 arithmetic and compiler conditions; no completed-run receipt was independently inspected. ProofAtlas has not rerun or audited that computation. The maximum in other non-prime-power composite dimensions remains unresolved by this release. No independent proof audit, confirmation of refereeing, or full-target formal verification is claimed.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Maximal Number of Mutually Unbiased Bases in Composite Dimensions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #387 · Maximal Number of Mutually Unbiased Bases in Composite Dimensions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 377–558.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  388. Rank388Aaronson-Ambainis ConjectureCan a polynomial of binary inputs, taking values in [0,1], spread its variation so evenly that no input bit has appreciable influence? The conjecture predicts a bit whose influence, or sensitivity to changing it, has a universal polynomial lower bound in inverse degree and output variance.Quantum information & computation

    Exact statement

    Prove or refute: for every degree-d polynomial f: {-1,1}^n -> [0,1], there exists a coordinate j whose influence satisfies Inf_j[f] >= poly(1/d, Var[f]).

    Why it matters

    The bound would guarantee a detectably influential coordinate whenever the polynomial has sufficient variance relative to its degree, independently of the number of coordinates. Failure would allow variation to be dispersed below every proposed bound of this form.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Aaronson-Ambainis Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #388 · Aaronson-Ambainis Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 359–570.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  389. Rank389Greenfield-Wallach-Katok Conjecture for Cohomology-Free Vector FieldsSuppose a smooth vector field X on a closed connected orientable manifold lets every smooth real function v be written as Xu + c, with u smooth and c constant. Must a smooth change of coordinates carry X to a constant Diophantine vector field on a torus? The required conjugacy preserves the flow's time parameter.Probability, ergodic theory & dynamicsCohomology-free flows · 1 problem

    Exact statement

    Let M be any closed connected orientable smooth manifold of dimension d>=1 and X a smooth vector field. Suppose that for every smooth real function v on M there exist a smooth real function u and a real constant c such that Xu=v-c. Must there be a smooth diffeomorphism h:M->T^d for which h_*X is a constant vector alpha satisfying |k dot alpha|>=C ||k||^(-tau) for every nonzero k in Z^d, for some C>0 and tau>=0? A proof for all such M,X or a counterexample resolves the question. This cohomology-free form is equivalent to the globally hypoelliptic Greenfield-Wallach form; they constitute one candidate.

    Why it matters

    This would classify all smooth flows with this universal solvability property by rigid torus models. A counterexample would reveal a cohomology-free flow outside that classification.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Cohomology-free flows · 1 problem in this edition. Family membership alone does not assert an implication.

    • #389 · Greenfield-Wallach-Katok Conjecture for Cohomology-Free Vector Fields (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 418–523.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  390. Rank390Lusztig's P1-P15 Conjectures for Bounded Weighted Coxeter GroupsFor every finite-rank Coxeter system with an integer-valued weight function positive on the generators, assume a uniform bound on the degrees of its Hecke algebra's standard-basis structure constants. Do all fifteen assertions P1–P15 in §14.2 of Lusztig's fixed 2014 edition hold? The boundedness condition is a hypothesis of the question.Algebra, representation & category theoryCoxeter groups and Hecke algebra cells · 1 problem

    Exact statement

    For every finite-rank Coxeter system (W,S) and integer-valued weight function L:W->Z, additive whenever Coxeter lengths add and positive on S, assume (W,L) is bounded in the sense of Section 13.2 of Lusztig, Hecke algebras with unequal parameters, arXiv:math/0208154v2 (2014): some N uniformly bounds the v-degrees of all standard-basis structure constants f_{x,y,z}. Do all fifteen assertions P1-P15 of Conjectures 14.2 in that fixed edition hold for its Hecke algebra? Establish the entire conjunction under these hypotheses or disprove at least one assertion for an admissible (W,L). Boundedness itself is a hypothesis, not an additional required theorem.

    Why it matters

    These assertions govern Kazhdan–Lusztig cells and Lusztig's asymptotic algebra. Proving the full package would establish that structure throughout the specified bounded, unequal-parameter setting; one failed assertion would refute the package.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Coxeter groups and Hecke algebra cells · 1 problem in this edition. Family membership alone does not assert an implication.

    • #390 · Lusztig's P1-P15 Conjectures for Bounded Weighted Coxeter Groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 413–527.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  391. Rank391Andrews-Curtis conjectureA group presentation gives generators and relations. If a finite presentation has equally many of each and describes the trivial group, can it always reach the standard trivial presentation by swapping or inverting relations, replacing one by its product with another, and conjugating individual relations?Algebra, representation & category theory

    Exact statement

    Prove or disprove that every balanced finite presentation of the trivial group can be transformed into the standard trivial presentation by Nielsen transformations of relators together with conjugations of individual relators.

    Why it matters

    A proof would show that the permitted changes always suffice to simplify these balanced presentations. A counterexample would reveal a presentation encoding the trivial group that remains inaccessible to the standard form through those moves.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Research on ProofAtlas

    Problem family

    Andrews-Curtis conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #391 · Andrews-Curtis conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 420–531.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  392. Rank392Pisot Substitution ConjectureA substitution system builds sequences by repeatedly replacing symbols with words. Under the irreducible unimodular Pisot conditions, must its spectrum be entirely discrete, so square-integrable observables decompose into eigenfunctions?Probability, ergodic theory & dynamics

    Exact statement

    Prove that every irreducible unimodular Pisot substitution dynamical system has pure discrete spectrum.

    Why it matters

    A proof would show that the stated substitution conditions force this spectral structure in every case. A counterexample would reveal additional spectral behavior compatible with all those conditions.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Pisot Substitution Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #392 · Pisot Substitution Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 411–530.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  393. Rank393Potential Density of Rational Points on K3 SurfacesOn every K3 surface defined over a number field, can a finite-degree enlargement of that field make the rational points Zariski dense—so they are not all confined to any proper algebraic closed subset of the surface?Number theory & arithmetic geometry

    Exact statement

    For every number field K and every K3 surface X/K (smooth, projective and geometrically integral, with trivial canonical bundle and H^1(X,O_X)=0), there exists a finite extension L/K for which X(L) is Zariski dense in X.

    Why it matters

    A proof would ensure that, after a suitable finite extension, the rational points escape every proper algebraic closed subset. Failure would identify a K3 surface for which no finite extension achieves this.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Related problems

    • #84 · Campana's Conjecture Characterizing Special Varieties by Potential Density of Rational PointsStronger assertionCampana's Conjecture Characterizing Special Varieties by Potential Density of Rational Points

      This K3 assertion follows from the specialness-to-potential-density direction of Campana's conjecture. Proving the K3 case alone does not establish Campana's full biconditional.

      Condition: The stronger assertion is proved; a negative answer to it need not settle the weaker assertion.

      Scope qualification: Only the positive-answer implication is recorded. No converse or equivalence of resolution outcomes is established here.

    Problem family

    Potential Density of Rational Points on K3 Surfaces · 1 problem in this edition. Family membership alone does not assert an implication.

    • #393 · Potential Density of Rational Points on K3 Surfaces (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 433–528.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  394. Rank394Liouville problem for stationary Navier–Stokes D-solutions in R^3Must a smooth, steady, unforced incompressible flow throughout three-dimensional space be completely at rest if its velocity tends uniformly to zero far away and the integral of its squared velocity gradient is finite?Analysis & PDE

    Exact statement

    Must every smooth real solution (u,p) of -Delta u+(u dot grad)u+grad p=0 and div u=0 on R^3, with integral_R^3 |grad u|^2 dx finite and u(x) tending uniformly to zero as |x| tends to infinity, satisfy u identically zero?

    Why it matters

    A proof would rule out all nonzero steady flows satisfying these decay and energy conditions. A counterexample would exhibit a persistent smooth flow with finite Dirichlet integral and vanishing velocity at infinity.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Liouville problem for stationary Navier–Stokes D-solutions in R^3 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #394 · Liouville problem for stationary Navier–Stokes D-solutions in R^3 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 400–557.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  395. Rank395Greek-Letter Redshift Conjecture in Algebraic K-TheoryGreek-letter families mark successive levels of structure in the stable homotopy groups of spheres. Whenever a commutative ring spectrum detects one such family, must taking its algebraic K-theory make the next family detectable?Algebra, representation & category theory

    Exact statement

    Prove that if a commutative ring spectrum R detects the n-th Greek-letter family in the stable homotopy groups of spheres, then its algebraic K-theory K(R) detects the (n+1)-st Greek-letter family.

    Why it matters

    A proof would establish this specific passage from one chromatic height to the next under algebraic K-theory. Failure would identify a limit of the Greek-letter detection criterion.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Greek-Letter Redshift Conjecture in Algebraic K-Theory · 1 problem in this edition. Family membership alone does not assert an implication.

    • #395 · Greek-Letter Redshift Conjecture in Algebraic K-Theory (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 396–568.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  396. Rank396Arveson-Douglas Essential Normality Conjecture for Homogeneous Quotient ModulesTake the Drury–Arveson space in any finite dimension and quotient by the closure of a proper homogeneous polynomial ideal I, including nonradical ideals. For its compressed coordinate operators S_j, must every commutator [S_i*, S_j] have p-summable singular values whenever p exceeds the complex dimension of the zero set Z(I)?AnalysisArveson–Douglas essential normality · 1 problem

    Exact statement

    For every integer d >= 1 and every proper homogeneous ideal I of C[z_1,...,z_d], let [I] be its closure in the Drury-Arveson space H_d^2 on the unit ball, and let S_j be multiplication by z_j compressed to Q_I=H_d^2 orthogonal-complement [I]. Is [S_i^*,S_j] in the Schatten class S^p for all i,j and every real p > dim_C Z(I)? A positive answer covers all such ideals, including nonradical ideals, in all finite dimensions; a negative answer exhibits a failing ideal and exponent in this range.

    Why it matters

    The conjecture ties how nearly these operators commute to the dimension of an algebraic zero set. A proof would give the proposed Schatten summability bound for every homogeneous ideal quotient in the stated setting.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Arveson–Douglas essential normality · 1 problem in this edition. Family membership alone does not assert an implication.

    • #396 · Arveson-Douglas Essential Normality Conjecture for Homogeneous Quotient Modules (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 434–529.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  397. Rank397Zariski Dense Orbit ConjectureLet a dominant rational self-map of an irreducible quasiprojective variety over an algebraically closed field of characteristic zero have no nonconstant invariant rational functions. Must some point have a forward orbit that is defined at every step and is Zariski dense in the variety?Number theory & arithmetic geometryZariski dense orbits · 1 problem

    Exact statement

    For every irreducible quasiprojective variety X over an algebraically closed field K of characteristic zero and dominant rational self-map f:X⇢X with K(X)^f=K, there exists x∈X(K) such that all forward iterates avoid the indeterminacy locus and the forward orbit is Zariski dense in X.

    Why it matters

    A proof would connect the absence of rational invariants with the existence of an orbit that escapes every proper algebraic subset. It asks whether that algebraic obstruction is the only one in this general setting.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Zariski dense orbits · 1 problem in this edition. Family membership alone does not assert an implication.

    • #397 · Zariski Dense Orbit Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 437–536.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  398. Rank398Capacity Region of General 3-Receiver Discrete Memoryless Broadcast ChannelsFor a discrete memoryless channel broadcasting to three receivers, which combinations of reliable communication rates are possible? Find a computable formula using one channel use, without assuming that one receiver's signal is a degraded version of another's.Cryptography, coding, information & optimization

    Exact statement

    Find a computable single-letter characterization of the capacity region for general 3-receiver discrete memoryless broadcast channels without degradedness assumptions.

    Why it matters

    Such a formula would characterize the simultaneously achievable communication rates from the data of a single channel use. It would apply to general three-receiver channels under the stated absence of degradedness assumptions.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Capacity Region of General 3-Receiver Discrete Memoryless Broadcast Channels · 1 problem in this edition. Family membership alone does not assert an implication.

    • #398 · Capacity Region of General 3-Receiver Discrete Memoryless Broadcast Channels (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 403–572.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  399. Rank399Positive Counting Rule for Type-A Schubert Structure ConstantsWhen Schubert polynomials are multiplied, can each ordinary type-A coefficient be expressed as the number of combinatorial objects passing one uniform, efficient check? Each object must have polynomial-size encoding and be checkable in polynomial time from the three input permutations.Algebra, representation & category theory

    Exact statement

    For the ordinary type-A Schubert-polynomial structure constants c^w_(u,v), defined by S_u S_v = sum_w c^w_(u,v) S_w for finite permutations, does there exist one uniform positive counting rule in the #P sense: a polynomial p and deterministic polynomial-time predicate R such that c^w_(u,v) equals the number of bit strings y of length p(|x|) with R(x,y)=1, for every explicitly one-line-encoded triple x=(u,v,w)?

    Why it matters

    A proof would place these multiplication coefficients in #P through a single counting predicate. It would express their values as counts of efficiently verifiable finite objects for every encoded input triple.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Positive Counting Rule for Type-A Schubert Structure Constants · 1 problem in this edition. Family membership alone does not assert an implication.

    • #399 · Positive Counting Rule for Type-A Schubert Structure Constants (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 441–544.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  400. Rank400Fujita Very Ampleness ConjectureFor an n-dimensional smooth complex projective variety, does adding n+2 copies of any ample divisor to the canonical divisor always provide enough coordinates to embed the variety in projective space?Geometry & topologyFujita’s adjoint bundle conjectures · 1 problem

    Exact statement

    For every smooth complex projective variety X of dimension n and every ample divisor L, prove that K_X + (n+2)L is very ample.

    Why it matters

    The formula would provide a uniform very-ampleness guarantee depending only on dimension. Failure would identify a variety and ample divisor for which the stated adjoint construction cannot give an embedding.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The source guarantees global generation also at m=n+2, so it supplies the basepoint-freeness part of the desired conclusion at the full very-ampleness threshold. Very ampleness additionally requires separation of points and tangent directions, which the published theorem does not state. The very-ampleness conjecture must therefore remain unresolved by this family. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Fujita’s adjoint bundle conjectures · 1 problem in this edition. Family membership alone does not assert an implication.

    • #400 · Fujita Very Ampleness Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 415–573.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  401. Rank401F-Conjecture for the Nef Cone of Mbar_(0,n)On the complex moduli space of stable n-pointed genus-zero curves, for every n≥3, is a divisor nef exactly when it has nonnegative intersection with every one-dimensional boundary stratum, or F-curve? Being nef means having nonnegative intersection with every curve.Geometry & topologyF-conjecture for pointed rational curves · 1 problem

    Exact statement

    For every integer n>=3 and every divisor class D on the complex moduli space Mbar_(0,n) of stable n-pointed genus-zero curves, D is nef if and only if D has nonnegative intersection with every one-dimensional boundary stratum (F-curve); equivalently, its closed Mori cone is generated by the F-curve classes.

    Why it matters

    A proof would reduce the nefness test from all curves on this moduli space to its F-curves. Equivalently, their classes would generate the closed cone of effective curves.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    F-conjecture for pointed rational curves · 1 problem in this edition. Family membership alone does not assert an implication.

    • #401 · F-Conjecture for the Nef Cone of Mbar_(0,n) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 435–563.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  402. Rank402Zilber's Quasiminimality Conjecture for the Complex Exponential FieldIf a subset of the complex numbers can be described by a logical formula using addition, multiplication, exponentiation, and fixed complex parameters, must either it or its complement be countable?Logic, foundations & set theoryThe complex exponential field · 2 problems

    Exact statement

    Prove that the complex exponential field C_exp = (C; +, *, exp) is quasiminimal: every subset of C definable (with parameters) in the expansion of the complex field by the complex exponential function is either countable or co-countable.

    Why it matters

    A proof would sharply restrict the sizes of definable subsets in the complex exponential field. Failure would permit a definable set and its complement to both be uncountable.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    The complex exponential field · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 437–559.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  403. Rank403Hall's Conjecture on Gaps Between Squares and CubesHow close can a positive integer square come to a distinct positive integer cube? Hall's conjecture predicts that the gap from x³ is always greater than a positive constant times x^(1/2−ε), for every ε>0, with the constant allowed to depend only on ε.Number theory & arithmetic geometry

    Exact statement

    For every epsilon > 0, prove that there is c(epsilon) > 0 such that all positive integers x and y with y^2 != x^3 satisfy |y^2-x^3| > c(epsilon) x^(1/2-epsilon).

    Why it matters

    The inequality would set a uniform lower scale for how closely a square can approach a distinct cube. Failure would yield pairs closer than that scale for some fixed positive ε.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Hall's Conjecture on Gaps Between Squares and Cubes · 1 problem in this edition. Family membership alone does not assert an implication.

    • #403 · Hall's Conjecture on Gaps Between Squares and Cubes (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 420–579.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  404. Rank404Voronoi Conjecture on ParallelohedraCan every convex polytope that tiles space using translations be turned, by an invertible affine change of coordinates, into a lattice's nearest-point cell—the region at least as close to one lattice point as to any other? The question covers every dimension.Combinatorics & discrete geometry

    Exact statement

    Prove or disprove that every parallelohedron (a convex polytope that tiles R^d by translations) is an affine image of the Dirichlet-Voronoi polytope of some lattice.

    Why it matters

    A proof would describe every such translational tile through a lattice nearest-point construction after an affine change of coordinates. A counterexample would exhibit a tiling shape beyond that representation.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Voronoi Conjecture on Parallelohedra · 1 problem in this edition. Family membership alone does not assert an implication.

    • #404 · Voronoi Conjecture on Parallelohedra (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 461–530.

    Source-fit opponent count: 37.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  405. Rank405Campana-Peternell ConjectureOn a smooth complex projective Fano variety, if the tangent bundle is nef—a weak positivity condition—must the variety be rational homogeneous, so its algebraic symmetries can move any point to any other?Geometry & topology

    Exact statement

    Prove that every smooth complex projective Fano variety whose tangent bundle is nef is a rational homogeneous variety.

    Why it matters

    A proof would make the tangent-bundle condition a classification criterion within smooth complex projective Fano varieties. A counterexample would exhibit a variety satisfying that condition outside the rational homogeneous class.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The contact-Fano result makes every smooth complex projective contact Fano manifold in its stated dimension range an adjoint variety, hence rational homogeneous. It therefore supplies the contact-Fano subclass of the full assertion whenever its nef-tangent hypothesis also holds. The full Campana–Peternell target includes all Fano manifolds with nef tangent, with no contact structure assumed; the source gives no reduction of all such manifolds to contact ones. The manuscript establishes rational homogeneity for smooth connected complex projective Fano varieties with nef tangent bundle in dimension six. This is the exact Campana-Peternell conclusion in that dimension, but the full target quantifies over every dimension. The additional universal-cover corollary has a dimension-minus-irregularity bound and does not remove this restriction. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Campana-Peternell Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #405 · Campana-Peternell Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 442–561.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  406. Rank406Congruence Subgroup Property for Automorphism Groups of Free GroupsFor every free group F_n of rank n ≥ 3, does every finite-index subgroup of Aut(F_n) contain the kernel of its action on some finite characteristic quotient of F_n?Group theory

    Exact statement

    For every integer n>=3, let F_n be the free group on n generators. Does every finite-index subgroup H of Aut(F_n) contain ker(Aut(F_n)->Aut(F_n/N)) for some characteristic finite-index subgroup N of F_n? Prove the universal assertion, or exhibit a rank n>=3 and finite-index H for which no such N exists. Characteristic means invariant under all automorphisms. Equivalently, ask injectivity of the natural map from the profinite completion of Aut(F_n) to Aut of the profinite completion of F_n. The known rank-two theorem is excluded from the unresolved quantifier.

    Also known as

    Aut(F_n) congruence subgroup problem for n>=3

    Why it matters

    This asks whether finite characteristic quotients detect all finite-index restrictions on these automorphism groups. Rank two is already settled and is outside the unresolved quantifier.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Congruence Subgroup Property for Automorphism Groups of Free Groups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #406 · Congruence Subgroup Property for Automorphism Groups of Free Groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 443–556.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  407. Rank407Auslander Conjecture on Affine Crystallographic GroupsWhen affine transformations—linear changes followed by translations—act properly discontinuously on Euclidean space with compact quotient, must their group contain a solvable subgroup with only finitely many cosets?Geometry & topology

    Exact statement

    Prove or disprove that every affine crystallographic group is virtually solvable, equivalently that the fundamental group of every compact complete affine manifold is virtually polycyclic.

    Why it matters

    A proof would impose the stated algebraic restrictions on all groups arising in this geometric setting. A counterexample would demonstrate a compact complete affine structure whose group escapes those restrictions.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Auslander Conjecture on Affine Crystallographic Groups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #407 · Auslander Conjecture on Affine Crystallographic Groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 442–561.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  408. Rank408Normality of PiDo π’s digits give every pattern of the same length the same long-run frequency, no matter which integer base, from base 2 onward, is used?Number theory & arithmetic geometry

    Exact statement

    Prove that pi is a normal number: in every integer base b >= 2, every block of k digits occurs in the base-b expansion of pi with asymptotic frequency b^{-k}.

    Why it matters

    A proof would establish uniform limiting frequencies for every finite digit pattern across every base. A counterexample would identify a specific base and block whose frequency differs or fails to exist.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Normality of Pi · 1 problem in this edition. Family membership alone does not assert an implication.

    • #408 · Normality of Pi (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 377–652.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  409. Rank409Voisin's Zero-Cycle Conjecture for Varieties of Geometric Genus OneFor every smooth projective complex n-fold with h^(n,0)=1 and h^(j,0)=0 for 0<j<n, do degree-zero zero-cycles with rational coefficients satisfy a × b = (−1)^n b × a modulo rational equivalence on X × X?Algebraic geometry

    Exact statement

    For every integer n>=1 and every smooth projective irreducible complex variety X of dimension n satisfying h^(j,0)(X)=0 for 0<j<n and h^(n,0)(X)=1, and every a,b in CH_0(X)_Q of degree zero, prove a x b = (-1)^n b x a in CH_0(X x X)_Q, where Chow groups mean algebraic cycles modulo rational equivalence with rational coefficients and x is the exterior product. A proof for all such X or a counterexample within this class resolves the target; verification for a particular family does not. No trivial-canonical-bundle hypothesis is imposed.

    Also known as

    Voisin signed exterior-product identity for degree-zero cycles; Voisin conjecture for self-products with geometric genus one

    Why it matters

    This predicts a symmetry of zero-cycles from the stated Hodge-number conditions. The assertion covers every variety in that class, without imposing a trivial canonical bundle.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Voisin's Zero-Cycle Conjecture for Varieties of Geometric Genus One · 1 problem in this edition. Family membership alone does not assert an implication.

    • #409 · Voisin's Zero-Cycle Conjecture for Varieties of Geometric Genus One (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 444–551.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  410. Rank410Infinitude of Fermat PrimesAre there infinitely many indices n≥0 for which the Fermat number 2^(2^n)+1 is prime, or do prime values eventually stop appearing?Number theory & arithmetic geometry

    Exact statement

    Determine whether infinitely many Fermat numbers F_n=2^(2^n)+1 are prime; equivalently, prove that the set of n≥0 with F_n prime is infinite or prove that it is finite.

    Why it matters

    Infinitude would guarantee prime Fermat numbers beyond every index bound. Finiteness would establish a bound after which every number in this particular sequence is composite.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Infinitude of Fermat Primes · 1 problem in this edition. Family membership alone does not assert an implication.

    • #410 · Infinitude of Fermat Primes (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 448–556.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  411. Rank411Existence of the Diagonal Ramsey Growth ConstantR(k,k) is the smallest complete-graph size for which every red/blue edge-coloring forces k vertices with all connecting edges one color. As k grows, does R(k,k) have a well-defined exponential growth base—that is, do its kth roots converge?Combinatorics & discrete geometry

    Exact statement

    Prove or disprove that the limit lim_{k->infinity} R(k,k)^(1/k) exists.

    Why it matters

    Existence of the limit would assign one exponential growth rate to these Ramsey numbers without requiring its value. Failure would mean that distinct subsequences retain different limiting growth rates.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Existence of the Diagonal Ramsey Growth Constant · 1 problem in this edition. Family membership alone does not assert an implication.

    • #411 · Existence of the Diagonal Ramsey Growth Constant (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 430–584.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  412. Rank412Polynomial-Time Quasioptimal Logarithmic-Energy Points on the 2-SphereCan an efficient algorithm place N points on a sphere so their logarithmic repulsion energy is within a fixed constant times log N of the best possible value? One algorithm must work for every N≥2 in the real-number model, using time polynomial in N.Applied & computational mathematics

    Exact statement

    Give a uniform real-number algorithm that, for every N >= 2, constructs N points on S^2 in time polynomial in N with logarithmic energy at most the minimum possible energy plus c log N for an absolute constant c.

    Why it matters

    The result would couple an efficient construction with a uniform guarantee of near-minimal energy. It requires one algorithm and one absolute error constant to work for every number of points.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Polynomial-Time Quasioptimal Logarithmic-Energy Points on the 2-Sphere · 1 problem in this edition. Family membership alone does not assert an implication.

    • #412 · Polynomial-Time Quasioptimal Logarithmic-Energy Points on the 2-Sphere (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 442–566.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  413. Rank413Maximum Number of k-sets and Complexity of k-levelsAmong all configurations of n points in the plane, what is the maximum number of k-point subsets that a straight line can separate from all remaining points?Combinatorics & discrete geometry

    Exact statement

    What is the maximum number of k-sets of an n-point set in the plane (or equivalently, the maximum complexity of a k-level in an arrangement of n lines or hyperplanes)?

    Why it matters

    Determining this maximum would quantify how many distinct subsets can be selected by linear separation. It would also determine the corresponding extremal complexity of k-levels in arrangements of lines.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The family summary improves the planar k-set upper bound to O(n(k+1)^(1/3-epsilon_0)) for 1<=k<=n/2, and the abstract supplies the halving-line case for sufficiently large even n and no three collinear. This improves an upper bound for the target extremal quantity but does not determine its asymptotic maximum or supply matching lower bounds. The constants and power saving are nonquantitative; no higher-dimensional maximum follows. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Maximum Number of k-sets and Complexity of k-levels · 1 problem in this edition. Family membership alone does not assert an implication.

    • #413 · Maximum Number of k-sets and Complexity of k-levels (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 432–582.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  414. Rank414Uniform epsilon-equilibrium in finite multiplayer stochastic gamesIn a finite stochastic game with bounded payoffs, can players choose randomized strategies so no one gains more than any chosen positive tolerance by changing strategy alone? The same strategies must work for all sufficiently long games, using expected average payoffs, from each given starting state.Applied & computational mathematics

    Exact statement

    For every finite stochastic game with bounded stage payoffs, every epsilon>0, and every initial state s_0, prove that there are a behavioral-strategy profile sigma and N such that for every horizon n>=N no player can improve that player's expected n-stage average payoff by more than epsilon through any unilateral behavioral deviation.

    Why it matters

    A proof would guarantee one approximate equilibrium that works uniformly across long horizons. Failure would mean that, in some game and initial state, no single profile maintains the required incentive bound as horizons vary.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Uniform epsilon-equilibrium in finite multiplayer stochastic games · 1 problem in this edition. Family membership alone does not assert an implication.

    • #414 · Uniform epsilon-equilibrium in finite multiplayer stochastic games (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 394–615.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  415. Rank415Hadwiger-Boltyanski Illumination ConjectureFor every convex body in n-dimensional Euclidean space, can its entire boundary be illuminated by at most 2^n external point sources or parallel beams?Combinatorics & discrete geometry

    Exact statement

    Prove that every convex body K in n-dimensional Euclidean space can be illuminated by at most 2^n external point sources or parallel beams of light.

    Why it matters

    A proof would give a dimension-dependent cap on the illumination needed for every convex body. A counterexample would show that some convex shape requires more than this number of sources or directions.

    Status

    Open

    October 6, 2026: A claimed fractional illumination bound 2^d and integral covering bound 2^d times a dimension-dependent polynomial factor do not prove the integral 2^d illumination target. The complete proof was not audited in this review.

    Status reviewed

    Sources

    Research on ProofAtlas

    Problem family

    Hadwiger-Boltyanski Illumination Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #415 · Hadwiger-Boltyanski Illumination Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 444–597.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  416. Rank416Residual Finiteness of All Artin-Tits GroupsIs every Artin–Tits group defined by a finite Coxeter matrix residually finite: can each nonidentity element be distinguished from the identity in some finite quotient?Geometry & topologyGeometry and topology of Artin groups · 2 problems

    Exact statement

    Decide whether every Artin-Tits group associated with a finite Coxeter matrix is residually finite: for each nonidentity element g, there exists a homomorphism to a finite group whose value on g is nonidentity. No spherical, FC, triangle, or other diagram restriction is imposed.

    Why it matters

    This asks whether finite groups can detect every nontrivial element of these groups. A positive answer would establish this separation property for all such groups, beyond the classes where it is already known.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Geometry and topology of Artin groups · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 444–575.

    Source-fit opponent count: 35.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  417. Rank417Leray's Problem for Stationary Navier-Stokes under the Total-Flux Condition in Three DimensionsIn a bounded, multiply connected three-dimensional region, is balanced total inflow and outflow enough for a steady incompressible Navier–Stokes flow to exist in the weak sense? Flow may enter through one boundary component and leave through another, with no symmetry assumed.Analysis & PDE

    Exact statement

    Let Omega be a bounded multiply connected domain in R^3 and let a be boundary data satisfying the total-flux compatibility condition integral_{partial Omega} a dot n = 0. Determine whether the stationary incompressible Navier-Stokes system in Omega has a weak solution without requiring zero flux on each connected boundary component or imposing symmetry.

    Why it matters

    The answer would decide whether global conservation of flow is enough for stationary weak solvability in this setting. It would allow, or expose an obstruction to, compensating inflow and outflow through different boundary components.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Leray's Problem for Stationary Navier-Stokes under the Total-Flux Condition in Three Dimensions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #417 · Leray's Problem for Stationary Navier-Stokes under the Total-Flux Condition in Three Dimensions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 458–564.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  418. Rank418Freyd's Generating HypothesisIf a map between finite spectra is zero on every stable homotopy group, must the map itself be null-homotopic?Geometry & topology

    Exact statement

    Prove that if f:X→Y is a map of finite spectra inducing the zero homomorphism on every stable homotopy group, then f is null-homotopic; equivalently, the natural map [X,Y]→Hom_{π_*S}(π_*X,π_*Y) is injective for all finite spectra X and Y.

    Why it matters

    Stable homotopy groups are fundamental invariants in topology. The question asks whether these invariants detect all maps between finite spectra, rather than only some of their effects.

    Status

    Open

    The question asks whether stable homotopy groups detect every nonzero map between finite spectra. It remains open in the stated generality.

    Status review date not recorded in this edition

    Sources

    Problem family

    Freyd's Generating Hypothesis · 1 problem in this edition. Family membership alone does not assert an implication.

    • #418 · Freyd's Generating Hypothesis (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 414–594.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  419. Rank419List edge-coloring conjectureSuppose a finite graph with no loops needs k colors on its edges so edges meeting at a vertex have different colors. If each edge instead receives its own arbitrary list of k allowed colors, can the graph always be colored from those lists? Parallel edges are allowed.Combinatorics & discrete geometry

    Exact statement

    Prove or disprove that every finite loopless multigraph G has list chromatic index equal to its ordinary chromatic index, that is, chi'_ell(G)=chi'(G).

    Why it matters

    A proof would show that allowing each edge its own color list never raises the required list size above the ordinary chromatic index. A counterexample would demonstrate an extra obstruction caused by the lists.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    List edge-coloring conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #419 · List edge-coloring conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 460–561.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  420. Rank420Combinatorial Boolean Matrix Multiplication in Near-Linear TimeCan two n-by-n matrices of zeros and ones be multiplied with AND and OR replacing multiplication and addition, in almost quadratic time in n? The algorithm must be purely combinatorial, taking n² times a subpolynomial factor, without Strassen-style algebraic tensor methods.Theoretical computer science

    Exact statement

    Determine whether there exists a purely combinatorial algorithm for boolean matrix multiplication running in time O(n^{2+o(1)}) without using algebraic Strassen-like tensor methods.

    Why it matters

    A positive answer would put this form of matrix multiplication within the stated almost-quadratic time bound using combinatorial methods. An impossibility result would establish a limitation of that algorithmic class.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Combinatorial Boolean Matrix Multiplication in Near-Linear Time · 1 problem in this edition. Family membership alone does not assert an implication.

    • #420 · Combinatorial Boolean Matrix Multiplication in Near-Linear Time (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 414–598.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  421. Rank421Dynamical Manin–Mumford Conjecture: Special-Subvariety FormulationFor every smooth projective variety X over a field K of characteristic zero, every polarizable endomorphism f:X→X and every subvariety Z⊆X containing a Zariski-dense set of f-preperiodic geometric points, Z is dynamically special: there exist an integer n≥1, a subvariety Y with Z⊆Y⊆X, and a polarizable endomorphism g:X→X such that f^n(Y)=Y=g(Y), f^n∘g=g∘f^n on Y, and g^a(Z)=g^b(Z) for some integers 0≤a<b. Polarizable means h*L≅L^⊗d for some ample line bundle L and integer d>1, separately for each endomorphism h; geometric points are taken over an algebraic closure.Number theory & arithmetic geometry

    Exact statement

    For every smooth projective variety X over a field K of characteristic zero, every polarizable endomorphism f:X→X and every subvariety Z⊆X containing a Zariski-dense set of f-preperiodic geometric points, Z is dynamically special: there exist an integer n≥1, a subvariety Y with Z⊆Y⊆X, and a polarizable endomorphism g:X→X such that f^n(Y)=Y=g(Y), f^n∘g=g∘f^n on Y, and g^a(Z)=g^b(Z) for some integers 0≤a<b. Polarizable means h*L≅L^⊗d for some ample line bundle L and integer d>1, separately for each endomorphism h; geometric points are taken over an algebraic closure.

    Why it matters

    A proof would describe dense preperiodic-point phenomena through invariant subvarieties and commuting polarized dynamics, including the known obstructions to the older assertion.

    A counterexample would reveal dense preperiodic points not explained by this commuting-endomorphism notion of special subvariety.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Dynamical Manin–Mumford Conjecture: Special-Subvariety Formulation · 1 problem in this edition. Family membership alone does not assert an implication.

    • #421 · Dynamical Manin–Mumford Conjecture: Special-Subvariety Formulation (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 459–567.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  422. Rank422Stolz Conjecture on Positive Ricci Curvature and the Witten GenusProve or disprove that every closed smooth string manifold M of positive dimension, meaning a spin manifold with the integral spin characteristic class p_1(M)/2 equal to zero, that admits a Riemannian metric with positive Ricci curvature has Witten genus zero.Geometry & topology

    Exact statement

    Prove or disprove that every closed smooth string manifold M of positive dimension, meaning a spin manifold with the integral spin characteristic class p_1(M)/2 equal to zero, that admits a Riemannian metric with positive Ricci curvature has Witten genus zero.

    Why it matters

    The Witten genus would obstruct positive Ricci curvature for every manifold in the stated string class.

    Exhibit a positive-dimensional closed string manifold carrying a positive-Ricci metric whose Witten genus is nonzero.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Stolz Conjecture on Positive Ricci Curvature and the Witten Genus · 1 problem in this edition. Family membership alone does not assert an implication.

    • #422 · Stolz Conjecture on Positive Ricci Curvature and the Witten Genus (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 457–572.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  423. Rank423Koblitz Conjecture for Prime Orders of Elliptic CurvesFor an elliptic curve over the rationals without complex multiplication, how often does good reduction modulo a prime produce a group with a prime number of points? When the corrected Koblitz constant C is positive, is the count up to x asymptotic to Cx/(log x)²?Number theory & arithmetic geometry

    Exact statement

    For every non-CM elliptic curve E over Q with corrected Koblitz constant C_E>0, prove that #{p≤x: p is of good reduction and #E(F_p) is prime} is asymptotic to C_E x/(log x)^2.

    Why it matters

    The formula would quantify how frequently reduction produces a group with a prime number of points. Its positive constant would in particular force infinitely many such primes for every curve in the stated class.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Koblitz Conjecture for Prime Orders of Elliptic Curves · 1 problem in this edition. Family membership alone does not assert an implication.

    • #423 · Koblitz Conjecture for Prime Orders of Elliptic Curves (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 458–570.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  424. Rank424The McKay-Navarro Conjecture for Finite GroupsFor every finite group and prime p, do its irreducible representations of dimension coprime to p match those of the Sylow p-subgroup's normalizer when counted according to Galois symmetries of character values? The target includes the blockwise refinement for odd primes.Algebra, representation & category theory

    Exact statement

    Prove the McKay-Navarro (Galois-McKay) conjecture: for a finite group G, a prime p and a Sylow p-subgroup P of G, the p'-degree irreducible characters of G and of N_G(P) can be counted compatibly with the action of Galois automorphisms on character values, generalizing the McKay conjecture blockwise for odd primes.

    Why it matters

    The result would relate character counts while preserving the specified arithmetic symmetries of character values. A counterexample would show that the group and its Sylow normalizer differ under one of the required Galois-sensitive comparisons.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    The McKay-Navarro Conjecture for Finite Groups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #424 · The McKay-Navarro Conjecture for Finite Groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 441–590.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  425. Rank425Spectral Determination of Convex Planar DomainsCan you hear the shape of a convex planar drum? If two bounded convex shapes have exactly the same Dirichlet vibration spectrum, with their boundaries held fixed, must they be identical up to translation, rotation, or reflection?Analysis & PDE

    Exact statement

    Determine whether two bounded convex planar domains with identical Dirichlet-Laplacian spectra must be congruent.

    Why it matters

    The answer would determine whether the full spectrum fixes a convex planar shape. A counterexample would give geometrically different convex domains that the entire list of Dirichlet eigenvalues cannot distinguish.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Spectral Determination of Convex Planar Domains · 1 problem in this edition. Family membership alone does not assert an implication.

    • #425 · Spectral Determination of Convex Planar Domains (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 428–643.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  426. Rank426Entropy Photon-Number InequalityFor every n>=1, eta in [0,1], and n-mode bosonic states rho_A,rho_B of finite von Neumann entropy, take independent inputs rho_A tensor rho_B and mix corresponding modes by c_i=sqrt(eta)a_i+sqrt(1-eta)b_i. Let rho_C be the retained n-mode output. With S in nats and g(x)=(x+1)ln(x+1)-x ln x for x>=0 (0 ln 0=0), prove or disprove S(rho_C)>=n*g(eta*g^{-1}(S(rho_A)/n)+(1-eta)*g^{-1}(S(rho_B)/n)). Correlations within each input block are allowed; S(rho_C)=+infinity satisfies this bound.Quantum information & computation

    Exact statement

    For every n>=1, eta in [0,1], and n-mode bosonic states rho_A,rho_B of finite von Neumann entropy, take independent inputs rho_A tensor rho_B and mix corresponding modes by c_i=sqrt(eta)a_i+sqrt(1-eta)b_i. Let rho_C be the retained n-mode output. With S in nats and g(x)=(x+1)ln(x+1)-x ln x for x>=0 (0 ln 0=0), prove or disprove S(rho_C)>=n*g(eta*g^{-1}(S(rho_A)/n)+(1-eta)*g^{-1}(S(rho_B)/n)). Correlations within each input block are allowed; S(rho_C)=+infinity satisfies this bound.

    Why it matters

    Prove the stated output-entropy lower bound for every n, eta and pair of finite-entropy n-mode product inputs.

    Construct an allowed n, eta and pair of finite-entropy input states and rigorously show that their output entropy is strictly below the stated finite right-hand side.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The manuscript reports the entropy photon-number inequality for every finite number of modes and all transmissivities, with independent input blocks and arbitrary correlations within each block, when both inputs have finite mean photon number. Finite von Neumann entropy alone does not imply that finite-energy condition. The stated continuity and truncation results retain finite or uniformly bounded energy, so finite-entropy inputs of infinite mean energy remain outside the theorem. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Entropy Photon-Number Inequality · 1 problem in this edition. Family membership alone does not assert an implication.

    • #426 · Entropy Photon-Number Inequality (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 456–581.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  427. Rank427Optimal Universal Constant in the Lieb–Oxford InequalityDetermine the exact least universal C_LO for which E_ind(P) ≥ -C_LO ∫_(R³) ρ_P(x)^(4/3) dx for every N≥1 and symmetric probability distribution P on (R³)^N with finite terms, where E_ind is expected pairwise Coulomb energy minus one half of the Coulomb self-energy of the one-particle density ρ_P. Completion requires a matching universal lower bound on energy and an optimality construction, not just a better numerical estimate.Mathematical physics

    Exact statement

    Determine the exact least universal C_LO for which E_ind(P) ≥ -C_LO ∫_(R³) ρ_P(x)^(4/3) dx for every N≥1 and symmetric probability distribution P on (R³)^N with finite terms, where E_ind is expected pairwise Coulomb energy minus one half of the Coulomb self-energy of the one-particle density ρ_P. Completion requires a matching universal lower bound on energy and an optimality construction, not just a better numerical estimate.

    Why it matters

    The exact quantity in the resolution contract would be determined, whatever its value.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Optimal Universal Constant in the Lieb–Oxford Inequality · 1 problem in this edition. Family membership alone does not assert an implication.

    • #427 · Optimal Universal Constant in the Lieb–Oxford Inequality (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 448–600.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  428. Rank428Ruzsa-Szemeredi (6,3) Extremal-Order ProblemAs n grows, what is the order of magnitude of the most distinct triples one can choose from n vertices if every set of at most six vertices contains at most two chosen triples?Combinatorics & discrete geometrySparse triple systems · 2 problems

    Exact statement

    Let f(n,6,3) be the maximum number of edges in a finite simple 3-uniform hypergraph on n vertices in which every set of at most six vertices spans at most two edges. Determine the asymptotic order of f(n,6,3) as n tends to infinity.

    Why it matters

    A sharp asymptotic answer would determine how densely triples can be packed while excluding that local pattern. Matching constructions and upper bounds would identify the scale forced by the six-vertex restriction.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Sparse triple systems · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 462–572.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  429. Rank429Irrationality of zeta(5)Is the infinite sum 1 + 1/2⁵ + 1/3⁵ + ⋯, called ζ(5), impossible to express exactly as a ratio of two integers?Number theory & arithmetic geometry

    Exact statement

    Prove that zeta(5) = sum_{n>=1} 1/n^5 is irrational.

    Why it matters

    A proof would establish irrationality for this particular odd zeta value. A rational evaluation would instead identify an exact arithmetic relation for ζ(5), without deciding the behavior of other zeta values.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Irrationality of zeta(5) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #429 · Irrationality of zeta(5) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 375–680.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  430. Rank430Witsenhausen Counterexample: Optimal Strategy and CostTwo controllers act in sequence: the first sees a Gaussian state, while the second sees the altered state through unit Gaussian noise. Which strategies achieve the exact least expected cost, balancing the weighted square of the first control against the squared final error?Applied & computational mathematics

    Exact statement

    Determine the optimal control strategies and exact minimum expected cost for the two-stage Witsenhausen problem x1 = x0 + u1, x2 = x1 + u2, where x0 ~ N(0, sigma^2), decision maker 1 observes x0 and incurs k u1^2, and decision maker 2 observes z2 = x1 + v with v ~ N(0,1) and incurs x2^2.

    Why it matters

    An exact solution would identify globally optimal control laws and their cost under the specified unequal information. It would quantify the best achievable balance between the first control action and the final residual state.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Witsenhausen Counterexample: Optimal Strategy and Cost · 1 problem in this edition. Family membership alone does not assert an implication.

    • #430 · Witsenhausen Counterexample: Optimal Strategy and Cost (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 411–652.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  431. Rank431Square Peg Problem (Toeplitz's Inscribed Square Conjecture)Must every continuous closed loop in the plane with no self-intersections contain four points forming a square, even when the loop is jagged or otherwise not smooth?Geometry & topology

    Exact statement

    Prove or disprove that every continuous simple closed (Jordan) curve in the plane contains four points forming the vertices of a square.

    Why it matters

    A proof would guarantee the same inscribed shape on every Jordan curve, even when its geometry is irregular. A counterexample would identify a continuous closed curve that avoids every square.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Square Peg Problem (Toeplitz's Inscribed Square Conjecture) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #431 · Square Peg Problem (Toeplitz's Inscribed Square Conjecture) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 441–603.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  432. Rank432Bounded Negativity Conjecture for Complex Projective SurfacesFor every smooth complex projective surface X, prove that there is a constant b(X) such that C^2 >= -b(X) for every reduced irreducible curve C on X.Number theory & arithmetic geometry

    Exact statement

    For every smooth complex projective surface X, prove that there is a constant b(X) such that C^2 >= -b(X) for every reduced irreducible curve C on X.

    Why it matters

    A proof would impose a global numerical constraint on negative curves across complex surface geometry.

    A counterexample would exhibit a single complex projective surface carrying curves with unboundedly negative self-intersection.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Bounded Negativity Conjecture for Complex Projective Surfaces · 1 problem in this edition. Family membership alone does not assert an implication.

    • #432 · Bounded Negativity Conjecture for Complex Projective Surfaces (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 460–571.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  433. Rank433Segre–Harbourne–Gimigliano–Hirschowitz ConjectureFor projective plane curves constrained at general-position points, does their family have unexpectedly large dimension exactly when its fixed part contains a repeated curve of self-intersection −1 on the blown-up plane?Geometry & topologyPlane curve linear systems · 1 problem

    Exact statement

    Prove that a linear system of plane curves in P^2 with base points in general position is special if and only if its fixed locus contains a multiple (-1)-curve; equivalently, every Cremona-reduced such linear system is non-special.

    Why it matters

    The equivalence would identify the precise geometric cause of speciality in these linear systems. Failure would show that their dimensions can deviate from the predicted behavior without matching this fixed-curve criterion.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. Nagata’s strict inequality constrains which plane-curve linear systems at very general points can be effective, supplying a weaker consequence/subproblem associated with SHGH. The full SHGH assertion classifies all special systems through multiple (-1)-curves or non-speciality after Cremona reduction. A degree/multiplicity obstruction and asymptotic Seshadri maximality do not compute those dimensions or give the stated specialness classification. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Related problems

    Problem family

    Plane curve linear systems · 1 problem in this edition. Family membership alone does not assert an implication.

    • #433 · Segre–Harbourne–Gimigliano–Hirschowitz Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 452–597.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  434. Rank434Chern Conjecture on the Euler Characteristic of Closed Affine ManifoldsIf a compact manifold without boundary has local coordinates whose changes are affine transformations—linear maps followed by translations—must its Euler characteristic be zero?Geometry & topologyOpen · disputed claim

    Exact statement

    Prove that every closed affine manifold, equivalently every compact boundaryless manifold whose tangent bundle admits a flat torsion-free connection, has Euler characteristic zero.

    Why it matters

    A proof would make vanishing Euler characteristic a necessary topological condition for a closed affine structure. A counterexample would show that such a structure can coexist with nonzero Euler characteristic.

    Status

    Open · disputed claim

    Status review date not recorded in this edition

    Sources

    Problem family

    Chern Conjecture on the Euler Characteristic of Closed Affine Manifolds · 1 problem in this edition. Family membership alone does not assert an implication.

    • #434 · Chern Conjecture on the Euler Characteristic of Closed Affine Manifolds (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 463–584.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  435. Rank435Kawaguchi–Silverman conjecture for dense point orbitsFor every smooth projective variety X of positive dimension over Qbar, every dominant rational self-map f:X⇢X over Qbar, every ample divisor H with logarithmic Weil height h_H, and every x∈X(Qbar) whose forward orbit is well defined and Zariski dense, the limit lim_{n→∞} max{1,h_H(f^n(x))}^{1/n} exists and equals δ_1(f)=lim_{n→∞}(((f^n)^*H)·H^(dim X−1))^{1/n}.Number theory & arithmetic geometry

    Exact statement

    For every smooth projective variety X of positive dimension over Qbar, every dominant rational self-map f:X⇢X over Qbar, every ample divisor H with logarithmic Weil height h_H, and every x∈X(Qbar) whose forward orbit is well defined and Zariski dense, the limit lim_{n→∞} max{1,h_H(f^n(x))}^{1/n} exists and equals δ_1(f)=lim_{n→∞}(((f^n)^*H)·H^(dim X−1))^{1/n}.

    Why it matters

    A proof establishes precisely the assertion stated in the formal statement; no further unsupported consequence is asserted.

    A rigorous negation establishes the failure of that assertion in its stated scope; no further unsupported consequence is asserted.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

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    Kawaguchi–Silverman conjecture for dense point orbits · 1 problem in this edition. Family membership alone does not assert an implication.

    • #435 · Kawaguchi–Silverman conjecture for dense point orbits (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 460–584.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  436. Rank436Weiss's Hyperfiniteness Question for Countable Amenable Group ActionsDetermine whether the orbit equivalence relation of every Borel action of a countable amenable group is hyperfinite.Logic, foundations & set theory

    Exact statement

    Determine whether the orbit equivalence relation of every Borel action of a countable amenable group is hyperfinite.

    Why it matters

    A positive resolution would make every orbit relation of a Borel action by a countable amenable group hyperfinite.

    A negative resolution would exhibit a countable amenable Borel action with a nonhyperfinite orbit relation.

    Status

    Open

    Status review date not recorded in this edition

    Sources

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    Weiss's Hyperfiniteness Question for Countable Amenable Group Actions · 1 problem in this edition. Family membership alone does not assert an implication.

    • #436 · Weiss's Hyperfiniteness Question for Countable Amenable Group Actions (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 474–585.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  437. Rank437p-adic Littlewood conjectureProve that for every prime p and every real number x, liminf_{q->infinity} q * |q|_p * ||q x|| = 0, where ||.|| denotes distance to the nearest integer and |.|_p the p-adic absolute value; equivalently, for every badly approximable real x and every prime p, inf_{q>=1} q*||q x||*|q|_p = 0.Probability, ergodic theory & dynamics

    Exact statement

    Prove that for every prime p and every real number x, liminf_{q->infinity} q * |q|_p * ||q x|| = 0, where ||.|| denotes distance to the nearest integer and |.|_p the p-adic absolute value; equivalently, for every badly approximable real x and every prime p, inf_{q>=1} q*||q x||*|q|_p = 0.

    Why it matters

    A proof would force the p-adically weighted Diophantine liminf to vanish for every real x and prime p.

    A counterexample would provide a prime p and badly approximable x with inf q||qx|| |q|_p strictly positive.

    Status

    Open

    Status review date not recorded in this edition

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    p-adic Littlewood conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #437 · p-adic Littlewood conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 466–612.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  438. Rank438Beauville–Voisin Conjecture for Projective Hyperkähler ManifoldsFor every smooth projective irreducible holomorphic symplectic complex variety X, the Betti cycle-class map is injective on the Q-subalgebra of CH*(X)_Q generated by all divisor classes and all Chern classes c_i(T_X); thus a homologically trivial polynomial in these classes is zero modulo rational equivalence with rational coefficients.Number theory & arithmetic geometry

    Exact statement

    For every smooth projective irreducible holomorphic symplectic complex variety X, the Betti cycle-class map is injective on the Q-subalgebra of CH*(X)_Q generated by all divisor classes and all Chern classes c_i(T_X); thus a homologically trivial polynomial in these classes is zero modulo rational equivalence with rational coefficients.

    Why it matters

    A proof would identify all cohomological relations among these generators with relations modulo rational equivalence with Q coefficients.

    A counterexample would separate a cohomological relation in the specified generators from rational equivalence, refuting this injectivity assertion.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Beauville–Voisin Conjecture for Projective Hyperkähler Manifolds · 1 problem in this edition. Family membership alone does not assert an implication.

    • #438 · Beauville–Voisin Conjecture for Projective Hyperkähler Manifolds (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 503–573.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  439. Rank439Evasiveness Conjecture for Graph PropertiesProve that every nontrivial monotone graph property is evasive, i.e. any deterministic decision-tree algorithm deciding it must, in the worst case, query all possible edges of the input graph.Theoretical computer science

    Exact statement

    Prove that every nontrivial monotone graph property is evasive, i.e. any deterministic decision-tree algorithm deciding it must, in the worst case, query all possible edges of the input graph.

    Why it matters

    Would prove that every nontrivial monotone graph property is evasive, i.e. any deterministic decision-tree algorithm deciding it must, in the worst case, query all possible edges of the input graph.

    Would exhibit a counterexample or obstruction to the assertion that every nontrivial monotone graph property is evasive, i.e. any deterministic decision-tree algorithm deciding it must, in the worst case, query all possible edges of the input graph.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Evasiveness Conjecture for Graph Properties · 1 problem in this edition. Family membership alone does not assert an implication.

    • #439 · Evasiveness Conjecture for Graph Properties (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 461–635.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  440. Rank440Zygmund's Conjecture on Maximal Functions along Lipschitz Vector FieldsLet v:R^2->S^1 be Lipschitz. Prove that there is epsilon_0 > 0 depending only on ||v||_Lip such that M_(v,epsilon_0)f(x) = sup_(0<t<epsilon_0) (1/(2t)) integral_(-t)^t |f(x-sv(x))| ds is of weak type (2,2): sup_(lambda>0) lambda^2 |{x:M_(v,epsilon_0)f(x)>lambda}| <= C ||f||_2^2.Analysis & PDE

    Exact statement

    Let v:R^2->S^1 be Lipschitz. Prove that there is epsilon_0 > 0 depending only on ||v||_Lip such that M_(v,epsilon_0)f(x) = sup_(0<t<epsilon_0) (1/(2t)) integral_(-t)^t |f(x-sv(x))| ds is of weak type (2,2): sup_(lambda>0) lambda^2 |{x:M_(v,epsilon_0)f(x)>lambda}| <= C ||f||_2^2.

    Why it matters

    A positive resolution would establish the stated norm-dependent locally truncated weak-type (2,2) maximal-averaging bound for every Lipschitz unit vector field in the plane.

    A negative resolution would disprove the full stated weak-type (2,2) maximal-averaging assertion, including its permitted norm-dependent cutoff, not merely fail at an arbitrary scale.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Zygmund's Conjecture on Maximal Functions along Lipschitz Vector Fields · 1 problem in this edition. Family membership alone does not assert an implication.

    • #440 · Zygmund's Conjecture on Maximal Functions along Lipschitz Vector Fields (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 482–589.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  441. Rank441Second Hardy-Littlewood Conjecture on Prime-Counting SubadditivityProve or disprove that pi(x+y) <= pi(x)+pi(y) for all integers x,y >= 2, where pi is the prime-counting function.Number theory & arithmetic geometry

    Exact statement

    Prove or disprove that pi(x+y) <= pi(x)+pi(y) for all integers x,y >= 2, where pi is the prime-counting function.

    Why it matters

    A proof would settle a century-old exact global inequality for the prime-counting function.

    A counterexample would resolve the conjecture negatively and clarify its historical tension with prime-tuple heuristics.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Second Hardy-Littlewood Conjecture on Prime-Counting Subadditivity · 1 problem in this edition. Family membership alone does not assert an implication.

    • #441 · Second Hardy-Littlewood Conjecture on Prime-Counting Subadditivity (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 466–607.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  442. Rank442Roughening Phase Transition for the Interface of the Three-Dimensional Ising ModelProve the existence of a roughening phase transition for the three-dimensional Ising model: show that the interface separating the plus and minus phases is rigid, with bounded fluctuations, below some roughening temperature and rough above it, and determine the threshold and the nature of the transition. Equivalently, determine the range of inverse temperatures for which non-translation-invariant Dobrushin states exist.Mathematical physics3D Ising · 3 problems

    Exact statement

    Prove the existence of a roughening phase transition for the three-dimensional Ising model: show that the interface separating the plus and minus phases is rigid, with bounded fluctuations, below some roughening temperature and rough above it, and determine the threshold and the nature of the transition. Equivalently, determine the range of inverse temperatures for which non-translation-invariant Dobrushin states exist.

    Why it matters

    A proof would identify a temperature threshold separating bounded interface fluctuations from a rough interface and characterize the transition.

    A contrary result would rule out the asserted rigid-to-rough separation or its equivalence with the stated Dobrushin-state range.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    3D Ising · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 416–674.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  443. Rank443Gyarfas-Sumner ConjectureFor every tree T, prove that the class of finite graphs containing no induced subgraph isomorphic to T is chi-bounded; equivalently, find f_T with chi(G)<=f_T(omega(G)) for every induced-T-free graph G.Combinatorics & discrete geometry

    Exact statement

    For every tree T, prove that the class of finite graphs containing no induced subgraph isomorphic to T is chi-bounded; equivalently, find f_T with chi(G)<=f_T(omega(G)) for every induced-T-free graph G.

    Also known as

    Gyárfás–Sumner conjecture; Gyarfas-Sumner conjecture; Sumner conjecture on induced-tree-free graphs

    Why it matters

    A proof would show that every induced-T-free graph class is chi-bounded for each fixed tree T.

    A counterexample would give a fixed tree T and induced-T-free graphs with bounded clique number but unbounded chromatic number.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Gyarfas-Sumner Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #443 · Gyarfas-Sumner Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 484–598.

    Source-fit opponent count: 37.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  444. Rank444Existence of Rational Curves on Every Simply Connected Calabi–Yau ThreefoldEvery smooth projective threefold X over C with K_X≅O_X, h¹(X,O_X)=0 and π₁(X)={1} admits a nonconstant morphism P¹_C→X.Geometry & topology

    Exact statement

    Every smooth projective threefold X over C with K_X≅O_X, h¹(X,O_X)=0 and π₁(X)={1} admits a nonconstant morphism P¹_C→X.

    Why it matters

    A proof would establish rational-curve existence without imposing the additional nef-boundary-divisor assumptions of the cited partial results.

    A counterexample would refute the universal existence expectation while leaving divisor-conditioned subcases intact.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Existence of Rational Curves on Every Simply Connected Calabi–Yau Threefold · 1 problem in this edition. Family membership alone does not assert an implication.

    • #444 · Existence of Rational Curves on Every Simply Connected Calabi–Yau Threefold (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 501–595.

    Source-fit opponent count: 35.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  445. Rank445Erdos-Turan Conjecture on Additive BasesProve or disprove that for every set A of nonnegative integers which is an additive basis of order 2 (i.e. A + A contains every sufficiently large integer), the ordered representation function r(n) = #{(a,b) in A x A : a + b = n} is unbounded as n tends to infinity.Combinatorics & discrete geometry

    Exact statement

    Prove or disprove that for every set A of nonnegative integers which is an additive basis of order 2 (i.e. A + A contains every sufficiently large integer), the ordered representation function r(n) = #{(a,b) in A x A : a + b = n} is unbounded as n tends to infinity.

    Why it matters

    A proof would show that every additive basis of order two has an unbounded ordered representation function.

    A counterexample would be an additive basis of order two whose ordered representation function stays bounded.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Erdos-Turan Conjecture on Additive Bases · 1 problem in this edition. Family membership alone does not assert an implication.

    • #445 · Erdos-Turan Conjecture on Additive Bases (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 449–629.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  446. Rank447Permanent Anti-Concentration and Robust Classical Hardness of BosonSamplingProve the Permanent Anti-Concentration Conjecture: there exists a polynomial p, positive on the indicated arguments, such that for every positive integer n and every delta > 0, Pr[|Per(X)| < sqrt(n!)/p(n,1/delta)] < delta, where X is an n x n matrix of independent standard complex Gaussian entries (independent real and imaginary parts, each N(0,1/2), with 1/2 denoting variance).Quantum information & computation

    Exact statement

    Prove the Permanent Anti-Concentration Conjecture: there exists a polynomial p, positive on the indicated arguments, such that for every positive integer n and every delta > 0, Pr[|Per(X)| < sqrt(n!)/p(n,1/delta)] < delta, where X is an n x n matrix of independent standard complex Gaussian entries (independent real and imaginary parts, each N(0,1/2), with 1/2 denoting variance).

    Why it matters

    A proof would establish the stated lower-tail anti-concentration bound for permanents of complex Gaussian matrices.

    A disproof would show that no polynomial p gives the stated uniform lower-tail bound for permanents of standard complex Gaussian matrices.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Permanent Anti-Concentration and Robust Classical Hardness of BosonSampling · 1 problem in this edition. Family membership alone does not assert an implication.

    • #447 · Permanent Anti-Concentration and Robust Classical Hardness of BosonSampling (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 490–608.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  447. Rank448Decidability of the First-Order Theory of F_p((t))For each prime p, determine whether the complete first-order theory Th(F_p((t))) in the pure language of rings {0,1,+,-,×}, with no named t or extra predicates, is decidable. A full determination settles the answer for every p: a decidable case requires a total correct algorithm on sentences, and an undecidable case requires a proof that none exists.Logic, foundations & set theory

    Exact statement

    For each prime p, determine whether the complete first-order theory Th(F_p((t))) in the pure language of rings {0,1,+,-,×}, with no named t or extra predicates, is decidable. A full determination settles the answer for every p: a decidable case requires a total correct algorithm on sentences, and an undecidable case requires a proof that none exists.

    Why it matters

    The exact quantity in the resolution contract would be determined, whatever its value.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Decidability of the First-Order Theory of F_p((t)) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #448 · Decidability of the First-Order Theory of F_p((t)) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 502–598.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  448. Rank449Trifference Problem: Exponential Rate of Ternary Perfect 3-Hash CodesFor n>=1 let T(n) be the largest cardinality of a set C subset {0,1,2}^n such that every three distinct words x,y,z in C have some coordinate i with {x_i,y_i,z_i}={0,1,2}. Determine R_3 = limsup as n tends to infinity of log_2(T(n))/n by proving matching asymptotic lower and upper bounds. Resolution requires identifying the value of R_3, not merely improving one side, computing finitely many T(n), or restating this variational definition. Codes are unrestricted, not necessarily linear. The chosen entity is the classical ternary k=3 problem, not the simultaneous determination of every perfect k-hash rate.Cryptography, coding, information & optimization

    Exact statement

    For n>=1 let T(n) be the largest cardinality of a set C subset {0,1,2}^n such that every three distinct words x,y,z in C have some coordinate i with {x_i,y_i,z_i}={0,1,2}. Determine R_3 = limsup as n tends to infinity of log_2(T(n))/n by proving matching asymptotic lower and upper bounds. Resolution requires identifying the value of R_3, not merely improving one side, computing finitely many T(n), or restating this variational definition. Codes are unrestricted, not necessarily linear. The chosen entity is the classical ternary k=3 problem, not the simultaneous determination of every perfect k-hash rate.

    Also known as

    Trifference asymptotic-rate problem; Asymptotic rate of ternary trifferent codes; Perfect 3-hashing rate; Zero-error list-size-two rate of the 3/2 channel

    Why it matters

    Determining the rate would establish the asymptotically optimal exponential size of ternary perfect-three-hash codes and the corresponding zero-error list-size-two coding rate, without requiring an efficient construction.

    Status

    Open

    Status reviewed

    Sources

    Problem family

    Trifference Problem: Exponential Rate of Ternary Perfect 3-Hash Codes · 1 problem in this edition. Family membership alone does not assert an implication.

    • #449 · Trifference Problem: Exponential Rate of Ternary Perfect 3-Hash Codes (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 478–608.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  449. Rank450Hadamard Maximal Determinant ProblemFor every positive integer n, determine D(n)=max{|det A|: A is an n by n matrix with every entry in {+1,-1}}.Applied & computational mathematics

    Exact statement

    For every positive integer n, determine D(n)=max{|det A|: A is an n by n matrix with every entry in {+1,-1}}.

    Why it matters

    A general determination would give the exact extremal determinant, equivalently the maximal parallelotope volume, for every matrix order.

    A counterexample to a proposed universal formula would establish necessary order-dependent exceptions and constrain any eventual exact classification.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Hadamard Maximal Determinant Problem · 1 problem in this edition. Family membership alone does not assert an implication.

    • #450 · Hadamard Maximal Determinant Problem (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 473–627.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  450. Rank451Higman's PORC Conjecture on the Number of Groups of Order p^nFor every positive integer n, let a_n(p) be the number of abstract isomorphism classes of ALL groups of order p^n. There exist a positive integer N_n and polynomials P_(n,r) in Q[X], with rational coefficients, for residues 0<=r<N_n, such that for every rational prime p, a_n(p)=P_(n,p mod N_n)(p). The modulus and polynomials may depend on n, but not on p. This is a statement on primes, not all positive integers or all prime powers.Algebra, representation & category theory

    Exact statement

    For every positive integer n, let a_n(p) be the number of abstract isomorphism classes of ALL groups of order p^n. There exist a positive integer N_n and polynomials P_(n,r) in Q[X], with rational coefficients, for residues 0<=r<N_n, such that for every rational prime p, a_n(p)=P_(n,p mod N_n)(p). The modulus and polynomials may depend on n, but not on p. This is a statement on primes, not all positive integers or all prime powers.

    Why it matters

    Would establish residue-polynomial structure for the total number of isomorphism classes of finite groups at each fixed prime-power exponent. It would not by itself produce an explicit classification, an efficient enumeration algorithm, or a modulus uniform in the exponent.

    Would show that the total enumeration function at some fixed prime-power exponent cannot be described by finitely many rational polynomials selected by prime residue classes. Non-PORC behavior of a restricted family alone does not imply this conclusion.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Higman's PORC Conjecture on the Number of Groups of Order p^n · 1 problem in this edition. Family membership alone does not assert an implication.

    • #451 · Higman's PORC Conjecture on the Number of Groups of Order p^n (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 512–589.

    Source-fit opponent count: 46.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  451. Rank452Moore's Conjecture on Homotopy Exponents and Rational EllipticityFor every simply connected finite CW-complex X, the following are equivalent: (a) ⊕_{i≥2}π_i(X)⊗Q is finite-dimensional; (b) for some prime p a power p^e annihilates all p-primary torsion in all π_i(X); (c) for every prime p there exists such a power p^{e(p)}.Geometry & topology

    Exact statement

    For every simply connected finite CW-complex X, the following are equivalent: (a) ⊕_{i≥2}π_i(X)⊗Q is finite-dimensional; (b) for some prime p a power p^e annihilates all p-primary torsion in all π_i(X); (c) for every prime p there exists such a power p^{e(p)}.

    Why it matters

    A proof would identify rational ellipticity exactly with bounded p-primary torsion exponents, uniformly in degree, at each prime.

    A counterexample would break that correspondence for a finite simply connected complex.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Moore's Conjecture on Homotopy Exponents and Rational Ellipticity · 1 problem in this edition. Family membership alone does not assert an implication.

    • #452 · Moore's Conjecture on Homotopy Exponents and Rational Ellipticity (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 503–599.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  452. Rank453Oppermann's Conjecture on Primes Between n(n-1), n^2 and n(n+1)For every integer n>1, prove or disprove that each of the open intervals (n(n-1),n^2) and (n^2,n(n+1)) contains a prime.Number theory & arithmetic geometryPrime gaps · 3 problemsOpen · disputed claim

    Exact statement

    For every integer n>1, prove or disprove that each of the open intervals (n(n-1),n^2) and (n^2,n(n+1)) contains a prime.

    Why it matters

    Every indicated interval would be certified to contain a prime; Legendre's conjecture would follow by the explicit interval inclusions.

    One empty prescribed interval would refute Oppermann; whether the larger consecutive-square interval still contains another prime would remain a separate question.

    Status

    Open · disputed claim

    Status review date not recorded in this edition

    Sources

    Problem family

    Prime gaps · 3 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 497–602.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  453. Rank454Simon's Decaying-Potential Absolutely Continuous Spectrum ConjectureLet V be a measurable real potential on R^nu, nu >= 2, satisfying int |x|^(-nu+1) |V(x)|^2 d^nu x < infinity. Prove that H = -Delta + V has absolutely continuous spectrum of infinite multiplicity on [0,infinity), under the standard self-adjoint realization of H.Mathematical physics

    Exact statement

    Let V be a measurable real potential on R^nu, nu >= 2, satisfying int |x|^(-nu+1) |V(x)|^2 d^nu x < infinity. Prove that H = -Delta + V has absolutely continuous spectrum of infinite multiplicity on [0,infinity), under the standard self-adjoint realization of H.

    Why it matters

    A proof would settle Simon's multidimensional decaying-potential absolutely-continuous-spectrum conjecture under the stated weighted L2 condition.

    A counterexample would sharply delimit the weighted-L2 decay threshold for absolutely continuous spectrum in multidimensional Schrodinger operators.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Simon's Decaying-Potential Absolutely Continuous Spectrum Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #454 · Simon's Decaying-Potential Absolutely Continuous Spectrum Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 470–623.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  454. Rank455Strong Chern Conjecture for Minimal HypersurfacesProve that every connected closed minimal hypersurface in a Euclidean unit sphere with constant scalar curvature is isoparametric.Geometry & topology

    Exact statement

    Prove that every connected closed minimal hypersurface in a Euclidean unit sphere with constant scalar curvature is isoparametric.

    Why it matters

    A proof would complete a major rigidity and classification program for minimal hypersurfaces in spheres.

    A counterexample would produce a new non-isoparametric constant-scalar-curvature minimal hypersurface.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Strong Chern Conjecture for Minimal Hypersurfaces · 1 problem in this edition. Family membership alone does not assert an implication.

    • #455 · Strong Chern Conjecture for Minimal Hypersurfaces (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 455–662.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  455. Rank456Pansu's isoperimetric conjecture in the first Heisenberg groupIn the standard first Heisenberg group H1 with Haar volume and horizontal perimeter, are the finite-perimeter minimizers of perimeter at each fixed positive finite volume exactly, up to null sets and left translation, the dilates of the region enclosed by a Pansu sphere?Geometry & topology

    Exact statement

    In the standard first Heisenberg group H1 with Haar volume and horizontal perimeter, are the finite-perimeter minimizers of perimeter at each fixed positive finite volume exactly, up to null sets and left translation, the dilates of the region enclosed by a Pansu sphere?

    Why it matters

    Show that Pansu bubble regions achieve the least horizontal perimeter at each volume and that every minimizer agrees almost everywhere with a translated dilate.

    Exhibit a volume for which a Pansu bubble is not minimizing, or an isoperimetric minimizer not equivalent to a translated dilate.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Pansu's isoperimetric conjecture in the first Heisenberg group · 1 problem in this edition. Family membership alone does not assert an implication.

    • #456 · Pansu's isoperimetric conjecture in the first Heisenberg group (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 490–625.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  456. Rank457Jones's Irreducible Index Realization Problem for Hyperfinite II_1 SubfactorsDetermine whether every finite Jones index value in {4 cos^2(pi/n) : n >= 3} union [4, infinity) occurs as the index of an irreducible finite-index subfactor of the hyperfinite II_1 factor R.Analysis & PDE

    Exact statement

    Determine whether every finite Jones index value in {4 cos^2(pi/n) : n >= 3} union [4, infinity) occurs as the index of an irreducible finite-index subfactor of the hyperfinite II_1 factor R.

    Why it matters

    A positive solution would determine the finite index spectrum of irreducible subfactors of the hyperfinite II_1 factor throughout the Jones-allowed range.

    A missing allowed value would reveal an irreducibility or hyperfiniteness obstruction not detected by the general Jones index theorem.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Jones's Irreducible Index Realization Problem for Hyperfinite II_1 Subfactors · 1 problem in this edition. Family membership alone does not assert an implication.

    • #457 · Jones's Irreducible Index Realization Problem for Hyperfinite II_1 Subfactors (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 514–608.

    Source-fit opponent count: 30.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  457. Rank458Decidability of conjugacy for shifts of finite typeDetermine whether there exists an algorithm which, given any two finite square matrices A and B with nonnegative integer entries, always terminates and correctly decides whether their two-sided edge shifts are topologically conjugate, equivalently whether A and B are strong shift equivalent over the nonnegative integers.Probability, ergodic theory & dynamics

    Exact statement

    Determine whether there exists an algorithm which, given any two finite square matrices A and B with nonnegative integer entries, always terminates and correctly decides whether their two-sided edge shifts are topologically conjugate, equivalently whether A and B are strong shift equivalent over the nonnegative integers.

    Why it matters

    A positive resolution supplies a total procedure deciding conjugacy for every pair of finite edge-shift presentations.

    A negative resolution establishes that no total algorithm can decide conjugacy from every pair of finite edge-shift presentations.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Decidability of conjugacy for shifts of finite type · 1 problem in this edition. Family membership alone does not assert an implication.

    • #458 · Decidability of conjugacy for shifts of finite type (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 500–617.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  458. Rank459Gromov's Vanishing Simplicial Volume and Euler Characteristic QuestionProve or disprove that every oriented closed connected aspherical manifold M with real simplicial volume ||M||=0 has Euler characteristic chi(M)=0. Here ||M|| is the l1-seminorm of the real fundamental class. The target concerns manifolds, not general aspherical complexes, and has no nonpositive-curvature or residual-finiteness hypothesis.Geometry & topology

    Exact statement

    Prove or disprove that every oriented closed connected aspherical manifold M with real simplicial volume ||M||=0 has Euler characteristic chi(M)=0. Here ||M|| is the l1-seminorm of the real fundamental class. The target concerns manifolds, not general aspherical complexes, and has no nonpositive-curvature or residual-finiteness hypothesis.

    Why it matters

    Vanishing real simplicial volume would force vanishing Euler characteristic throughout the closed aspherical manifold class.

    A counterexample would separate these two vanishing properties for an actual closed aspherical manifold.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Gromov's Vanishing Simplicial Volume and Euler Characteristic Question · 1 problem in this edition. Family membership alone does not assert an implication.

    • #459 · Gromov's Vanishing Simplicial Volume and Euler Characteristic Question (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 509–621.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  459. Rank460Exact Growth Constant of Maximal Cap Sets in F_3^nLet r(n) be the maximum cardinality of a subset A of F_3^n having no three distinct elements x,y,z with x+y+z=0. Determine the exact exponential growth constant c=lim_{n->infinity} r(n)^(1/n)=sup_{n>=1} r(n)^(1/n). Cartesian products make r supermultiplicative and the limit exists.Combinatorics & discrete geometry

    Exact statement

    Let r(n) be the maximum cardinality of a subset A of F_3^n having no three distinct elements x,y,z with x+y+z=0. Determine the exact exponential growth constant c=lim_{n->infinity} r(n)^(1/n)=sup_{n>=1} r(n)^(1/n). Cartesian products make r supermultiplicative and the limit exists.

    Why it matters

    Fix the sharp exponential growth rate of maximal cap sets in F_3^n.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Exact Growth Constant of Maximal Cap Sets in F_3^n · 1 problem in this edition. Family membership alone does not assert an implication.

    • #460 · Exact Growth Constant of Maximal Cap Sets in F_3^n (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 515–617.

    Source-fit opponent count: 31.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  460. Rank461Bounded Excess Nuclear Charge in Large AtomsFor the nonrelativistic Coulomb atom with a fixed nucleus of charge Z and fermionic electrons, prove that the maximal bound electron number N_0(Z) satisfies N_0(Z)-Z = O(1) as Z tends to infinity.Mathematical physics

    Exact statement

    For the nonrelativistic Coulomb atom with a fixed nucleus of charge Z and fermionic electrons, prove that the maximal bound electron number N_0(Z) satisfies N_0(Z)-Z = O(1) as Z tends to infinity.

    Why it matters

    A proof would settle the constant-excess maximum-ionization conjecture for atoms and sharpen known asymptotic neutrality bounds.

    A counterexample would show that neutral atoms can bind an unbounded number of excess electrons, reshaping the expected large-Z ionization picture.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The manuscript reports a universal constant excess-charge bound for strictly bound atomic sectors, defined by E_n<E_(n−1), in the antisymmetric two-spin nonrelativistic Coulomb model. It explicitly leaves threshold ground states untreated. The maximal bound-electron formulation here does not settle whether threshold eigenstates count, so the strict-binding result does not close that broader formulation. The accompanying energy and radius results concern neutral atoms and do not supply the missing threshold bound. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Bounded Excess Nuclear Charge in Large Atoms · 1 problem in this edition. Family membership alone does not assert an implication.

    • #461 · Bounded Excess Nuclear Charge in Large Atoms (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 438–728.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  461. Rank462Linear Independence of Distinct Positive Zeta-Zero OrdinatesLet Gamma={gamma>0: zeta(beta+i gamma)=0 for some real beta with 0<beta<1}, as a set of distinct real values, not an indexed list counting multiplicities. For every finite set of distinct gamma_1,...,gamma_k in Gamma and rational c_1,...,c_k, the equality sum_i c_i gamma_i=0 implies c_1=...=c_k=0. No Riemann-hypothesis assumption is imposed.Number theory & arithmetic geometry

    Exact statement

    Let Gamma={gamma>0: zeta(beta+i gamma)=0 for some real beta with 0<beta<1}, as a set of distinct real values, not an indexed list counting multiplicities. For every finite set of distinct gamma_1,...,gamma_k in Gamma and rational c_1,...,c_k, the equality sum_i c_i gamma_i=0 implies c_1=...=c_k=0. No Riemann-hypothesis assumption is imposed.

    Why it matters

    Prove that every finite rational relation among distinct elements of Gamma is trivial, with Gamma containing all positive ordinates of nontrivial zeta zeros and no RH hypothesis.

    Prove that some finite collection of distinct elements of Gamma has a rational linear relation with coefficients not all zero. No explicit numerical exhibition is an additional requirement.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Linear Independence of Distinct Positive Zeta-Zero Ordinates · 1 problem in this edition. Family membership alone does not assert an implication.

    • #462 · Linear Independence of Distinct Positive Zeta-Zero Ordinates (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 508–628.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  462. Rank463Existence of EFX Allocations for Additive Indivisible GoodsFor every finite set of agents and indivisible goods with additive nonnegative valuations, determine whether there is a complete allocation such that for every pair of agents i,j and every good g in j's bundle, agent i values its own bundle at least as much as j's bundle after removing g.Theoretical computer science

    Exact statement

    For every finite set of agents and indivisible goods with additive nonnegative valuations, determine whether there is a complete allocation such that for every pair of agents i,j and every good g in j's bundle, agent i values its own bundle at least as much as j's bundle after removing g.

    Why it matters

    A proof would guarantee a complete EFX allocation for every additive nonnegative fair-division instance.

    A counterexample would demonstrate that complete EFX allocations do not always exist under additive nonnegative valuations.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Existence of EFX Allocations for Additive Indivisible Goods · 1 problem in this edition. Family membership alone does not assert an implication.

    • #463 · Existence of EFX Allocations for Additive Indivisible Goods (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 465–660.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  463. Rank464Symplectic Isotopy Conjecture for Smooth Surfaces in CP²Every smooth closed embedded symplectic surface in complex projective two-space with its standard Fubini–Study symplectic form is isotopic, through smooth embedded symplectic surfaces, to a smooth complex algebraic curve of the same degree.Geometry & topology

    Exact statement

    Every smooth closed embedded symplectic surface in complex projective two-space with its standard Fubini–Study symplectic form is isotopic, through smooth embedded symplectic surfaces, to a smooth complex algebraic curve of the same degree.

    Why it matters

    A proof would identify the symplectic isotopy class of each smooth plane curve with the algebraic class of the same degree.

    A counterexample would exhibit a smooth symplectic plane curve whose isotopy class is absent from smooth complex algebraic curves.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Symplectic Isotopy Conjecture for Smooth Surfaces in CP² · 1 problem in this edition. Family membership alone does not assert an implication.

    • #464 · Symplectic Isotopy Conjecture for Smooth Surfaces in CP² (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 526–620.

    Source-fit opponent count: 28.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  464. Rank465Exact value of the critical activity for the hard-core model on Z^2Determine the exact critical activity lambda_c(Z^2) for the hard-core Gibbs model on the square lattice and prove uniqueness of the infinite-volume Gibbs measure for lambda below it and phase coexistence for lambda above it.Probability, ergodic theory & dynamics

    Exact statement

    Determine the exact critical activity lambda_c(Z^2) for the hard-core Gibbs model on the square lattice and prove uniqueness of the infinite-volume Gibbs measure for lambda below it and phase coexistence for lambda above it.

    Why it matters

    An exact determination would close the gap between the rigorous uniqueness and coexistence ranges and establish the square-lattice phase diagram at every activity away from the threshold.

    A rigorous threshold different from the numerical value near 3.796 would refute the conjectured location while still resolving the exact-value and phase-separation target.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Exact value of the critical activity for the hard-core model on Z^2 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #465 · Exact value of the critical activity for the hard-core model on Z^2 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 501–650.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  465. Rank466Rohrlich Conjecture on Multiplicative Relations among Gamma Values at Rational PointsProve or disprove that every relation pi^(b/2) product_a Gamma(a)^m_a in Qbar, for integers b and finitely supported integers m_a at rational 0<a<1, follows multiplicatively, modulo nonzero algebraic factors, from the translation, reflection and multiplication formulas for the classical gamma function.Number theory & arithmetic geometry

    Exact statement

    Prove or disprove that every relation pi^(b/2) product_a Gamma(a)^m_a in Qbar, for integers b and finitely supported integers m_a at rational 0<a<1, follows multiplicatively, modulo nonzero algebraic factors, from the translation, reflection and multiplication formulas for the classical gamma function.

    Why it matters

    All multiplicative relations in the specified classical gamma-value class would be generated by the standard identities.

    Exhibit such a multiplicative relation and prove that it is not generated by the standard gamma identities modulo nonzero algebraic factors.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Rohrlich Conjecture on Multiplicative Relations among Gamma Values at Rational Points · 1 problem in this edition. Family membership alone does not assert an implication.

    • #466 · Rohrlich Conjecture on Multiplicative Relations among Gamma Values at Rational Points (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 501–642.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  466. Rank467Lang-Trotter Conjecture for Supersingular Primes of Non-CM Elliptic CurvesFor every elliptic curve E over Q without complex multiplication, prove or refute the Lang-Trotter asymptotic that the number of primes p < X at which E has supersingular reduction is asymptotic to C_E sqrt(X)/log X for a positive constant C_E depending on E.Number theory & arithmetic geometry

    Exact statement

    For every elliptic curve E over Q without complex multiplication, prove or refute the Lang-Trotter asymptotic that the number of primes p < X at which E has supersingular reduction is asymptotic to C_E sqrt(X)/log X for a positive constant C_E depending on E.

    Why it matters

    Would determine the expected quantitative distribution of supersingular reduction primes for non-CM elliptic curves over Q.

    Would disprove the Lang-Trotter heuristic in the supersingular t=0 case by finding a different asymptotic or no asymptotic.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Lang-Trotter Conjecture for Supersingular Primes of Non-CM Elliptic Curves · 1 problem in this edition. Family membership alone does not assert an implication.

    • #467 · Lang-Trotter Conjecture for Supersingular Primes of Non-CM Elliptic Curves (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 519–635.

    Source-fit opponent count: 42.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  467. Rank468Geometric Simple Connectivity of Closed Simply Connected Smooth Four-ManifoldsDetermine whether every closed, connected, simply connected smooth four-manifold admits a smooth handle decomposition with no 1-handles.Geometry & topology

    Exact statement

    Determine whether every closed, connected, simply connected smooth four-manifold admits a smooth handle decomposition with no 1-handles.

    Why it matters

    Simple connectivity would suffice for existence of a smooth handle decomposition without 1-handles in the closed four-dimensional case.

    A counterexample would separate simple connectivity from geometric simple connectivity for a closed smooth four-manifold.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Geometric Simple Connectivity of Closed Simply Connected Smooth Four-Manifolds · 1 problem in this edition. Family membership alone does not assert an implication.

    • #468 · Geometric Simple Connectivity of Closed Simply Connected Smooth Four-Manifolds (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 528–632.

    Source-fit opponent count: 25.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  468. Rank469Measure Zero of Newtonian Noncollision SingularitiesFor every integer N >= 5 and every fixed N-tuple of positive masses, prove that the subset of collision-free initial conditions in R^{6N} whose forward Newtonian solution has a finite maximal time and a noncollision singularity has Lebesgue measure zero.Mathematical physics

    Exact statement

    For every integer N >= 5 and every fixed N-tuple of positive masses, prove that the subset of collision-free initial conditions in R^{6N} whose forward Newtonian solution has a finite maximal time and a noncollision singularity has Lebesgue measure zero.

    Why it matters

    A proof that the stated set of collision-free initial conditions producing finite-time noncollision singularities has Lebesgue measure zero for every N and mass tuple in scope.

    A fixed N and positive mass tuple together with a proof that the stated set of initial conditions has positive Lebesgue measure.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Measure Zero of Newtonian Noncollision Singularities · 1 problem in this edition. Family membership alone does not assert an implication.

    • #469 · Measure Zero of Newtonian Noncollision Singularities (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 514–645.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  469. Rank470Liouville ergodicity of closed rank-one nonpositively curved geodesic flowsFor every closed connected smooth Riemannian manifold M of dimension at least two with nonpositive sectional curvature and rank one, is the geodesic flow on T1M ergodic with respect to normalized Liouville measure?Probability, ergodic theory & dynamics

    Exact statement

    For every closed connected smooth Riemannian manifold M of dimension at least two with nonpositive sectional curvature and rank one, is the geodesic flow on T1M ergodic with respect to normalized Liouville measure?

    Why it matters

    Prove that every invariant measurable subset of T1M has Liouville measure zero or one for all manifolds in the stated class.

    Construct a closed nonpositively curved rank-one manifold whose geodesic flow has an invariant measurable subset of intermediate Liouville measure.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Liouville ergodicity of closed rank-one nonpositively curved geodesic flows · 1 problem in this edition. Family membership alone does not assert an implication.

    • #470 · Liouville ergodicity of closed rank-one nonpositively curved geodesic flows (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 532–628.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  470. Rank471Output-Polynomial Enumeration of Minimal Hypergraph TransversalsDetermine whether all inclusion-minimal transversals of an arbitrary finite hypergraph can be enumerated in time polynomial in the combined size of the input hypergraph and the complete output list.Theoretical computer science

    Exact statement

    Determine whether all inclusion-minimal transversals of an arbitrary finite hypergraph can be enumerated in time polynomial in the combined size of the input hypergraph and the complete output list.

    Why it matters

    An output-polynomial algorithm would place unrestricted minimal-transversal generation within the stated output-sensitive time bound and would also solve the equivalent duality test within polynomial time.

    A qualifying lower bound would exclude output-polynomial enumeration under its stated assumption and delimit the unrestricted case from known polynomial restricted classes.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Output-Polynomial Enumeration of Minimal Hypergraph Transversals · 1 problem in this edition. Family membership alone does not assert an implication.

    • #471 · Output-Polynomial Enumeration of Minimal Hypergraph Transversals (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 494–650.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  471. Rank472Decidability of the Dot-Depth Hierarchy Membership RelationDoes there exist an algorithm which, given a nonnegative half-integer h and a deterministic finite automaton for a language L over a finite alphabet A, decides whether L belongs to B_h? Here B_0 is the class of finite or cofinite languages over A*, B_(n+1/2)=Pol(B_n), where Pol consists of finite unions of marked products L_0 a_1 L_1 ... a_k L_k with L_i in B_n and a_i in A, and B_(n+1) is the Boolean closure of B_(n+1/2).Theoretical computer science

    Exact statement

    Does there exist an algorithm which, given a nonnegative half-integer h and a deterministic finite automaton for a language L over a finite alphabet A, decides whether L belongs to B_h? Here B_0 is the class of finite or cofinite languages over A*, B_(n+1/2)=Pol(B_n), where Pol consists of finite unions of marked products L_0 a_1 L_1 ... a_k L_k with L_i in B_n and a_i in A, and B_(n+1) is the Boolean closure of B_(n+1/2).

    Why it matters

    A positive solution provides an effective test for each input language and dot-depth level, exactly as the uniform membership assertion requires.

    A negative solution establishes that the joint level-membership relation cannot be decided by a total algorithm, without claiming a particular fixed level is undecidable.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Decidability of the Dot-Depth Hierarchy Membership Relation · 1 problem in this edition. Family membership alone does not assert an implication.

    • #472 · Decidability of the Dot-Depth Hierarchy Membership Relation (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 529–632.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  472. Rank473Nonrecursive Combinatorial Interpretation of Kazhdan-Lusztig PolynomialsFor every Coxeter system and u less than or equal to v, construct an explicit finite combinatorial set M(u,v) and statistic s with P_{u,v}(q)=sum over a in M(u,v) of q raised to s(a), without a recursive definition of M.Algebra, representation & category theory

    Exact statement

    For every Coxeter system and u less than or equal to v, construct an explicit finite combinatorial set M(u,v) and statistic s with P_{u,v}(q)=sum over a in M(u,v) of q raised to s(a), without a recursive definition of M.

    Why it matters

    A construction would realize every Kazhdan-Lusztig polynomial as the generating function of an explicit finite nonrecursive combinatorial model.

    A nonexistence result would rule out such a finite nonrecursive model and statistic for all Coxeter intervals.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Nonrecursive Combinatorial Interpretation of Kazhdan-Lusztig Polynomials · 1 problem in this edition. Family membership alone does not assert an implication.

    • #473 · Nonrecursive Combinatorial Interpretation of Kazhdan-Lusztig Polynomials (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 490–685.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  473. Rank474Tutte's 4-Flow Conjecture for Petersen-Minor-Free GraphsProve or disprove that every finite bridgeless graph with no Petersen minor admits a nowhere-zero 4-flow.Combinatorics & discrete geometry

    Exact statement

    Prove or disprove that every finite bridgeless graph with no Petersen minor admits a nowhere-zero 4-flow.

    Why it matters

    A proof would settle this excluded-minor flow conjecture and strengthen the graph-coloring consequences beyond known planar and proper-minor cases.

    A counterexample would identify a bridgeless Petersen-minor-free graph with no nowhere-zero 4-flow.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Tutte's 4-Flow Conjecture for Petersen-Minor-Free Graphs · 1 problem in this edition. Family membership alone does not assert an implication.

    • #474 · Tutte's 4-Flow Conjecture for Petersen-Minor-Free Graphs (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 453–696.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  474. Rank475Gaussian Value Distribution for Eigenfunctions (Marklof, source fragment)Let X be a generic compact negatively curved Riemannian manifold with normalized volume mu, and let (phi_i) be an L2(mu)-normalized basis of Laplace eigenfunctions ordered by increasing eigenvalue. Prove that for every real a < b, mu({x in X : a <= phi_i(x) <= b}) converges to the standard Gaussian probability of [a,b] as i tends to infinity.Probability, ergodic theory & dynamics

    Exact statement

    Let X be a generic compact negatively curved Riemannian manifold with normalized volume mu, and let (phi_i) be an L2(mu)-normalized basis of Laplace eigenfunctions ordered by increasing eigenvalue. Prove that for every real a < b, mu({x in X : a <= phi_i(x) <= b}) converges to the standard Gaussian probability of [a,b] as i tends to infinity.

    Why it matters

    A proof would establish convergence of the eigenfunction pushforward measures to the standard Gaussian law for the stated generic negatively curved setting.

    A disproof would provide a qualifying genericity framework, manifold, eigenbasis, or interval for which the asserted Gaussian value-distribution limit fails.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Gaussian Value Distribution for Eigenfunctions (Marklof, source fragment) · 1 problem in this edition. Family membership alone does not assert an implication.

    • #475 · Gaussian Value Distribution for Eigenfunctions (Marklof, source fragment) (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 516–653.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  475. Rank476Plurality Is Stablest: Balanced Positive-Correlation ClauseFor every integer q>=3, 0<rho<1 and epsilon>0, prove or disprove that some tau>0 works for all n and all balanced functions f:[q]^n->Delta_q with Inf_i(f_j)=E[Var_{x_i}(f_j)]<=tau for every i,j: Stab_rho(f)<=lim_{m->infinity}Stab_rho(PLUR_{q,m})+epsilon. Here X is uniform on [q]^n, Y independently retains each coordinate with probability rho and otherwise resamples it uniformly, Stab_rho(f)=E[<f(X),f(Y)>], balance means E f_j=1/q, and PLUR chooses the plurality winner with symmetric tie treatment.Theoretical computer science

    Exact statement

    For every integer q>=3, 0<rho<1 and epsilon>0, prove or disprove that some tau>0 works for all n and all balanced functions f:[q]^n->Delta_q with Inf_i(f_j)=E[Var_{x_i}(f_j)]<=tau for every i,j: Stab_rho(f)<=lim_{m->infinity}Stab_rho(PLUR_{q,m})+epsilon. Here X is uniform on [q]^n, Y independently retains each coordinate with probability rho and otherwise resamples it uniformly, Stab_rho(f)=E[<f(X),f(Y)>], balance means E f_j=1/q, and PLUR chooses the plurality winner with symmetric tie treatment.

    Why it matters

    A proof establishes precisely the assertion stated in the formal statement; no further unsupported consequence is asserted.

    A rigorous negation establishes the failure of that assertion in its stated scope; no further unsupported consequence is asserted.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Plurality Is Stablest: Balanced Positive-Correlation Clause · 1 problem in this edition. Family membership alone does not assert an implication.

    • #476 · Plurality Is Stablest: Balanced Positive-Correlation Clause (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 530–636.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  476. Rank477Klingenberg Conjecture on Anosov Geodesic FlowsProve that every closed manifold admitting a Riemannian metric whose geodesic flow is Anosov also admits a Riemannian metric of negative sectional curvature.Geometry & topology

    Exact statement

    Prove that every closed manifold admitting a Riemannian metric whose geodesic flow is Anosov also admits a Riemannian metric of negative sectional curvature.

    Why it matters

    A proof would connect a dynamical hyperbolicity condition to the topology of manifolds admitting negative sectional curvature.

    A counterexample would separate Anosov geodesic dynamics from the existence of any negatively curved metric on the manifold.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Klingenberg Conjecture on Anosov Geodesic Flows · 1 problem in this edition. Family membership alone does not assert an implication.

    • #477 · Klingenberg Conjecture on Anosov Geodesic Flows (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 511–645.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  477. Rank478Stable Bernstein Problem for Six-Dimensional Minimal Hypersurfaces in R^7Prove or disprove that every connected, complete, two-sided, stable minimally immersed hypersurface M^6 in R^7 is a hyperplane.Geometry & topology

    Exact statement

    Prove or disprove that every connected, complete, two-sided, stable minimally immersed hypersurface M^6 in R^7 is a hyperplane.

    Why it matters

    A positive resolution would force every complete two-sided stable minimal M^6 in R^7 to be a hyperplane.

    A negative resolution would construct a connected complete nonflat stable minimal hypersurface M^6 in R^7.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Stable Bernstein Problem for Six-Dimensional Minimal Hypersurfaces in R^7 · 1 problem in this edition. Family membership alone does not assert an implication.

    • #478 · Stable Bernstein Problem for Six-Dimensional Minimal Hypersurfaces in R^7 (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 524–650.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  478. Rank479Coven–Meyerowitz Tiling Conjecture for Finite Sets of IntegersFor every nonempty finite A⊂Z that tiles Z by translations (there exists B⊂Z such that every integer has a unique representation a+b with a∈A and b∈B), translate A so min A=0 and set A(X)=Σ_{a∈A}X^a. Let S_A={p^k: p prime, k≥1, Φ_{p^k}(X) divides A(X)}. For every selection s_1,…,s_m∈S_A that are powers of pairwise different primes, prove or disprove that Φ_{s_1⋯s_m}(X) divides A(X).Analysis & PDE

    Exact statement

    For every nonempty finite A⊂Z that tiles Z by translations (there exists B⊂Z such that every integer has a unique representation a+b with a∈A and b∈B), translate A so min A=0 and set A(X)=Σ_{a∈A}X^a. Let S_A={p^k: p prime, k≥1, Φ_{p^k}(X) divides A(X)}. For every selection s_1,…,s_m∈S_A that are powers of pairwise different primes, prove or disprove that Φ_{s_1⋯s_m}(X) divides A(X).

    Why it matters

    Together with the established T1 necessity and T1/T2 sufficiency, a proof yields a cyclotomic characterization of finite integer tiles; the cited theorem also yields the bounded real-tile spectral consequence.

    A counterexample disproves necessity of T2 for finite integer tiles; this alone would not disprove the full Fuglede equivalence.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Coven–Meyerowitz Tiling Conjecture for Finite Sets of Integers · 1 problem in this edition. Family membership alone does not assert an implication.

    • #479 · Coven–Meyerowitz Tiling Conjecture for Finite Sets of Integers (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 514–655.

    Source-fit opponent count: 26.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  479. Rank480Exact value of the connective constant of the square latticeLet c_n be the number of n-step nearest-neighbor self-avoiding walks from the origin in the square lattice Z^2, and set mu = lim as n tends to infinity of c_n^(1/n). Determine mu exactly.Probability, ergodic theory & dynamics

    Exact statement

    Let c_n be the number of n-step nearest-neighbor self-avoiding walks from the origin in the square lattice Z^2, and set mu = lim as n tends to infinity of c_n^(1/n). Determine mu exactly.

    Why it matters

    An exact evaluation would determine the exponential growth rate of square-lattice self-avoiding walks.

    A proof that a proposed expression is wrong would eliminate that expression but would not alone determine mu exactly.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Exact value of the connective constant of the square lattice · 1 problem in this edition. Family membership alone does not assert an implication.

    • #480 · Exact value of the connective constant of the square lattice (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 456–734.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  480. Rank481Infinitude of Sophie Germain PrimesAre there infinitely many primes p for which doubling p and adding one also gives a prime, forming a pair of primes p and 2p+1?Number theory & arithmetic geometry

    Exact statement

    Determine whether there are infinitely many Sophie Germain primes, that is, primes p such that 2p + 1 is also prime.

    Why it matters

    A resolution would decide whether this linked prime pattern occurs at arbitrarily large scales. A positive answer would establish infinitude without requiring a count or density formula for the pairs.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Infinitude of Sophie Germain Primes · 1 problem in this edition. Family membership alone does not assert an implication.

    • #481 · Infinitude of Sophie Germain Primes (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 490–679.

    Source-fit opponent count: 35.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  481. Rank482Fully Dynamic All-Pairs Shortest Paths in Subquadratic Update TimeDesign a fully dynamic data structure for all-pairs shortest paths in directed weighted graphs supporting edge updates in worst-case or amortized subquadratic time O(n^{2-epsilon}) while maintaining exact distance queries in O(1) time.Theoretical computer science

    Exact statement

    Design a fully dynamic data structure for all-pairs shortest paths in directed weighted graphs supporting edge updates in worst-case or amortized subquadratic time O(n^{2-epsilon}) while maintaining exact distance queries in O(1) time.

    Why it matters

    A data structure would give exact directed all-pairs distances with subquadratic updates and constant-time queries.

    A lower bound would rule out that update-query combination for exact fully dynamic directed shortest paths.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Fully Dynamic All-Pairs Shortest Paths in Subquadratic Update Time · 1 problem in this edition. Family membership alone does not assert an implication.

    • #482 · Fully Dynamic All-Pairs Shortest Paths in Subquadratic Update Time (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 528–651.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  482. Rank483Edge Reconstruction Conjecture for GraphsProve or disprove that every finite simple undirected graph G with at least four edges is determined up to isomorphism by its edge deck: the multiset of isomorphism classes of graphs G-e, one for each edge e, obtained by deleting only that edge and retaining every vertex, including isolated vertices. Thus if two such graphs have equal edge decks, they must be isomorphic. No labels identifying the removed edges are supplied.Combinatorics & discrete geometryGraph reconstruction · 2 problems

    Exact statement

    Prove or disprove that every finite simple undirected graph G with at least four edges is determined up to isomorphism by its edge deck: the multiset of isomorphism classes of graphs G-e, one for each edge e, obtained by deleting only that edge and retaining every vertex, including isolated vertices. Thus if two such graphs have equal edge decks, they must be isomorphic. No labels identifying the removed edges are supplied.

    Why it matters

    A proof establishes uniqueness of the original isomorphism class from every allowed edge deck, without adding a running-time bound.

    A verified nonisomorphic pair establishes that the edge-deck data do not always identify the original finite simple graph.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Related problems

    • #298 · Graph reconstruction conjectureVariantGraph reconstruction conjecture

      A proof of #298 would also establish this assertion.

      Scope qualification: The recorded direction concerns positive assertions only; no converse or whole binary-completion equivalence is asserted.

    Problem family

    Graph reconstruction · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 533–643.

    Source-fit opponent count: 27.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  483. Rank484Uniqueness of Finite Projective Planes of Prime Order pProve that every finite projective plane of prime order p is isomorphic to the Desarguesian projective plane PG(2,p) over the field F_p.Cryptography, coding, information & optimizationFinite projective planes · 2 problems

    Exact statement

    Prove that every finite projective plane of prime order p is isomorphic to the Desarguesian projective plane PG(2,p) over the field F_p.

    Why it matters

    A proof would classify every prime-order finite projective plane as the field plane PG(2,p).

    A disproof would construct a non-Desarguesian finite projective plane of prime order.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Finite projective planes · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 537–644.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  484. Rank485Lonely Runner ConjectureProve that for every integer n>=1 and every set of n distinct positive integer speeds s_1,...,s_n, there exists a real time t such that ||s_i t|| >= 1/(n+1) for every i, where ||x|| is distance to the nearest integer. Equivalently, among n+1 runners with pairwise distinct constant real speeds starting together on the unit circle, each runner is at some time at circular distance at least 1/(n+1) from every other runner.Combinatorics & discrete geometry

    Exact statement

    Prove that for every integer n>=1 and every set of n distinct positive integer speeds s_1,...,s_n, there exists a real time t such that ||s_i t|| >= 1/(n+1) for every i, where ||x|| is distance to the nearest integer. Equivalently, among n+1 runners with pairwise distinct constant real speeds starting together on the unit circle, each runner is at some time at circular distance at least 1/(n+1) from every other runner.

    Why it matters

    A proof would guarantee a time at which all stated fractional parts are at least 1/(n+1) from an integer.

    A counterexample would be a finite distinct speed set for which no time achieves the required simultaneous separation.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Lonely Runner Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #485 · Lonely Runner Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 529–657.

    Source-fit opponent count: 23.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  485. Rank486Etingof-Nikshych-Ostrik Conjecture that Every Fusion Category Admits a Pivotal StructureDetermine whether every fusion category over an algebraically closed field of characteristic zero admits a pivotal structure, that is, a monoidal natural isomorphism from the identity functor to the double-dual functor.Algebra, representation & category theory

    Exact statement

    Determine whether every fusion category over an algebraically closed field of characteristic zero admits a pivotal structure, that is, a monoidal natural isomorphism from the identity functor to the double-dual functor.

    Why it matters

    Every fusion category in this scope would admit a pivotal structure.

    Construct a fusion category over such a field for which no monoidal natural isomorphism Id => (−)** exists.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Etingof-Nikshych-Ostrik Conjecture that Every Fusion Category Admits a Pivotal Structure · 1 problem in this edition. Family membership alone does not assert an implication.

    • #486 · Etingof-Nikshych-Ostrik Conjecture that Every Fusion Category Admits a Pivotal Structure (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 534–653.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  486. Rank487Decidability of the Word Problem for General Artin-Tits GroupsDecide whether every Artin-Tits group associated with a finite Coxeter matrix has a decidable word problem: for each such fixed group there is an algorithm which, given a finite word in its standard generators and their inverses, decides whether the word is the identity. No Coxeter-type restriction is imposed. A uniform algorithm from the matrix and word would suffice, but uniformity is not imposed beyond the classical every-group formulation.Geometry & topologyGeometry and topology of Artin groups · 2 problems

    Exact statement

    Decide whether every Artin-Tits group associated with a finite Coxeter matrix has a decidable word problem: for each such fixed group there is an algorithm which, given a finite word in its standard generators and their inverses, decides whether the word is the identity. No Coxeter-type restriction is imposed. A uniform algorithm from the matrix and word would suffice, but uniformity is not imposed beyond the classical every-group formulation.

    Why it matters

    A positive resolution establishes the full assertion in the reviewed target and all of its stated special cases.

    Exhibit one Artin-Tits group associated with a finite Coxeter matrix and prove that no algorithm decides its identity-word language.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Geometry and topology of Artin groups · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 537–653.

    Source-fit opponent count: 22.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  487. Rank488Density of Diffeomorphisms with Trivial CentralizerFor every compact manifold M and r ≥ 1, determine whether each C^r diffeomorphism of M can be C^r-approximated by a diffeomorphism T whose centralizer consists exactly of the iterates {T^k : k ∈ Z}.Probability, ergodic theory & dynamics

    Exact statement

    For every compact manifold M and r ≥ 1, determine whether each C^r diffeomorphism of M can be C^r-approximated by a diffeomorphism T whose centralizer consists exactly of the iterates {T^k : k ∈ Z}.

    Why it matters

    Would establish the affirmative mathematical outcome specified for Density of Diffeomorphisms with Trivial Centralizer: For every compact manifold M and r ≥ 1, determine whether each C^r diffeomorphism of M can be C^r-approximated by a diffeomorphism T whose centralizer consists exactly of the iterates {T^k : k ∈ Z}.

    Would establish the negative outcome for Density of Diffeomorphisms with Trivial Centralizer by a counterexample or impossibility proof, without asserting more than the failed clause.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Density of Diffeomorphisms with Trivial Centralizer · 1 problem in this edition. Family membership alone does not assert an implication.

    • #488 · Density of Diffeomorphisms with Trivial Centralizer (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 464–702.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  488. Rank489Lehmer's Conjecture on the Nonvanishing of the Ramanujan Tau FunctionProve that Ramanujan's tau function never vanishes: tau(n) != 0 for every positive integer n, where tau(n) are the Fourier coefficients of the weight-12 cusp form Delta(z) = q prod_{k>=1} (1-q^k)^24.Number theory & arithmetic geometry

    Exact statement

    Prove that Ramanujan's tau function never vanishes: tau(n) != 0 for every positive integer n, where tau(n) are the Fourier coefficients of the weight-12 cusp form Delta(z) = q prod_{k>=1} (1-q^k)^24.

    Why it matters

    A proof would establish nonvanishing of every Ramanujan tau coefficient at positive index.

    A counterexample would identify a positive integer n with tau(n)=0.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Lehmer's Conjecture on the Nonvanishing of the Ramanujan Tau Function · 1 problem in this edition. Family membership alone does not assert an implication.

    • #489 · Lehmer's Conjecture on the Nonvanishing of the Ramanujan Tau Function (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 483–728.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  489. Rank490Reed Conjecture on Chromatic NumberEvery graph G satisfies chi(G)<=ceil((Delta(G)+1+omega(G))/2), where chi is chromatic number, Delta is maximum degree, and omega is clique number.Combinatorics & discrete geometry

    Exact statement

    Every graph G satisfies chi(G)<=ceil((Delta(G)+1+omega(G))/2), where chi is chromatic number, Delta is maximum degree, and omega is clique number.

    Why it matters

    A proof would establish the conjectured universal interpolation between the maximum-degree and clique lower bounds on chromatic number.

    A counterexample would show that maximum degree and clique number do not control chromatic number at Reed's proposed midpoint.

    Status

    Open · subcases solved

    October 7, 2026: OpenAI’s October 6 release reports partial progress. For each fixed forbidden clique K_r, the stated ordinary-coloring consequence O_r(Delta/log Delta) is eventually below Delta/2 and hence implies Reed's inequality for sufficiently large Delta within that fixed-clique class. The threshold depends on r, and the source gives no all-Delta bound when clique size grows with Delta. This limited asymptotic consequence does not resolve Reed's universal inequality or claim novelty over existing subcase bounds. No new claim of priority for this implication is made. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Reed Conjecture on Chromatic Number · 1 problem in this edition. Family membership alone does not assert an implication.

    • #490 · Reed Conjecture on Chromatic Number (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 489–695.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  490. Rank491Decidability of Multiplicative-Exponential Linear LogicDetermine whether provability is decidable for unit-free propositional multiplicative-exponential linear logic (MELL): given a finite sequent built from propositional atoms and their linear negations using tensor, par, !, and ?, decide whether it has a derivation in the standard MELL sequent calculus.Logic, foundations & set theoryOpen · disputed claim

    Exact statement

    Determine whether provability is decidable for unit-free propositional multiplicative-exponential linear logic (MELL): given a finite sequent built from propositional atoms and their linear negations using tensor, par, !, and ?, decide whether it has a derivation in the standard MELL sequent calculus.

    Why it matters

    A positive resolution would establish decidability of provability for this exact unit-free MELL calculus, without an efficiency guarantee.

    A negative resolution would establish undecidability of provability for this exact unit-free MELL calculus, not merely delimit a fragment or refute one proposed algorithm.

    Status

    Open · disputed claim

    Status review date not recorded in this edition

    Sources

    Problem family

    Decidability of Multiplicative-Exponential Linear Logic · 1 problem in this edition. Family membership alone does not assert an implication.

    • #491 · Decidability of Multiplicative-Exponential Linear Logic (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 551–646.

    Source-fit opponent count: 40.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  491. Rank492Broué–Malle–Michel Canonical Symmetrizing-Trace ConjectureFor every finite irreducible complex reflection group W, let B(W) be its braid group and R=Z[u_(s,j)^(+/-1)] the generic Laurent parameter ring, with parameters identified for conjugate distinguished reflections s and j=1,...,e_s. Let H(W) be R[B(W)] modulo product_j(T_s-u_(s,j))=0 for each distinguished braid reflection. Is there an R-linear symmetrizing trace tau:H(W)->R, meaning tau(xy)=tau(yx) and the pairing (x,y)->tau(xy) identifies H(W) with its R-dual, such that specialization u_(s,j)=exp(2*pi*i*j/e_s) gives the identity-coefficient trace on the group algebra, and tau(T_(b^-1))^* tau(T_pi)=tau(T_(b*pi)) for every braid b? Here * inverts all parameters, z is the standard positive generator of the braid-group center, and pi=z^|Z(W)| is the full twist. The specialization makes tau(T_pi) nonzero, so the last identity is equivalent to the usual quotient normalization. Prove existence for all W or give a counterexample. Reducible reflection groups are covered by tensor products; no explicit lifted basis or algorithm is additionally demanded.Representation theory

    Exact statement

    For every finite irreducible complex reflection group W, let B(W) be its braid group and R=Z[u_(s,j)^(+/-1)] the generic Laurent parameter ring, with parameters identified for conjugate distinguished reflections s and j=1,...,e_s. Let H(W) be R[B(W)] modulo product_j(T_s-u_(s,j))=0 for each distinguished braid reflection. Is there an R-linear symmetrizing trace tau:H(W)->R, meaning tau(xy)=tau(yx) and the pairing (x,y)->tau(xy) identifies H(W) with its R-dual, such that specialization u_(s,j)=exp(2*pi*i*j/e_s) gives the identity-coefficient trace on the group algebra, and tau(T_(b^-1))^* tau(T_pi)=tau(T_(b*pi)) for every braid b? Here * inverts all parameters, z is the standard positive generator of the braid-group center, and pi=z^|Z(W)| is the full twist. The specialization makes tau(T_pi) nonzero, so the last identity is equivalent to the usual quotient normalization. Prove existence for all W or give a counterexample. Reducible reflection groups are covered by tensor products; no explicit lifted basis or algorithm is additionally demanded.

    Also known as

    BMM symmetrising trace conjecture; Canonical trace for generic complex-reflection Hecke algebras

    Why it matters

    The generic Hecke algebras would have the specified canonical symmetric-algebra structure and normalization throughout the complex-reflection setting.

    A generic Hecke algebra would fail to have that canonical normalized trace, despite the separate freeness theorem.

    Status

    Open · subcases solved

    Status reviewed

    Sources

    Problem family

    Broué–Malle–Michel Canonical Symmetrizing-Trace Conjecture · 1 problem in this edition. Family membership alone does not assert an implication.

    • #492 · Broué–Malle–Michel Canonical Symmetrizing-Trace Conjecture (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 544–652.

    Source-fit opponent count: 24.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  492. Rank493Strict Convexity of the Planar First-Passage Percolation Limit ShapeFor nearest-neighbor first-passage percolation on Z^2, let edge weights be iid nonnegative real variables with a common law absolutely continuous with respect to Lebesgue measure and E[exp(a t_e)]<infinity for some a>0. Let B be the deterministic asymptotic limit shape from the shape theorem. Assertion: for every such law, B is strictly convex, meaning that its boundary contains no nontrivial line segment.Probability, ergodic theory & dynamics

    Exact statement

    For nearest-neighbor first-passage percolation on Z^2, let edge weights be iid nonnegative real variables with a common law absolutely continuous with respect to Lebesgue measure and E[exp(a t_e)]<infinity for some a>0. Let B be the deterministic asymptotic limit shape from the shape theorem. Assertion: for every such law, B is strictly convex, meaning that its boundary contains no nontrivial line segment.

    Why it matters

    Prove that the limit shape has no nontrivial boundary segment for every in-scope nonnegative absolutely-continuous law with a finite exponential moment.

    Prove existence of one in-scope law whose deterministic limit shape has a nontrivial boundary segment; no explicit density or formula is a further success requirement.

    Status

    Open

    October 7, 2026: OpenAI’s October 6 release reports partial progress. The full convexity assertion covers every absolutely continuous nonnegative weight law having some exponential moment. Exponential weights form a valid subcase, and the source proves strict convexity and C1 boundary for that subcase. Its Gamma extension only states differentiability, which does not imply strict convexity for arbitrary Gamma laws or for the target's full distribution class. No independent proof audit or Lean verification was performed for this update.

    Status reviewed

    Sources

    Problem family

    Strict Convexity of the Planar First-Passage Percolation Limit Shape · 1 problem in this edition. Family membership alone does not assert an implication.

    • #493 · Strict Convexity of the Planar First-Passage Percolation Limit Shape (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 558–657.

    Source-fit opponent count: 33.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  493. Rank494Stable Fields ConjectureProve that every infinite field whose first-order theory in the language of rings is stable is separably closed, or exhibit an infinite stable field that is not separably closed.Logic, foundations & set theoryNIP fields · 2 problems

    Exact statement

    Prove that every infinite field whose first-order theory in the language of rings is stable is separably closed, or exhibit an infinite stable field that is not separably closed.

    Why it matters

    A proof would classify every infinite stable field as separably closed, extending the known subclass results to the unrestricted statement.

    A counterexample would provide an infinite field with stable ring theory and a nontrivial finite separable algebraic extension.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    NIP fields · 2 problems in this edition. Family membership alone does not assert an implication.

    Ranking uncertainty

    90% source-fit bootstrap rank range: 534–675.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  494. Rank495Finite-Time Interface Singularities for the 2D Muskat Problem in the Stable RegimeDetermine whether the two-dimensional Muskat problem on the whole plane admits a smooth Rayleigh-Taylor stable initial interface whose evolution develops a finite-time interface-curve singularity while in the stable regime.Analysis & PDE

    Exact statement

    Determine whether the two-dimensional Muskat problem on the whole plane admits a smooth Rayleigh-Taylor stable initial interface whose evolution develops a finite-time interface-curve singularity while in the stable regime.

    Why it matters

    An existence proof would produce whole-plane stable-regime data and a finite-time mechanism for loss of interface regularity under the Muskat evolution.

    A nonexistence proof would establish a whole-plane regularity obstruction preventing finite-time interface singularities throughout the stable regime.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Finite-Time Interface Singularities for the 2D Muskat Problem in the Stable Regime · 1 problem in this edition. Family membership alone does not assert an implication.

    • #495 · Finite-Time Interface Singularities for the 2D Muskat Problem in the Stable Regime (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 514–689.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  495. Rank496Total Coloring Conjecture on the Total Chromatic NumberProve that for every simple graph G with maximum degree Delta(G), the total chromatic number chi''(G), the least number of colors needed to color vertices and edges so that adjacent or incident elements receive distinct colors, satisfies chi''(G) <= Delta(G) + 2.Combinatorics & discrete geometry

    Exact statement

    Prove that for every simple graph G with maximum degree Delta(G), the total chromatic number chi''(G), the least number of colors needed to color vertices and edges so that adjacent or incident elements receive distinct colors, satisfies chi''(G) <= Delta(G) + 2.

    Why it matters

    A positive resolution would color all vertices and edges of every simple graph with at most Delta+2 colors.

    A negative resolution would exhibit a simple graph whose total chromatic number exceeds Delta+2.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Total Coloring Conjecture on the Total Chromatic Number · 1 problem in this edition. Family membership alone does not assert an implication.

    • #496 · Total Coloring Conjecture on the Total Chromatic Number (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 539–689.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  496. Rank497Birch-Tate Conjecture for Totally Real Number FieldsFor every totally real number field F, prove the Birch-Tate formula |zeta_F(-1)| = |K_2(O_F)|/w_2(F).Number theory & arithmetic geometry

    Exact statement

    For every totally real number field F, prove the Birch-Tate formula |zeta_F(-1)| = |K_2(O_F)|/w_2(F).

    Why it matters

    A proof would complete the special-value formula relating Dedekind zeta values to the tame kernel K_2.

    A counterexample would overturn the expected integral 2-primary relation between zeta values and algebraic K-theory.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Birch-Tate Conjecture for Totally Real Number Fields · 1 problem in this edition. Family membership alone does not assert an implication.

    • #497 · Birch-Tate Conjecture for Totally Real Number Fields (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 562–665.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  497. Rank498Kelvin Problem on Least-Area Partitions of Space into Unit-Volume CellsDetermine a partition of three-dimensional Euclidean space into cells of equal unit volume that minimizes the average surface area of the interfaces (perimeter density), and in particular decide whether the Weaire-Phelan structure is optimal.Geometry & topology

    Exact statement

    Determine a partition of three-dimensional Euclidean space into cells of equal unit volume that minimizes the average surface area of the interfaces (perimeter density), and in particular decide whether the Weaire-Phelan structure is optimal.

    Why it matters

    An optimality proof would identify a unit-volume space partition of minimum perimeter density, potentially certifying Weaire-Phelan.

    A lower-area partition would disprove Weaire-Phelan optimality and establish a smaller attainable perimeter density.

    Status

    Open · subcases solved

    Status review date not recorded in this edition

    Sources

    Problem family

    Kelvin Problem on Least-Area Partitions of Space into Unit-Volume Cells · 1 problem in this edition. Family membership alone does not assert an implication.

    • #498 · Kelvin Problem on Least-Area Partitions of Space into Unit-Volume Cells (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 482–776.

    Source-fit opponent count: 20.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Also in the Top 100 most famous open problems.

    Link to this problem
  498. Rank499Ivanov's Conjecture on Virtual First Homology of Mapping Class GroupsFor every closed connected orientable surface of genus g≥3 with n≥0 punctures, and every finite-index subgroup Γ of its orientation-preserving mapping class group Mod_(g,n), prove H_1(Γ;Q)=0, equivalently that Γ has no surjection onto Z.Geometry & topology

    Exact statement

    For every closed connected orientable surface of genus g≥3 with n≥0 punctures, and every finite-index subgroup Γ of its orientation-preserving mapping class group Mod_(g,n), prove H_1(Γ;Q)=0, equivalently that Γ has no surjection onto Z.

    Why it matters

    The full universal assertion would hold under exactly the stated hypotheses.

    A rigorous failure would determine the limit of the stated universal assertion.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Ivanov's Conjecture on Virtual First Homology of Mapping Class Groups · 1 problem in this edition. Family membership alone does not assert an implication.

    • #499 · Ivanov's Conjecture on Virtual First Homology of Mapping Class Groups (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 562–656.

    Source-fit opponent count: 29.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
  499. Rank500Markus Conjecture on Completeness of Compact Affine Manifolds with Parallel VolumeDetermine whether every compact connected smooth manifold without boundary carrying a flat torsion-free affine connection and a nowhere-vanishing parallel volume form is geodesically complete for that affine connection.Geometry & topology

    Exact statement

    Determine whether every compact connected smooth manifold without boundary carrying a flat torsion-free affine connection and a nowhere-vanishing parallel volume form is geodesically complete for that affine connection.

    Why it matters

    A positive resolution would establish that parallel volume forces completeness for the entire specified compact affine class.

    A negative resolution would establish that parallel volume is insufficient for completeness even within the specified compact affine class.

    Status

    Open

    Status review date not recorded in this edition

    Sources

    Problem family

    Markus Conjecture on Completeness of Compact Affine Manifolds with Parallel Volume · 1 problem in this edition. Family membership alone does not assert an implication.

    • #500 · Markus Conjecture on Completeness of Compact Affine Manifolds with Parallel Volume (this problem)
    Ranking uncertainty

    90% source-fit bootstrap rank range: 539–687.

    Source-fit opponent count: 21.

    The range describes variation in the ranking procedure; it is not uncertainty about the truth of the problem.

    Link to this problem
Resolved problem history154 problems · historical ranks retained

These exact targets have source-reported resolutions and are outside the current open directory. Ranks are labeled by their original website or source edition; a source rank does not imply that a problem appeared in the website’s Top 500. Their permanent links remain available here. The status update does not grant acceptance to a ProofAtlas proof or formalization package.

  1. Historical rank022Hilbert's Tenth Problem over the RationalsIs there an algorithm that always halts and correctly decides whether any given integer-coefficient polynomial, in any finite number of variables, has a rational solution?Logic, foundations & set theorySource-reported disproved

    Historical rank #22 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    There exists a total algorithm which, given any finite encoding of a multivariate polynomial f in Z[x_1,...,x_n] with arbitrary finite n and integer coefficients, halts and correctly decides whether there is q in Q^n with f(q)=0. Standard computably interconvertible polynomial encodings are equivalent; no uniform running-time complexity bound is part of the assertion.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 004: The target is the existence of a total decision algorithm for arbitrary finite-variable integer-coefficient polynomials having rational zeros. The principal abstract denies precisely that algorithm, explicitly making the number of variables part of the input. Its negative answer disproves the target assertion without changing coefficient field, solution field, or input arity.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A positive answer would supply a universal decision procedure for rational solvability, with no efficiency guarantee. A negative answer would establish that no algorithm can settle every polynomial input while always terminating.

    Earlier status note

    Open

    Hilbert's tenth problem over Q remains open. Results for finitely generated rings or languages with extra height predicates do not give a decision/undecidability result for ordinary rational Diophantine solvability.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  2. Historical rank029Unique Games ConjectureIn a graph-labeling game, each edge requires the two labels to match through a one-to-one rule. Is distinguishing games where almost all rules can be satisfied from those where only a tiny fraction can be satisfied NP-hard, however small these two error thresholds are, with a fixed label set chosen for each pair of thresholds?Theoretical computer scienceSource-reported proved

    Historical rank #29 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every fixed pair of real constants epsilon, delta > 0 with epsilon + delta < 1, there exists an integer k >= 2, depending only on epsilon and delta, such that the following promise problem is NP-hard under deterministic polynomial-time many-one reductions. An instance is a finite directed constraint graph G=(V,E) with E nonempty and, for each edge e=(v,w), a permutation pi_e of [k]. For a labeling L:V->[k], an edge is satisfied if pi_e(L(v))=L(w); val(G) is the largest fraction of satisfied edges over all labelings. Distinguish val(G) >= 1-epsilon from val(G) <= delta, with no requirement on instances between these thresholds. The graph is the growing input; epsilon, delta and k are fixed constants. NP-hardness means that every language in NP has such a reduction mapping yes inputs to the first promised set and no inputs to the second.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 102: The manuscript asserts deterministic polynomial-time reductions from 3SAT to fixed-alphabet Unique Games with completeness 1-epsilon and soundness delta for every fixed epsilon,delta in (0,1/2). This covers the ranked range epsilon,delta>0, epsilon+delta<1: choose smaller positive parameters below 1/2 and below the requested values, giving stronger completeness and soundness promises. The alphabet remains fixed independently of the graph input. This is the ordinary UGC target, not merely a conditional approximation consequence.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would establish this hardness gap and support approximation thresholds obtained through Unique Games reductions. Claims of efficient-algorithm impossibility from NP-hardness would still require the assumption P ≠ NP.

    Earlier status note

    Open

    Classical UGC remains open. The accessible Hallelujah preprint's graphwise ratio strictly below 2 does not give the fixed uniform 2-epsilon improvement needed for its claimed contradiction. The linked journal full text was inaccessible; independent acceptance of a full disproof was not established.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Related problems

    • #234 · Unconditional 2−epsilon Inapproximability of Minimum Vertex CoverSource-reported proved · historical rankWeaker assertionUnconditional 2−epsilon Inapproximability of Minimum Vertex Cover

      The Unique Games Conjecture would imply the stated fixed-gap hardness of approximating vertex cover, by the Khot–Regev result.

      Scope qualification: The converse is not established here. A negative answer to Unique Games need not settle this vertex-cover hardness assertion.

    Research on ProofAtlas

    Formerly in the Top 100 most famous open problems.

    Link to this problem
  3. Historical rank048Nonexistence of Landau–Siegel zeros for quadratic Dirichlet L-functionsCan real zeros of primitive quadratic Dirichlet L-functions be kept uniformly away from one on the logarithmic scale? For conductor q, the measure of the character’s arithmetic complexity, is there one positive constant δ excluding zeros between 1 − δ/log q and 1 for every q ≥ 3?Number theory & arithmetic geometrySource-reported proved

    Historical rank #48 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Does there exist an absolute constant δ>0 such that, for every primitive quadratic Dirichlet character χ of conductor q≥3, L(β,χ)≠0 for every real β with 1−δ/log q<β<1? The logarithm is natural and the same δ must work for all conductors.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 003: The companion explicitly supplies one absolute positive c with (1-beta) log q >= c for every real zero of each primitive nonprincipal real Dirichlet L-function of conductor q>=3. Primitive quadratic characters are included, and setting delta=c excludes exactly the open real interval in the target, uniformly in conductor.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would remove the exceptional-real-zero possibility from this uniform region. It would eliminate that obstruction in applicable prime, class-number and L-value estimates while retaining their other hypotheses.

    Earlier status note

    Open

    The Landau–Siegel zero problem remains open. Recent special L-function and finite-range results are not universal Dirichlet zero exclusion. The September 14 review inspected Zhang v18; the related claim is now v19 dated September 14, so its new version was not independently adjudicated here and must not be represented as covered by the old assessment.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Related problems

    Formerly in the Top 100 most famous open problems.

    Link to this problem
  4. Historical rank052The Baum-Connes Conjecture for the K-Theory of Reduced Group C*-AlgebrasCan the topology of a group’s proper actions recover the K-theory invariants of its reduced operator algebra? The Baum–Connes conjecture asks whether the assembly map gives this exact match for every second-countable locally compact group.Algebra, representation & category theorySource-reported disproved

    Historical rank #52 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove the Baum-Connes conjecture for every second-countable locally compact group G: the assembly map K_top(G) → K(C*_r(G)) from the topological K-theory of the group, built from the classifying space for proper actions, to the K-theory of the reduced group C*-algebra is an isomorphism.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 285: The target asserts that coefficient-free reduced Baum-Connes assembly is an isomorphism for every second-countable locally compact group. A finitely generated torsion-free discrete group is in that class, and its infinite-order degree-zero kernel directly refutes injectivity. The companion irrational-trace example separately refutes surjectivity for a finitely generated discrete group. No coefficient-bearing or maximal-assembly substitution is used.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would recover the specified operator-algebra K-theory from topological group data through this map. A counterexample would show either information lost by the map or target classes it cannot reach.

    Earlier status note

    Open

    Coefficient-free Baum–Connes remains open. Corrected coefficient formulations and failed weakened permanence hypotheses are different statements. Finn–Sell's lacunary-group preprint was withdrawn May 12, 2026 for a norm-control gap and is not accepted subcase support.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  5. Historical rank078Abundance Conjecture for the minimal model programA nef divisor has nonnegative degree on every curve. For a minimal projective pair with Kawamata log terminal singularities, must some positive multiple of its canonical-plus-boundary divisor have enough global sections to define a map everywhere, with no common zero?Geometry & topologySource-reported proved

    Historical rank #78 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that for every minimal projective Kawamata log terminal pair (X,Delta), the nef adjoint divisor K_X+Delta is semiample.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 034: Coordinator convention ruling: the named classical projective abundance conjecture is interpreted in its characteristic-zero rational-boundary, Q-Cartier formulation. The positive-Cartier-multiple meaning of semiampleness in this contract is the rational formulation. No positive-characteristic or distinct real-divisor extension is added to its abbreviated historical wording. The unconditional projective Theorem 1.1 covers every klt minimal pair in that scope; the conditional Kahler companion is not the basis for closure. Statement reviewer context: Theorem 1.1 of the projective companion is unconditional as stated: over every algebraically closed characteristic-zero field, a normal projective log canonical pair with effective rational boundary, Q-Cartier nef adjoint, has a positive Cartier multiple generated by global sections. This includes all projective klt pairs in that rational-boundary characteristic-zero convention, and being minimal supplies the required nefness. There is no residual dimension, bigness, numerical-dimension, or nonvanishing assumption in the displayed theorem. Thus the conventional rational projective abundance conjecture is fully covered. The supplied historical dossier, however, repeats the shorthand target without defining coefficients, the Q-Cartier convention, or the ground field; its citation alone does not supply those omitted definitions. The remaining uncertainty is therefore an exact-target convention issue, not a missing rational-boundary abundance statement. If the coordinator confirms the classical rational characteristic-zero convention, disposition should be resolve/proved; if real-boundary semiampleness or positive characteristic was intended, that is a different scope requiring its own statement alignment. The separate compact-Kahler theorem explicitly assumes logarithmic Iitaka subadditivity and is unnecessary for the unconditional projective match.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would turn the nefness condition into semiampleness, producing the associated pluricanonical morphism. It would let every minimal pair in scope determine a geometric map through a suitable multiple of its adjoint divisor.

    Earlier status note

    Open

    Abundance remains open, with known restricted cases. Chern-class equality cases, boundary-restricted results and reductions of Kaehler abundance to projective abundance do not prove semiampleness for every stated projective klt pair.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  6. Historical rank080Hadwiger ConjectureIf a graph has no complete graph on t + 1 vertices as a minor, can its vertices always be colored with t colors so adjacent vertices have different colors, for every positive integer t? A minor is obtained by deleting vertices or edges and contracting edges.Combinatorics & discrete geometrySource-reported disproved

    Historical rank #80 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that every simple graph with no K_{t+1} minor is t-colourable for every integer t >= 1.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 157: The first paper gives finite simple graphs with chromatic number greater than their Hadwiger number h(G), even with fractional chromatic number greater than h(G). Taking t=h(G), each example has no K_{t+1} minor but is not t-colourable, directly contradicting the ranked all-t assertion. The separate linear list-coloring bound with an unspecified universal multiplier C does not restore the conjectured coefficient one.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make exclusion of a complete minor sufficient to control chromatic number. It would connect a graph’s coloring requirements with the complete structures obtainable by deleting and contracting its edges.

    Earlier status note

    Open

    Ordinary Hadwiger remains open. The September 9 Liu–Luo preprint improves an approximate chromatic bound, not the exact t-1 threshold; odd-minor counterexamples concern a stronger variant. October 6, 2026: Norin and Steiner (4 October) claim Ct-colourability for K_t-minor-free graphs, with an absolute C, and credit GPT-6 Astra. This linear bound does not establish the sharp t-1 colours required by ordinary Hadwiger. The complete proof was not audited in this review.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Research on ProofAtlas

    Formerly in the Top 100 most famous open problems.

    Link to this problem
  7. Historical rank087Bose-Einstein Condensation in Interacting Continuous Bose GasesCan Bose–Einstein condensation be proved for an interacting gas of bosons moving in continuous three-dimensional space at low temperatures? The target is persistent off-diagonal long-range order as particle number and volume grow together at fixed density.Mathematical physicsSource-reported proved

    Historical rank #87 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove mathematically that an interacting continuous Bose gas in three dimensions exhibits non-zero off-diagonal long-range order (Bose-Einstein condensation) in the thermodynamic limit at low temperatures.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 267: The ranked statement asks for an interacting three-dimensional continuum gas with thermodynamic-limit condensation at low temperatures. The bounded repulsive finite-range potential theorem supplies a constant-orbital condensate fraction uniformly positive at fixed small density for every temperature from zero to rho^2; the hard-sphere companion separately supplies a positive fixed temperature. These are interacting continuum systems with the thermodynamic limit taken before dilution, rather than a finite box or mean-field scaling. The mapping uses the target's existential interaction wording, not an assertion for all possible interactions.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would establish Bose–Einstein condensation in the stated interacting continuum setting. It would show that the relevant quantum correlations persist over arbitrarily large distances in the thermodynamic limit.

    Earlier status note

    Open

    Interacting continuum BEC remains provisionally open with an explicit claim/scope warning. Suto's 2023 high-density condensation claim potentially overlaps the existential wording, but acceptance and equivalence to fixed-density low-temperature thermodynamic ODLRO were not established. The independent April 2026 manuscript still calls interacting ODLRO open; interaction class and scaling need expert reconciliation.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Link to this problem
  8. Historical rank092Entanglement Area Law for 2D Gapped Quantum SystemsFor locally interacting quantum spins on a square lattice, with a unique ground state and a size-independent positive energy gap, must every region’s entanglement entropy be at most a constant times the number of edges crossing its boundary? The constant may depend on the fixed local parameters and gap, but not on the region or lattice size.Quantum information & computationSource-reported proved

    Historical rank #92 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove the two-dimensional ground-state area-law conjecture for finite-dimensional lattice spins: for families of finite 2D square-lattice Hamiltonians with fixed on-site dimension, uniformly bounded finite-range local interactions, a unique ground state, and a positive spectral-gap lower bound independent of system size, the ground-state entanglement entropy S(rho_A) is at most C times the boundary size |partial A| for every region A. The constant C may depend on the fixed local parameters and gap lower bound, but not on the system size or region. Here rho_A is the reduced pure-ground-state density operator and boundary size counts lattice edges crossing from A to its complement.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 265: The theorem uses precisely the target's fixed local dimension, finite interaction range, bounded strength, unique ground state, and a full-system gap uniform in size. It bounds entropy for every set of sites by the number of crossing lattice edges, with a size-independent constant, on arbitrary finite square-lattice domains. No frustration-free or local-gap condition is added.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would limit entanglement by boundary size throughout this class, including frustrated Hamiltonians. The constant could depend on fixed interaction parameters and the gap, while remaining valid as the system and chosen region grow.

    Earlier status note

    Open

    The universal two-dimensional gapped ground-state area law remains open. The 2026 AGSP theorem adds conditions available in restricted frustration-free locally gapped cases, not every globally gapped unique-ground-state Hamiltonian.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Link to this problem
  9. Historical rank114Iitaka conjecture C_{n,m} on subadditivity of Kodaira dimensionWhen one smooth projective variety is mapped algebraically onto another, is its Kodaira dimension—a measure of algebraic complexity—at least the sum of the base's Kodaira dimension and that of a general fibre?Geometry & topologySource-reported proved

    Historical rank #114 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove the Iitaka conjecture C_{n,m}: for a surjective morphism f: X -> Y between smooth projective varieties with general fibre F, kappa(X) >= kappa(Y) + kappa(F).

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 033: The ordinary-Iitaka companion's Theorem 1.1 is the ranked inequality verbatim for smooth connected projective varieties over any algebraically closed field of characteristic zero, with connected fibers and geometric generic fiber F. This resolves the named classical characteristic-zero Iitaka C_{n,m} target. The paper explicitly defines an algebraic fiber space as a surjective projective morphism with connected fibers. If the frozen abbreviated wording is read as allowing disconnected general fibers, the usual Stein-factorization reduction is harmless in characteristic zero: factor through a normal finite cover of Y, resolve the intermediate base and total space, apply the connected-fiber theorem, and use birational invariance of Kodaira dimension and nondecrease under the generically finite base map. The general smooth fiber has the same Kodaira dimension as the relevant geometric generic component; this yields the displayed inequality for the original morphism. Cases with a negative-infinity summand are vacuous. No positive-characteristic version is included. The publication result, rather than the much narrower linked Lean statement, supplies the resolution.

    Status reviewed

    Resolution sources

    Earlier ranking context

    Kodaira dimension measures the growth of pluricanonical forms. The inequality would let information about the base and typical fibre give a lower bound for this measure on the total space.

    Earlier statement and context sources

    Link to this problem
  10. Historical rank119The Borel Conjecture on Topological Rigidity of Aspherical ManifoldsFor closed manifolds whose universal covers contract to a point, can every homotopy equivalence—an equivalence up to continuous deformation—itself be deformed into a homeomorphism, a continuous bijection with continuous inverse?Geometry & topologySource-reported disproved

    Historical rank #119 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that every homotopy equivalence between closed aspherical manifolds is homotopic to a homeomorphism.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 320: The target asserts that every homotopy equivalence between closed aspherical manifolds is homotopic to a homeomorphism, with no dimension exclusion. The constructed closed connected aspherical topological four-manifolds are homotopy equivalent but not homeomorphic. Any equivalence between this pair therefore contradicts the universal target. Failure even of homeomorphism existence is sufficient; word-hyperbolicity of the common fundamental group adds a property and does not weaken the counterexample.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make homotopy information sufficient to recover the topology of these manifolds, while also controlling the particular map that relates them. A counterexample would expose a distinction invisible to homotopy equivalence.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  11. Historical rank124Aspherical Manifolds Admit No Metric of Positive Scalar CurvatureCan a closed manifold whose universal cover contracts to a point ever carry a Riemannian metric with scalar curvature positive everywhere?Geometry & topologySource-reported proved

    Historical rank #124 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that no closed aspherical manifold admits a Riemannian metric of positive scalar curvature.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 335: The companion asserts rational inessentiality for every closed connected oriented smooth positive-scalar-curvature manifold, without spin, group or upper-dimension hypotheses, and explicitly deduces the Gromov-Lawson aspherical conjecture. It further says every nonnegative-scalar-curvature metric on a closed aspherical manifold is flat, which is stronger than exclusion of strictly positive scalar curvature. Nonorientable manifolds can be handled through their oriented finite cover, preserving closedness, asphericity and positivity. The integral simplicial-volume inequality alone would not exclude aspherical manifolds of zero simplicial volume.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make asphericity a universal topological obstruction to positive scalar curvature. A counterexample would exhibit a manifold where this topology and everywhere positive scalar curvature can coexist.

    Earlier statement and context sources

    Link to this problem
  12. Historical rank136Erdős problem 3: divergent harmonic series and long arithmetic progressionsIf the reciprocals of the positive integers in a set have an infinite sum, must the set contain arbitrarily long sequences of equally spaced integers?Number theory & arithmetic geometrySource-reported proved

    Historical rank #136 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether every subset A of the natural numbers whose reciprocal sum diverges must contain arbitrarily long arithmetic progressions.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 159: The first assertion covers every set of positive integers with divergent reciprocal sum and yields arithmetic progressions of every finite length. This is precisely the frozen reciprocal-sum conjecture, with no positive-density hypothesis or restriction to a single progression length. The quantitative r_k(N) estimate supplies a stronger finite-range statement but is not needed to identify the exact affirmative resolution.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would turn the divergence of one numerical sum into a guarantee of arbitrarily long additive patterns. A counterexample would separate this measure of abundance from the existence of long arithmetic progressions.

    Earlier statement and context sources

    Research on ProofAtlas

    Formerly in the Top 100 most famous open problems.

    Link to this problem
  13. Historical rank139Cannon's conjectureIf a hyperbolic group's boundary at infinity is a two-dimensional sphere, must the group act by distance-preserving symmetries of hyperbolic three-space, properly discontinuously and with compact quotient?Geometry & topologySource-reported proved

    Historical rank #139 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that every Gromov-hyperbolic group whose boundary at infinity is homeomorphic to the two-sphere acts properly discontinuously, cocompactly, and isometrically on hyperbolic three-space, equivalently is virtually a cocompact Kleinian group.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 246: Both source and target quantify over every word/Gromov-hyperbolic group with boundary homeomorphic to S^2 and conclude a proper cocompact isometric action on H^3. The finite kernel permitted by the source is compatible with properness of a discrete group action; the target does not require faithful action. This is the full stated Cannon target, not merely the torsion-free three-manifold corollary.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would let the topology of a group's boundary determine the existence of a three-dimensional hyperbolic geometric realization. A counterexample would show that a spherical boundary alone does not force that realization.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  14. Historical rank145Residual Finiteness of Hyperbolic GroupsFor every word-hyperbolic group and each element other than its identity, is there a homomorphism into some finite group that keeps that element distinct from the identity?Algebra, representation & category theorySource-reported disproved

    Historical rank #145 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every word-hyperbolic group G and every nonidentity element g in G, prove that there is a finite group F and a homomorphism φ:G→F with φ(g)≠1.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 246: The family explicitly identifies each torsion-free S^2-boundary hyperbolic group as a closed hyperbolic three-manifold group. Such a group has a faithful finite-dimensional characteristic-zero linear representation and is finitely generated, so standard residual finiteness of finitely generated linear groups yields residual finiteness for this subclass. This implication does not assert residual finiteness for arbitrary hyperbolic groups or infer it solely from a finite-kernel action. Family 252 independently gives a counterexample to the universal target. Family 252: The target asserts that every nonidentity element of every word-hyperbolic group survives in some finite quotient. The source constructs a torsion-free word-hyperbolic group that is not residually finite, which negates exactly that universal assertion. Torsion-freeness is an additional feature of the counterexample. Nonlinearity over every field is not needed for this target and is not substituted for non-residual-finiteness. Family 258: Virtual compact specialness implies residual finiteness, so the second manuscript yields residual finiteness for every word-hyperbolic one-relator group. The target quantifies over all word-hyperbolic groups; the one-relator restriction is substantive and cannot be dropped. This is a consequence for a named subclass, not a full resolution.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would ensure that finite group images can detect every nonidentity element, although different elements may require different images. A counterexample would exhibit an element invisible in every finite quotient.

    Earlier statement and context sources

    Link to this problem
  15. Historical rank159Randomized Integer Word-RAM 3SUM ConjectureTheoretical computer scienceSource-reported disproved

    Historical rank #159 in the Oct 2, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every epsilon>0, there is no uniform Las Vegas randomized integer word-RAM algorithm that, on every set S of n distinct integers in [-n^3,n^3], decides whether three distinct elements sum to zero in worst-case expected O(n^(2-epsilon)) time. Use the standard logarithmic-word model of Abboud–Vassilevska Williams Conjecture 1, with enough Theta(log n) bits for the input universe, unit-cost word arithmetic, comparisons, Boolean operations, shifts and indirect memory access; no real-number oracle or free input-dependent preprocessing is permitted. Las Vegas means an always-correct answer with almost-sure termination, expectation over internal randomness, and a bound uniform over all valid inputs. Constants may depend on the algorithm and epsilon. This is not a literal quadratic lower bound.

    Status update

    Source-reported disproved

    The ranking source’s coordinator reports: Alman and Vassilevska Williams, arXiv:2610.06783v1 (5 October 2026), Theorem 22, gives a deterministic O(n^1.9992) integer word-RAM algorithm. The exact n^3 universe, distinctness, uniformity and worst-case bound match the ranked target; deterministic algorithms are always-correct Las Vegas algorithms using no random bits. Public Lean 4 formalization at the pinned commit passed Lean and Nanoda in the archived CI run; the coordinator rebuilt Solution.EndStatement and checked an exponent-3/injective-input scope bridge locally. This closes the lower-bound conjecture by disproof, not by proving the conjecture. Journal refereeing is not asserted.

    Status reviewed

    Resolution sources

    Link to this problem
  16. Historical rank161BPL = L (Derandomizing Bounded-Error Logspace)Can randomness always be removed from a polynomial-time decision algorithm that uses only logarithmic working memory and has error probability at most one-third, without increasing its memory beyond a logarithmic amount?Cryptography, coding, information & optimizationSource-reported proved

    Historical rank #161 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that every language decidable by a probabilistic Turing machine using O(log n) workspace, polynomial time, and bounded two-sided error is decidable by a deterministic Turing machine using O(log n) workspace.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 103: The catalogue states L=RL=BPL for polynomial-time randomized logarithmic-space machines, including bounded two-sided error, with an effective deterministic logarithmic-space compiler. These are the time, space, and error conditions of the ranked BPL=L assertion. No advice, nonuniformity, extra work space, or restricted input family is stated.

    Status reviewed

    Resolution sources

    Earlier ranking context

    Equality would remove randomness from every computation in this restricted-memory class without increasing the asymptotic space bound. A separation would show that randomness increases what logarithmic memory can decide.

    Earlier statement and context sources

    Related problems

    • #260 · RL versus LSource-reported proved · historical rankVariantRL versus L

      A proof of this assertion would also establish #260.

    Link to this problem
  17. Historical rank169The Free Group Factor Isomorphism ProblemCan the von Neumann algebras associated with free groups on different numbers of generators be isomorphic, in particular those associated with two and three generators?Algebra, representation & category theorySource-reported determined

    Historical rank #169 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether L(F_m) and L(F_n) are isomorphic when m != n, especially whether L(F_2) is isomorphic to L(F_3).

    Status update

    Source-reported determined

    Resolved (determined) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 287: The exact target asks whether free group factors with different generator counts are isomorphic, especially L(F_2) and L(F_3). The manuscript asserts an isomorphism of those tracial von Neumann algebras and explicitly invokes the free-group-factor alternative to include every interpolated factor and L(F_infinity). This determines the whole stated isomorphism question affirmatively, rather than only an amplification or embedding.

    Status reviewed

    Resolution sources

    Earlier ranking context

    The answer would determine whether these operator algebras retain the number of free generators or can lose that distinction. It tests what information survives the passage from a free group to its associated factor.

    Earlier statement and context sources

    Link to this problem
  18. Historical rank179Arnold Conjecture on Hamiltonian Fixed PointsFor Hamiltonian motion on a closed connected symplectic manifold, must its time-one transformation fix at least as many points as the fewest critical points a smooth real-valued function on that manifold can have?Geometry & topologySource-reported disproved

    Historical rank #179 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that every Hamiltonian diffeomorphism of an arbitrary closed connected symplectic manifold has at least Crit(M) fixed points, where Crit(M) is the minimum number of critical points of a smooth function on M.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 347: The ranked assertion is the unrestricted Crit(M) fixed-point bound, not merely its nondegenerate Morse-number or stable-Morse variants. The quadric threefold is closed and symplectic, and the stated Hamiltonian diffeomorphism has three fixed points although Crit(M)=4. Degeneracy is allowed by the target, so this is an exact counterexample. The other four manuscripts concern distinct stronger nondegenerate bounds and are not needed for this implication.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would turn a constraint on smooth functions into a universal lower bound on points left fixed by Hamiltonian motion. The minimum allows all smooth functions, so the precise critical-point count matters.

    Earlier statement and context sources

    Link to this problem
  19. Historical rank181Polyakov Conjecture: Absence of a Phase Transition for Planar O(n) Models with n >= 3In the nearest-neighbor O(n) model on a square lattice, spins are unit vectors with n components. For every n≥3 and every positive temperature, must correlations between distant spins decay exponentially with their separation?Mathematical physicsSource-reported proved

    Historical rank #181 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every integer n >= 3 and every finite inverse temperature beta > 0, prove that two-point correlations in the nearest-neighbor O(n) spin model on Z^2 decay exponentially with distance.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 215: The final manuscript quantifies over every n>=3 and every finite positive inverse temperature for the classical nearest-neighbor O(n) model on the square lattice and proves exponential decay of two-point correlations. This exactly matches the ranked Polyakov target. The separate O(3) continuum construction and O(4) mass asymptotics are not needed to supply missing n or temperature cases.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would give a finite correlation-length scale at every finite inverse temperature in these models. A counterexample would reveal a temperature and spin dimension where the specified exponential decay fails.

    Earlier statement and context sources

    Link to this problem
  20. Historical rank183Smooth anisotropic Calderón uniqueness from full boundary dataDo all boundary voltage measurements and their normal-derivative responses determine a smooth manifold's interior geometry up to a smooth change of coordinates fixing its boundary, for every compact connected manifold of dimension at least three with smooth nonempty boundary?Analysis & PDESource-reported proved

    Historical rank #183 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every compact connected smooth manifold M of dimension n>=3 with smooth nonempty boundary and every pair of smooth Riemannian metrics g1,g2 on that same M, let Lambda_gj f be the outward metric-unit normal derivative of the unique solution of Delta_gj u=0 with u|boundary=f. If Lambda_g1=Lambda_g2 as operators C-infinity(boundary M) to C-infinity(boundary M), compared under the identity marking on the whole boundary, must there be a smooth diffeomorphism Phi:M->M fixing the boundary pointwise with g2=Phi^*g1?

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 365: The scalar companion determines a smooth n>=3 metric from the zero-frequency energy form with data on any common open boundary patch. Choosing the entire boundary gives the required boundary-pointwise-fixed diffeomorphism. The target uses the metric-unit-normal DN operator: equality of these smooth operators also determines the same boundary metric (its principal symbol), so integration against the common induced boundary measure gives equal energy forms. The joint connection theorem is additional strength; no connection unknown is needed for this target.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make complete boundary measurements determine the interior metric up to this unavoidable freedom of coordinates. The uniqueness conclusion alone would not provide a reconstruction algorithm or stability under noisy measurements.

    Earlier statement and context sources

    Link to this problem
  21. Historical rank193Nonvanishing conjecture for projective log canonical pairsFor a complex projective log canonical pair with rational boundary, suppose its log canonical divisor lies in the closure of the cone of effective divisor classes. Must some positive multiple actually be linearly equivalent to an effective divisor?Geometry & topologySource-reported proved

    Historical rank #193 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Let (X, Delta) be a projective log canonical pair over the complex numbers, with Delta a Q-divisor and K_X+Delta Q-Cartier. If K_X+Delta is pseudo-effective, prove that kappa(K_X+Delta) >= 0; equivalently, some sufficiently divisible positive multiple of K_X+Delta is linearly equivalent to an effective divisor.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 034: The ordinary-pair Corollary 1.2 in family 036 applies to any projective log canonical Q-pair over an algebraically closed characteristic-zero field, with no Q-factoriality hypothesis. For a pseudo-effective adjoint D=K_X+Delta it gives a projective lc model (Y,Delta_Y), extracting no divisors, with D_Y nef and nondecreasing discrepancies. Theorem 1.1 of the projective characteristic-zero abundance manuscript in family 034 applies exactly to that model: its effective rational boundary and Q-Cartier nef adjoint have a positive Cartier multiple generated by global sections. The model comparison supplies the simple scope implication needed here: on a common resolution p:Z->X, q:Z->Y, the discrepancy inequalities give p*D=q*D_Y+E with E effective and q-exceptional; hence sufficiently divisible plurilog-canonical sections are preserved, and in particular a nonzero section on Y pulls back, multiplies by the canonical section of mE, and descends to a nonzero section of mD on X. A generated line bundle on the nonempty projective Y has a nonzero global section. Thus the two published statements together give kappa(D)>=0 for every pair in the frozen complex projective rational-boundary target. No arbitrary-MMP termination, generalized-nef-part semiampleness, or conditional Kahler theorem is required. Family 036: The ordinary-pair Corollary 1.2 in family 036 applies to any projective log canonical Q-pair over an algebraically closed characteristic-zero field, with no Q-factoriality hypothesis. For a pseudo-effective adjoint D=K_X+Delta it gives a projective lc model (Y,Delta_Y), extracting no divisors, with D_Y nef and nondecreasing discrepancies. Theorem 1.1 of the projective characteristic-zero abundance manuscript in family 034 applies exactly to that model: its effective rational boundary and Q-Cartier nef adjoint have a positive Cartier multiple generated by global sections. The model comparison supplies the simple scope implication needed here: on a common resolution p:Z->X, q:Z->Y, the discrepancy inequalities give p*D=q*D_Y+E with E effective and q-exceptional; hence sufficiently divisible plurilog-canonical sections are preserved, and in particular a nonzero section on Y pulls back, multiplies by the canonical section of mE, and descends to a nonzero section of mD on X. A generated line bundle on the nonempty projective Y has a nonzero global section. Thus the two published statements together give kappa(D)>=0 for every pair in the frozen complex projective rational-boundary target. No arbitrary-MMP termination, generalized-nef-part semiampleness, or conditional Kahler theorem is required.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would convert the numerical condition of pseudo-effectivity into an actual nonzero plurilog-canonical section. It would bridge a limiting positivity condition and the existence of an effective representative in the specified setting.

    Earlier statement and context sources

    Link to this problem
  22. Historical rank198Long-Range Order in the Quantum Heisenberg FerromagnetFor the nearest-neighbor isotropic quantum Heisenberg ferromagnet on a lattice of dimension at least three, can a translation-invariant equilibrium state have nonzero magnetization without an external field at every sufficiently low positive temperature, for every positive integer or half-integer spin?Mathematical physicsSource-reported proved

    Historical rank #198 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For the nearest-neighbor isotropic spin-S quantum Heisenberg ferromagnet on Z^d with Hamiltonian H=-sum_{<x,y>} S_x dot S_y, assert: for every integer d>=3 and every S in{1/2,1,3/2,...}, there is a finite beta_0(d,S)>0 such that for each finite beta>=beta_0 the infinite-volume zero-field model admits a translation-invariant equilibrium Gibbs state with nonzero spin expectation. Nonzero expectation refers to spontaneous magnetization in a symmetry-breaking state, not the rotation-invariant mixture. Coupling J=1 fixes temperature units; no pointwise-correlation equivalence is asserted.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 271: The exact target is spontaneous magnetization in a translation-invariant zero-field equilibrium state for every d>=3 and every positive quantum spin at all sufficiently low positive temperatures. The cited manuscript asserts those same quantifiers, the ferromagnetic sign and nearest-neighbor isotropic model, and a KMS state with magnetization at least S/4. No conversion from a rotation-invariant mixture or from a free-energy asymptotic is needed.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would establish a translation-invariant equilibrium state with nonzero magnetization despite zero applied field throughout the stated range. It would make collective magnetic order a consequence of this precise quantum model.

    Earlier statement and context sources

    Link to this problem
  23. Historical rank206Kaplansky's Zero-Divisor Conjecture for Torsion-Free GroupsTake a group with no nonidentity element of finite order, and form finite linear combinations of its elements with coefficients in any field. Can two nonzero combinations ever multiply to zero?Algebra, representation & category theorySource-reported disproved

    Historical rank #206 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that for every torsion-free group G and every field K, the group ring K[G] has no non-trivial zero-divisors, i.e. K[G] is a domain.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 196: The target universally quantifies over fields and torsion-free groups. The stated finitely presented torsion-free group with nonzero zero divisors in its ordinary scalar F2 group algebra is already a counterexample within that scope. Finite presentation and the two-dimensional classifying space are additional properties, not obstacles to the implication. Family 197: A domain is directly finite, so the announced failure of direct finiteness for a torsion-free F2 scalar group algebra entails nonzero zero divisors and refutes the universal zero-divisor target. The implication uses the torsion-free first manuscript; a failure for the torsion-containing odd-characteristic group alone would not refute this target.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would ensure that passing from a torsion-free group to its group ring creates no nontrivial zero divisors. A counterexample would display cancellation strong enough to annihilate the product of two nonzero elements.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  24. Historical rank211Hilbert-Smith ConjectureIf a locally compact topological group acts continuously on a connected manifold, with distinct group elements giving distinct symmetries, must it be a Lie group—a group with a finite-dimensional smooth structure compatible with its operations?Geometry & topologySource-reported proved

    Historical rank #211 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that every locally compact topological group acting faithfully on a connected topological manifold is a Lie group; equivalently, prove that Z_p admits no faithful action by homeomorphisms on a connected manifold.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 304: The manuscript forbids faithful jointly continuous actions of a non-Lie locally compact second-countable Hausdorff group on connected finite-dimensional standard topological manifolds without boundary. In particular, it excludes the compact second-countable non-Lie group Z_p in every finite dimension. The frozen target explicitly states this Z_p action formulation as equivalent, so the claimed theorem covers the canonical Hilbert-Smith target without extrapolating to nonstandard non-second-countable manifolds.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would strongly restrict which locally compact groups can serve as faithful symmetry groups of manifolds. It would rule out the specified p-adic actions even though only topological regularity is assumed.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  25. Historical rank213Falconer distance set conjectureIf a compact set in d-dimensional space has fractal (Hausdorff) dimension greater than d/2, must the distances between its points form a set of positive total length, for every d≥2?Combinatorics & discrete geometrySource-reported proved

    Historical rank #213 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that for every integer d>=2 and every compact set E in R^d with Hausdorff dimension greater than d/2, the distance set Delta(E)={|x-y|: x,y in E} has positive one-dimensional Lebesgue measure.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 073: The abstract agrees with every quantifier of the frozen Falconer target: every integer dimension d at least two, every compact subset of Euclidean d-space with Hausdorff dimension strictly greater than d/2, and positive Lebesgue measure of its Euclidean distance set. No pinned-distance, endpoint-equality or extra regularity hypothesis is substituted.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would turn a lower bound on fractal dimension into a guarantee of a substantial set of distance values. It would apply to distances between all pairs, without requiring any one point to realize that measure.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  26. Historical rank224Wall's Poincare Duality Group ConjectureA Poincaré duality group has the algebraic duality expected of a closed manifold. Must every finitely presented integral example of dimension at least three actually be the fundamental group of a closed topological manifold whose universal cover is contractible, with matching orientation character?Geometry & topologySource-reported disproved

    Historical rank #224 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every n≥3 and every finitely presented integral Poincaré duality group G of dimension n, prove that there is a closed aspherical topological n-manifold M with π1(M)≅G and matching orientation character.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 305: The target quantifies over every finitely presented integral PD_n group for n>=3 and asks for a closed aspherical topological n-manifold realization with matching orientation character. The manuscript provides an integral PD_4 group that is finitely presented and even has a finite classifying space, but is not the fundamental group of any such closed aspherical topological four-manifold. A single n=4 example refutes the universal target; no orientation-character match can repair total nonrealizability.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would realize each group satisfying these algebraic duality conditions as a manifold group of the prescribed kind. A counterexample would expose an additional obstruction to this geometric realization.

    Earlier statement and context sources

    Link to this problem
  27. Historical rank225K(pi,1) conjecture for Artin groupsThe quotient Salvetti construction builds a standard space from an Artin group's defining Coxeter system. For every such system with finitely many generators, is this space's universal cover contractible, making it a classifying space for the group?Geometry & topologySource-reported proved

    Historical rank #225 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every Coxeter system (W,S) with S finite and arbitrary Coxeter labels, let A_(W,S) be the associated Artin group and X_(W,S) its quotient Salvetti CW complex, with one |T|-cell for each T subset of S whose parabolic group W_T is finite and the standard attaching maps in Paolini Theorem 3.3 or Boyd's Salvetti construction modulo W. Prove that X_(W,S) is aspherical, hence a K(A_(W,S),1). Equivalently, prove the Deligne complex of cosets aA_T with W_T finite, ordered by inclusion, contractible. If an arrangement model is used, it is the W-quotient of the reflection-arrangement complement restricted using the interior of the Tits cone as in those sources, not an unrestricted complexified arrangement for infinite W. No spherical, affine, FC or other diagram-type restriction is imposed on W.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 254: The target is asphericity of the standard quotient Salvetti complex for every finite Coxeter matrix, including infinite Coxeter labels. The manuscript states asphericity for every Artin group with finitely many standard generators, without diagram-type restrictions. This matches the finite-rank canonical target; no infinite-rank assertion is needed.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would validate this explicit topological model for every Artin group in the stated class. Its universal cover would be contractible, leaving the fundamental group as its only nontrivial homotopy group.

    Earlier statement and context sources

    Link to this problem
  28. Historical rank234Unconditional 2−epsilon Inapproximability of Minimum Vertex CoverFor every fixed rational ε between 0 and 1, is it NP-hard to distinguish graphs with a vertex cover of size at most k from graphs whose smallest cover has size greater than (2−ε)k, using deterministic polynomial-time reductions and no Unique Games assumption?Theoretical computer scienceSource-reported proved

    Historical rank #234 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Also known as

    Vertex Cover hardness within every constant factor below2; Minimum Vertex Cover approximation threshold2

    Exact statement covered by this update

    For every fixed rational epsilon with 0<epsilon<1, prove that the following promise problem is NP-hard under deterministic polynomial-time many-one reductions: given a finite simple undirected graph G and a positive integer k, distinguish tau(G)<=k from tau(G)>(2-epsilon)k, where tau(G) is the minimum vertex-cover size. Equivalently, for every such fixed epsilon supply a polynomial-time gap-producing reduction from an NP-complete language, with a proof of completeness and soundness. The time polynomial may depend on epsilon. No Unique Games or Small-Set Expansion assumption is allowed. The assertion may instead be refuted by proving its negation. A uniform polynomial-time (2-epsilon0)-approximation algorithm would contradict the assertion under P!=NP; without that separation assumption it alone would not logically prove the promise problem non-NP-hard.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 102: The direct Vertex Cover manuscript asserts NP-hardness within every fixed factor below two on simple unweighted graphs, using established PCP/Label Cover hardness rather than a UGC assumption. This matches the ranked fixed rational 2-epsilon gap-hardness target on finite simple graphs. The family also independently asserts UGC, but the Vertex Cover mapping rests on its direct unconditional hardness manuscript. Family 106: The independent-set gap yields a genuine but weaker Vertex Cover gap: three-colorability gives alpha(G)>=n/3 and hence tau(G)<=2n/3, while soundness alpha(G)<delta*n gives tau(G)>(1-delta)*n. The resulting ratio approaches 3/2 as delta tends to zero, rather than two. Thus the stated reduction supports fixed-factor lower bounds below 3/2 and does not supply the ranked hardness for every factor 2-epsilon.

    Status reviewed

    Resolution sources

    Earlier ranking context

    This would establish hardness arbitrarily close to approximation factor two using ordinary NP-hardness alone. An approximation algorithm by itself refutes the hardness assertion only with an additional separation assumption such as P ≠ NP.

    Earlier statement and context sources

    Related problems

    • #29 · Unique Games ConjectureSource-reported proved · historical rankStronger assertionUnique Games Conjecture

      This fixed-gap vertex-cover hardness assertion follows from the Unique Games Conjecture. The vertex-cover target asks for the hardness unconditionally.

      Scope qualification: The converse is not established here. Proving this hardness assertion alone would not, by the relation recorded here, establish Unique Games.

    Link to this problem
  29. Historical rank239Universal Optimality of the Hexagonal Lattice in Dimension TwoAmong all closed discrete point sets of density one in the plane, does the triangular lattice minimize the lower-limit average interaction energy per point for every potential that is completely monotone in squared distance, meaning its successive derivatives alternate in sign, allowing zero?Combinatorics & discrete geometrySource-reported proved

    Historical rank #239 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Let A2 be the triangular lattice scaled to density1 in R^2. A configuration C is a discrete closed subset with lim(R->infinity)|C intersect B_R(0)|/area(B_R(0))=1. For p(r)=f(r^2), where f:(0,infinity)->R is C-infinity and (-1)^k f^(k)>=0 for every integer k>=0, define lower energy by liminf(R->infinity) |C intersect B_R|^(-1) sum_{x,y in C intersect B_R,x!=y} p(|x-y|), allowing +infinity. Assertion: for every such f and C, this lower energy is at least the corresponding lattice energy E_p(A2), defined by the limit of the same averages (extended values allowed).

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 090: The frozen target uses density-one discrete closed planar configurations, centered-disk counting density, the liminf of centered pair-energy averages, and nonnegative completely monotone functions of squared distance with extended energies allowed. The family explicitly states these same conventions for locally finite planar configurations and the density-one triangular lattice. Thus the energy comparison covers the ranked universal-optimality assertion rather than only competing lattices or periodic configurations.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would establish one optimal arrangement across the entire stated family of interactions. It would compare the lattice with arbitrary density-one configurations using the same large-disk energy averages.

    Earlier statement and context sources

    Link to this problem
  30. Historical rank246De Giorgi Conjecture for the Allen-Cahn Equation in Dimensions 4 to 8For the Allen–Cahn equation Δu = u³−u on all of Euclidean space, consider a smooth solution with values between −1 and 1 that increases strictly in one direction. In dimensions four through eight, must it vary along just one direction, so its level sets are parallel hyperplanes?Analysis & PDESource-reported proved

    Historical rank #246 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For each ambient dimension N with 4 <= N <= 8, prove that every bounded entire C^2 solution u from R^N to (-1,1) of the Allen-Cahn equation Laplacian u = u^3 - u that is strictly monotone in one direction is one-dimensional, that is, u(x) = g(a . x) for some unit vector a and one-variable profile g.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 375: The dimension-eight theorem has the same bounded entire C2 Allen-Cahn equation and strictly positive directional derivative as the endpoint of the target. Each target solution in dimension 4 through 7 extends constantly in the added coordinates to a bounded entire monotone solution on R^8. Applying the asserted endpoint classification makes that extension one-dimensional; independence of the added coordinates then gives the original lower-dimensional classification. Thus all dimensions 4<=N<=8 are covered without any extra energy-growth assumption.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would force planar level sets from the equation, boundedness, and directional monotonicity alone. A counterexample would exhibit a genuinely multidimensional profile satisfying all those restrictions in one of the specified dimensions.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  31. Historical rank254Exact satisfiability threshold for random k-SAT for small k (k = 3)Choose a random Boolean formula made of clauses, each an OR of three distinct literals, and require every clause to hold. What exact ratio of clauses to variables marks the transition from almost certainly having a satisfying assignment to almost certainly having none, as the number of variables grows?Probability, ergodic theory & dynamicsSource-reported determined

    Historical rank #254 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    In the uniform random 3-SAT model, choose a 3-CNF formula with n variables and m clauses uniformly from formulas whose clauses contain three distinct literals. Determine the exact constant r_3 such that, as n tends to infinity with m/n tending to r, satisfiability has probability tending to 1 for r < r_3 and tending to 0 for r > r_3.

    Status update

    Source-reported determined

    Resolved (determined) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 235: Theorem 1.1 of Computing the Random 3-SAT Threshold gives a unique positive finite alpha3, strict-side satisfiability limits, and one finite deterministic Turing machine which halts at every requested precision with a rational q_r within 2^(-r), using no oracle, advice or noncomputable constant. This is an effective exact specification (a certified Cauchy name) of r3, rather than mere existence; the frozen target and historical dossier impose no closed-form, cavity-formula identification, or efficiency requirement. Hence it determines the requested real constant in that permitted sense. The clause law is independent uniform proper signed clauses on three distinct variables. At m=O(n), repeated whole clauses have probability O(m^2/n^3)=O(1/n), so sampling without replacement gives the same limiting threshold. If three distinct literals is read literally to allow opposite signs of the same variable, such clauses are tautologies of probability O(1/n); discarding their o(n) count in probability preserves the clause density and leaves the uniform proper law. Monotonicity sandwiches m/n tending to r between fixed densities on the same strict side, giving the frozen sequence formulation. The theorem intentionally does not settle behavior at r3 itself, which is not requested.

    Status reviewed

    Resolution sources

    Earlier ranking context

    An exact threshold would locate the limiting transition between compatible and incompatible random Boolean constraints. It would determine which fixed clause densities fall on each side as the number of variables increases.

    Earlier status note

    Open

    October 6, 2026: Gaia Carenini’s ECCC TR26-229 claims a scaling window O(n^(1/2+1/k)) and existence of a limiting threshold for every fixed k ≥ 3. For k = 3, the claimed window is O(n^(5/6)); the paper does not determine the exact threshold constant required here. This is claimed partial progress, with no complete proof audit or formal-verification claim. The author states that none of the paper’s mathematical arguments was generated by a large language model.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Link to this problem
  32. Historical rank260Existence of a complex structure on the six-sphereGeometry & topologySource-reported solved

    Historical rank #260 in the Sep 19, 2026 edition. This entry is outside the current open ranking.

    Also known as

    S^6 complex structure problem; Complex structure on S6; Hopf problem; Hopf problem on the six-sphere

    Exact statement covered by this update

    Determine whether the six-dimensional sphere S^6 admits an integrable complex structure.

    Status update

    Source-reported solved

    Resolved affirmatively: the six-sphere admits an integrable complex structure. Philip Engel provides a self-contained proof and states that he independently verified its mathematical content. Google DeepMind’s Formal Conjectures project also classifies the problem as solved and links a Lean formalization. The earlier Open label came from older evidence. This status review did not include a local Lean replay or establish journal acceptance.

    Status reviewed

    Resolution sources

    Link to this problem
  33. Historical rank260RL versus LCan randomness be removed from every always-halting decision algorithm that uses only logarithmic working memory, never says yes when the answer is no, and says yes with probability at least two-thirds when the answer is yes? The replacement must also use only logarithmic working memory.Theoretical computer scienceSource-reported proved

    Historical rank #260 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Let L be the languages of binary strings decided by deterministic Turing machines using O(log(N+2)) work space on length-N inputs. Let RL use the same read-only input and read-write work-tape convention but allow fresh unbiased read-once random bits, retaining an old bit only by storing it in counted work space. The machine must halt for every input and every random tape, use O(log(N+2)) work space, accept each yes input with probability at least2/3, and accept each no input with probability0. Is every language in RL in L, equivalently RL=L?

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 103: The asserted L=RL=BPL includes the ranked one-sided-error equality RL=L. The target uses fresh read-once random bits, counted storage, and halting on every random tape. In this standard logspace model, an always-halting machine has polynomially many internal configurations and cannot revisit one along a possible execution, since repeating the intervening random choices would yield a nonhalting tape. Thus the target's all-tapes halting convention is compatible with the catalogue's polynomial-time RL scope.

    Status reviewed

    Resolution sources

    Earlier ranking context

    Equality would eliminate randomness from every decision problem in this one-sided-error model without changing the logarithmic space bound. A separation would establish a workspace advantage for randomness under these exact rules.

    Earlier statement and context sources

    Related problems

    Link to this problem
  34. Historical rank261Kadison Similarity Problem for C*-AlgebrasCan every bounded representation of a C*-algebra on a Hilbert space be transformed by one invertible bounded change of coordinates into a representation that preserves adjoints?Algebra, representation & category theorySource-reported proved

    Historical rank #261 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove whether every bounded representation of a C*-algebra on a Hilbert space is similar to a *-representation.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 288: The source proves similarity to a *-homomorphism for every bounded complex-linear unital algebra homomorphism from any unital complex C*-algebra on an arbitrary Hilbert space. This is the standard unital form of the ranked similarity problem; bounded representations of nonunital algebras are covered by unitization, and a nonunital representation of a unital algebra reduces using its range idempotent. No separability, nuclearity or finite-factor restriction is present.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would show that bounded representations can always be put into a form respecting the C*-algebra involution. A counterexample would reveal a representation whose incompatibility with adjoints cannot be removed by similarity.

    Earlier statement and context sources

    Link to this problem
  35. Historical rank267Sidorenko Conjecture on Bipartite Homomorphism DensitiesAt a fixed edge density, do random-like graphs minimize the density of every bipartite pattern? Patterns are counted by edge-preserving maps, allowing different pattern vertices to map to the same vertex; the conjecture is stated for all graphons, the limiting objects for dense graphs.Combinatorics & discrete geometrySource-reported disproved

    Historical rank #267 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that for every bipartite graph H and every graphon W, t(H,W) >= t(K_2,W)^{|E(H)|}; equivalently, quasirandom graphs minimize H-homomorphism density at fixed edge density.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 161: The target quantifies over every bipartite H and every graphon W. The supplied abstract states a bipartite H on 35 vertices and 66 edges with homomorphism density below the conjectured bound in a finite simple graph. Its associated step graphon gives the same homomorphism-density violation, so one such example disproves the universal target. The separate forcing conclusion is not required for this mapping.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would show that quasirandom structure minimizes every bipartite homomorphism density at fixed edge density. A counterexample would identify a bipartite pattern that can be suppressed below this benchmark.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  36. Historical rank268Shafarevich Holomorphic-Convexity ConjectureDoes the universal cover of every connected smooth complex projective variety have enough global holomorphic functions to keep the holomorphic hull of each compact set compact? The hull consists of points where no such function exceeds its maximum magnitude on the original set.Geometry & topologySource-reported disproved

    Historical rank #268 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every connected smooth complex projective variety X, its universal covering complex manifold X-tilde is holomorphically convex: for every compact K contained in X-tilde, the set {z in X-tilde: |f(z)|<=sup_{w in K}|f(w)| for every global holomorphic function f on X-tilde} is compact. This is the classical universal-cover conjecture, not the stronger assertion that X-tilde is Stein and not a statement restricted to linear fundamental groups.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 046: The surface construction is a smooth connected projective complex variety whose universal cover is not holomorphically convex, directly negating the universal ranked assertion. The fourfold construction additionally rules out the large-fundamental-group restriction. The disproof uses actual failure of holomorphic convexity; it does not confuse mere failure to be Stein with failure of the ranked property. Family 058: For a smooth projective variety whose universal cover has the stated semialgebraic realization, the classification makes it D x C^m x F with D bounded symmetric and F compact projective. Projection to the Stein factor D x C^m is proper, giving holomorphic convexity for this restricted class. The ranked Shafarevich assertion quantifies over all smooth projective varieties without a semialgebraicity hypothesis; the family does not establish that general statement.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would give universal covers this specific compactness property governed by holomorphic functions. A counterexample would exhibit a compact set whose holomorphic hull escapes every compact region.

    Earlier statement and context sources

    Link to this problem
  37. Historical rank270Kuznetsov Rationality Conjecture for Cubic FourfoldsIs a smooth complex cubic fourfold rational—birational to projective four-space—exactly when its Kuznetsov category, the K3-like part of its derived category, is equivalent to the untwisted derived category of a smooth projective K3 surface?Geometry & topologySource-reported disproved

    Historical rank #270 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every smooth cubic fourfold X contained in P^5_C, let Ku(X) be the full right-orthogonal subcategory of D^b(coh X) consisting of objects E with Hom(O_X(i),E[j])=0 for i=0,1,2 and all integers j. The Kuznetsov rationality conjecture asserts: X is rational, meaning birational over C to P^4_C, if and only if there exists a smooth projective complex K3 surface S and an exact C-linear triangulated equivalence Ku(X) ~= D^b(coh S). The K3 category is untwisted: no nonzero chosen Brauer twist is permitted; this does not require Br(S)=0. No polarization, marking, chosen kernel, explicit algorithm or separately imposed Fourier-Mukai condition is added. The assertion quantifies over ALL smooth complex cubic fourfolds, not only general points of a special divisor.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 054: For sufficiently large admissible Hassett discriminants the source supplies very general smooth complex cubic fourfolds that are irrational even though their Kuznetsov components are equivalent to ordinary, untwisted derived categories of projective K3 surfaces. Such examples violate the sufficiency direction of the exact ranked biconditional. Its universal quantification is refuted by this subclass; an ineffective threshold does not prevent the existential counterexample, and no Brauer-twisted category is being substituted.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would characterize rationality using the fourfold's specified category of sheaf complexes. A counterexample would expose a failure of one direction of this link between birational geometry and untwisted K3 categories.

    Earlier statement and context sources

    Link to this problem
  38. Historical rank271All groups are good: the four-dimensional disc-embedding conjectureA capped grope is a disc-like assembly of surfaces and caps. In four dimensions, for every such assembly of height 1.5 and any homomorphism from its fundamental group to another group, can its framed boundary be filled inside the assembly by an immersed disc—one allowed to cross itself—whose crossing loops all become trivial in that group?Geometry & topologySource-reported disproved

    Historical rank #271 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Every group Gamma is good in the four-dimensional disc-embedding sense: for every based height1.5 disc-like capped grope G^c (with its standard smooth four-dimensional thickening and framed attaching region) and every homomorphism phi:pi_1(G^c)->Gamma, there exists an immersed disc D->G^c whose framed boundary coincides with that attaching region and whose double-point loops, based by paths in G^c, all map to the identity under phi. The universal assertion is equivalently goodness of the rank-two free group F2, and equivalently goodness of every finitely presented group. Its disc-embedding consequence is in the locally flat topological category, not arbitrary smooth disc embedding.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 305: The ranked target explicitly identifies universal goodness with goodness of F_2 in the Freedman-Quinn disc-embedding sense. The boundary-only manuscript expressly concludes that F_2 is not good and gives discs with framed algebraic dual spheres but no disjoint locally flat spanning discs. The obstruction remains without prescribed output framings or relative homotopy classes, so it is not merely a smooth-category counterexample to the topological ranked contract.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would remove the good-group restriction from the associated four-dimensional topological disc-embedding theorem and its surgery applications. Its consequence concerns locally flat topological discs, without asserting arbitrary smooth disc embedding.

    Earlier statement and context sources

    Link to this problem
  39. Historical rank284Nearby Lagrangian ConjectureA cotangent bundle is a phase space whose zero section is the original space at zero momentum. Can every closed exact Lagrangian submanifold in it be moved onto that zero section by Hamiltonian motion?Geometry & topologySource-reported disproved

    Historical rank #284 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that every closed exact Lagrangian submanifold of the cotangent bundle T^*N is Hamiltonian isotopic to the zero section N.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 340: The indexed target is universal over cotangent bundles and closed exact Lagrangian submanifolds, demanding Hamiltonian isotopy to the zero section. The constructed closed exact smoothly embedded Lagrangian in T*(S9 x S^(N-1)) for some sufficiently large even N satisfies those hypotheses and is not Hamiltonian isotopic to the zero section. One permitted base and one counterexample suffice to refute the universal statement. Being diffeomorphic to the base does not meet the stronger Hamiltonian-isotopy conclusion.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would place all these Lagrangians in one Hamiltonian-isotopy class. A counterexample would reveal a closed exact Lagrangian carrying an obstruction to deformation into the zero section by Hamiltonian motion.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  40. Historical rank285Cartan-Hadamard isoperimetric conjectureIn a complete simply connected space with nonpositive sectional curvature, must every bounded region of finite perimeter have at least as much boundary as a Euclidean ball of the same volume? Is equality possible only when the region is isometric to such a ball?Geometry & topologySource-reported proved

    Historical rank #285 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that every bounded finite-perimeter region in every n-dimensional Cartan-Hadamard manifold satisfies the Euclidean isoperimetric inequality, with equality only for Euclidean balls; equivalently, extend the classical Euclidean isoperimetric lower bound to all complete simply connected Riemannian manifolds of nonpositive sectional curvature.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 337: Taking kappa=0 in the first manuscript covers all complete simply connected smooth manifolds with nonpositive sectional curvature and all finite-volume finite-perimeter sets in every dimension. It therefore covers each bounded finite-perimeter region in the target and supplies the required Euclidean equality rigidity for positive-volume bounded sets. Allowing kappa<0 and unbounded finite-volume sets strengthens the scope. The second manuscript separately extends sharp fillings to proper CAT(0) spaces for integral cycles of dimension at least two; the first supplies the target's full dimension and equality clauses.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make Euclidean space the sharp perimeter benchmark throughout this curvature class. It would rule out enclosing a fixed volume with less boundary than the Euclidean ball requires.

    Earlier status note

    Open · subcases solved

    October 6, 2026: A dimension-five Cartan-Hadamard isoperimetric claim addresses one dimension; the ranked target quantifies over every dimension. Retain the all-dimensional open target. The complete proof was not audited in this review.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Link to this problem
  41. Historical rank289Sharp Upper Bound in Yau's Conjecture on Nodal Sets of Laplace EigenfunctionsFor a standing wave on a smooth closed Riemannian manifold of dimension at least two, how large can the set where it vanishes be? Must its hypersurface measure be at most a constant times the square root of its Laplace eigenvalue, with the constant depending only on the manifold and metric?Analysis & PDESource-reported disproved

    Historical rank #289 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every smooth closed n-dimensional Riemannian manifold M with n >= 2, prove that every Laplace eigenfunction phi_lambda satisfying Delta phi_lambda + lambda phi_lambda = 0 obeys H^{n-1}({phi_lambda = 0}) <= C sqrt(lambda), where C depends only on M and its metric.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 350: The target requires the C sqrt(lambda) nodal-measure bound on every fixed smooth closed manifold in every dimension n>=2. The dimension-three and dimension-four manuscript explicitly gives sequences of exact eigenfunctions on one fixed metric with unbounded normalized nodal measure. One such metric refutes the universal target. The surface theorem proves its n=2 subcase, and the dimension-five paper provides a stronger quantitative violation.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would uniformly control the size of nodal sets by oscillation frequency for each smooth metric. A counterexample sequence would show that zero-set measure can outgrow every such bound on one manifold.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  42. Historical rank290Kaplansky's Idempotent ConjectureForm finite linear combinations of elements of a torsion-free group, using coefficients in any field. Are zero and one the only combinations that stay unchanged when squared? Torsion-free means that no nonidentity group element has finite order.Algebra, representation & category theorySource-reported disproved

    Historical rank #290 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every field K of arbitrary characteristic and every torsion-free group G, prove that each idempotent e in the ordinary scalar group ring K[G] is 0 or 1. Here K[G] consists of finite formal K-linear combinations of elements of G, with the group multiplication extended bilinearly, and idempotent means e^2=e. The group need not be finitely generated. This is not a claim that matrix rings over K[G], completed operator algebras or integral quandle rings have no nontrivial idempotents.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 197: Failure of direct finiteness in the ordinary scalar F2 group algebra of a torsion-free group forces the nontrivial scalar idempotent 1-ba when ab=1 and ba differs from 1. Thus this is a counterexample to the arbitrary-characteristic scalar idempotent target, not merely to a matrix-ring version or to a group with torsion. The torsion-free first manuscript is essential; the odd-characteristic companion does not supply that hypothesis. Family 207: The source proves absence of nontrivial scalar idempotents for every torsion-free group over every commutative unital characteristic-zero domain, so it covers all characteristic-zero fields in the target and even more coefficient rings there. The ranked target quantifies over fields of arbitrary characteristic; positive characteristic is not covered by this family. Family 197 proposes a characteristic-two counterexample separately.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would forbid nontrivial scalar idempotents uniformly across these group rings. A counterexample would also produce nonzero zero divisors, since a nontrivial idempotent satisfies e(1−e)=0.

    Earlier statement and context sources

    Link to this problem
  43. Historical rank291Calderón uniqueness for bounded measurable real isotropic conductivitiesInside a bounded connected region with Lipschitz boundary in dimension three or higher, can two different bounded, uniformly positive scalar conductivities produce exactly the same boundary voltage-to-current measurements? No smoothness of the conductivities is assumed.Analysis & PDESource-reported disproved

    Historical rank #291 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every integer n>=3, every bounded connected Lipschitz domain Omega in R^n and every pair of real scalar conductivities gamma_1,gamma_2 in L-infinity(Omega) with 0<c<=gamma_j<=C<infinity almost everywhere for some constants c,C, equality Lambda_gamma_1=Lambda_gamma_2 of the full weak Dirichlet-to-Neumann operators H^(1/2)(boundary Omega)->H^(-1/2)(boundary Omega) implies gamma_1=gamma_2 almost everywhere. For trace f, u is the unique H^1 solution of div(gamma grad u)=0 with trace f, and <Lambda_gamma f,g>=integral_Omega gamma grad u dot grad v for any H^1 extension v of test trace g. No coefficient derivatives, boundary agreement, connected-complement, smallness or nonzero-frequency hypothesis is imposed.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 365: The ranked uniqueness assertion is universal over n>=3 Lipschitz domains and uniformly positive bounded measurable real scalar conductivities with full weak zero-frequency DN data. Two distinct such conductivities on a three-dimensional ball, with the same full DN operator and even agreement near the boundary, give a counterexample in its permitted class. No anisotropy, nonzero frequency, or smoothness assumption is being imported from the other manuscripts.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would establish uniqueness even without derivative assumptions on conductivity. A counterexample would exhibit distinct admissible media with identical ideal boundary data; uniqueness alone would not supply stable recovery from noisy measurements.

    Earlier statement and context sources

    Link to this problem
  44. Historical rank292Global Attractor Conjecture for Complex-Balanced Mass-Action SystemsIn every finite mass-action reaction network with positive rate constants and a strictly positive complex-balanced equilibrium, must every strictly positive initial state converge to the unique positive equilibrium in its stoichiometric compatibility class? Complex balance equates incoming and outgoing flux at each reaction complex; the whole compatibility class need not be bounded.Applied & computational mathematicsSource-reported proved

    Historical rank #292 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every finite chemical reaction network with positive rate constants and mass-action kinetics that admits a strictly positive complex-balanced equilibrium, let S be the span of its reaction vectors. For every x0 in the strictly positive orthant, prove that the solution starting at x0 converges as t tends to infinity to the unique positive equilibrium in (x0+S) intersected with the strictly positive orthant; or construct such a system and positive initial condition for which this convergence fails. Complex balance means that at each complex total incoming reaction flux equals total outgoing reaction flux at the equilibrium. No boundedness assumption is imposed on the whole stoichiometric compatibility class.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 149: Complex-balanced mass-action systems are weakly reversible, so the classwise permanence theorem supplies global positive trajectories in a compact interior absorbing set even for unbounded classes. Combined with the standard complex-balanced free-energy Lyapunov function, whose zero-dissipation set in a class is its unique positive equilibrium, this yields convergence by the invariance principle. This is a scope implication using that established Lyapunov fact, not a claim that the catalogue explicitly states convergence for every weakly reversible system.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would turn complex balance into a universal guarantee of eventual relaxation to equilibrium for positive initial concentrations. A counterexample would identify a limitation of that stability principle.

    Earlier statement and context sources

    Related problems

    • #484 · Persistence Conjecture for Weakly Reversible Mass-Action SystemsSource-reported proved · historical rankRelated problemPersistence Conjecture for Weakly Reversible Mass-Action Systems

      The persistence question assumes a bounded stoichiometric compatibility class; the global-attractor question allows unbounded classes but assumes complex balance. Their complete stated scopes are different.

      Scope qualification: The persistence-to-convergence argument covers the bounded-class overlap. No implication between these two complete targets is established here.

    Link to this problem
  45. Historical rank295Zariski Multiplicity Conjecture for Hypersurface SingularitiesIf a continuous change of surrounding coordinates with continuous inverse identifies two reduced complex hypersurfaces near a point, must their multiplicities—the lowest degrees in their local defining equations—agree?Geometry & topologySource-reported disproved

    Historical rank #295 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For reduced holomorphic germs f,g:(C^n,0)->(C,0), prove that if the embedded hypersurface germs (V(f),0) and (V(g),0) are ambient topologically equivalent, then their multiplicities at the origin are equal; equivalently, prove that multiplicity is an invariant of embedded topological type for reduced complex analytic hypersurface singularities.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 059: The source constructs reduced holomorphic hypersurface germs with ambiently homeomorphic zero-set germs but unequal multiplicities. In particular the C4 example has isolated critical points and multiplicities four and five, directly violating equality in the universally quantified frozen target. The separate higher-dimensional examples also refute it; no deformation-family or nonembedded equivalence substitution is required.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would let the topology of a singularity determine this basic algebraic measure of its complexity. A counterexample would separate multiplicity from the shape preserved by an ambient homeomorphism.

    Earlier statement and context sources

    Link to this problem
  46. Historical rank297Amenability of Thompson's Group FDoes Thompson’s group F admit an averaging rule for all bounded functions on the group that gives the same result after any group translation? This is the property called amenability.Algebra, representation & category theorySource-reported determined

    Historical rank #297 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether Richard Thompson's group F of piecewise-linear orientation-preserving homeomorphisms of the unit interval is amenable.

    Status update

    Source-reported determined

    Resolved (determined) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 248: The target asks whether Thompson's interval group F is amenable. The source names precisely Thompson's group F and states it is nonamenable, resolving the amenability decision in the negative. No switch to Thompson T, V, a subgroup, a different coefficient system or a quantitative growth property is involved.

    Status reviewed

    Resolution sources

    Earlier ranking context

    This would decide amenability for F by determining whether group translation can preserve such an average. A negative answer would show that no invariant averaging operation exists for this group.

    Earlier statement and context sources

    Link to this problem
  47. Historical rank298Griffiths Positivity Conjecture for Ample Vector BundlesDoes every ample holomorphic vector bundle on a smooth complex projective variety admit a metric with Griffiths-positive curvature? This asks whether algebraic positivity can always be realized geometrically as positive curvature in every nonzero tangent and fiber direction.Geometry & topologySource-reported disproved

    Historical rank #298 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every ample holomorphic vector bundle E on a smooth complex projective variety X, prove that E admits a smooth Hermitian metric h whose Chern curvature is Griffiths positive: for every point x and nonzero vectors v in T_x^(1,0)X and e in E_x, the Hermitian pairing of Theta_(E,h)(v, conjugate(v))e with e is strictly positive.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 050: The source gives ample rank-two holomorphic bundles on the smooth complex projective quadric surface P1 x P1 admitting no smooth strictly Griffiths-positive Hermitian metric for sufficiently large indicated powers. These bundles satisfy the hypotheses and violate exactly the metric-existence conclusion of the universally quantified ranked conjecture. One such bundle is enough; no all-ranks or all-bases construction is needed.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would realize algebraic ampleness through this pointwise curvature condition. A counterexample would show that an ample bundle can lack every smooth metric with Griffiths-positive curvature.

    Earlier status note

    Resolution claim under review

    Claimed disproof pending refereeing. Du and Xie’s September 22, 2026 paper claims an ample rank-two bundle on an abelian surface with no smooth Griffiths-semipositive metric. Such a bundle would refute this target. A single-model manual review of Sections 1–4 on October 6 reported no concrete fatal gap, passed finite exact-arithmetic checks, and found that the inspected background theorems’ hypotheses matched. That review is not independent specialist acceptance or formal certification; the claim remains pending. No public proof-assistant formalization was located as of October 6. The authors disclose generative-AI assistance without naming a model.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Link to this problem
  48. Historical rank305Birkhoff Conjecture (Near-Boundary Continuous Invariant Foliation)For a smooth planar billiard table whose boundary has positive curvature everywhere, suppose all trajectories sufficiently close to grazing the boundary are organized by a continuous foliation of invariant closed curves, each winding once around phase space. Must the table be an ellipse, including a circle?Probability, ergodic theory & dynamicsSource-reported proved

    Historical rank #305 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every bounded planar billiard domain Omega whose boundary is a C-infinity simple closed curve with everywhere positive curvature, let T be its usual specular-reflection billiard map on the open phase cylinder M=(R/LZ) x (0,pi), where s is boundary arclength and delta is the incoming angle with the tangent. Suppose an open annular neighborhood U of the grazing end delta=0 is foliated by continuous simple closed T-invariant curves, each essential in M and winding once around the cylinder; U and its foliation are invariant under T. Here neighborhood of that end means U contains (R/LZ) x (0,eta) for some eta>0, with no fixed lower bound on eta or on the rotation numbers reached by the foliation. Assert that the boundary of Omega is an ellipse, with circles included. No central symmetry, closeness to an ellipse, period-four caustic or extension of the foliation to the whole phase cylinder is assumed. The invariant-curve formulation is selected directly; an unproved equivalence to a caustic, rational, measurable or polynomial-integral formulation is not part of the target.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 147: The first manuscript has precisely the frozen invariant-curve hypothesis: a smooth strictly convex planar billiard with positive curvature and a continuous foliation of a full grazing annulus by individually invariant essential curves. Its conclusion, using the companion collar-rigidity theorem, is that the table is an ellipse. The explicit phase-foliation manuscript supplies the bridge, so this mapping does not assume an unstated equivalence from a physical caustic collar alone.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would recover the table’s shape from this precise local organization of trajectories, without requiring a foliation of all phase space. A counterexample would provide a nonelliptic table with the same near-boundary property.

    Earlier status note

    Open · subcases solved

    October 6, 2026: Pierce-Birkhoff concerns lattice combinations of piecewise polynomials; our Birkhoff entry concerns smooth convex billiards. The complete proof was not audited in this review.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  49. Historical rank306Secret key from all entangled statesCan every entangled quantum state shared by two parties yield secret shared bits at a positive rate from many copies, using only local operations and public classical communication?Quantum information & computationSource-reported disproved

    Historical rank #306 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether every bipartite entangled quantum state rho has strictly positive distillable secret key K_D(rho) under asymptotic local operations and public classical communication.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 272: The source explicitly includes ordinary finite local-instrument protocols with finite classical records, joint processing of all copies, local auxiliaries/private randomness and unrestricted authenticated two-way public communication; the input is the only shared private resource and Eve retains a purification and the full public transcript. These are the conventional protocols used in asymptotic distillable secret key. Every ordinary terminating finite protocol is completed almost surely, so that condition does not remove any such protocol at a fixed blocklength; the source additionally allows countable processes. The stated rate is the supremum of liminf ell_n/n for outputs whose trace-norm error from an ideal uniform shared key tends to zero. Theorem 1.1 constructs an entangled 10-by-10 density operator and gives a uniform unhalved trace-norm distance at least 1/5 from any ideal single secret bit for every n and every admitted completed bit-output protocol. Any asymptotically secure positive-rate key sequence would eventually output at least one bit, and discarding its other bits is an allowed local operation that cannot increase the trace-norm error; it would contradict this uniform one-bit gap. Thus the state has zero key under the frozen conventional asymptotic operations and refutes the universal positive-key assertion.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make entanglement sufficient for extracting secret key at a positive asymptotic rate under these operations. A counterexample would separate the presence of entanglement from usable distillable secrecy.

    Earlier statement and context sources

    Link to this problem
  50. Historical rank311Mahler Conjecture on the Volume Product of a Convex BodyDoes every centrally symmetric convex body have volume product with its polar at least as large as the cube's, namely 4^n/n! in dimension n? The polar consists of vectors whose inner product with every point of the body is at most one.Combinatorics & discrete geometrySource-reported proved

    Historical rank #311 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that the product of the volume of a centrally symmetric convex body K in R^n and the volume of its polar body is at least 4^n/n!, with equality attained by the cube (and its polar).

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 087: The frozen target is the centrally symmetric Mahler lower bound 4^n/n! with equality attained by a cube and its polar. The symmetric manuscript gives that bound in every dimension and classifies all equality cases as invertible linear images of Hanner polytopes, which include those examples. This is stronger than the requested equality-attainment assertion. The nonsymmetric theorem and polar-product symplectic-width theorem are distinct additional results, not needed to replace the symmetric target.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would establish the proposed minimum volume product in every dimension. It would limit how small a convex body and its dual polar can be simultaneously.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  51. Historical rank321Finiteness of Finitely Presented Periodic GroupsMust a group with finitely many generators and defining relations have only finitely many elements if repeatedly multiplying any element by itself eventually gives the identity?Algebra, representation & category theorySource-reported disproved

    Historical rank #321 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that every finitely presented group in which every element has finite order is finite; equivalently, decide whether an infinite finitely presented periodic group exists.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 247: The target asks whether every ordinarily finitely presented periodic group is finite. The source constructs an infinite group with an ordinary finite presentation and every element of finite order, and the companion strengthens it to a residually finite infinite 2-group. This is a counterexample to precisely the universal finiteness assertion; neither profinite presentation nor finite generation alone is being substituted.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would show that finite presentation and elementwise finite order together force a finite group. A counterexample would give an infinite group satisfying both restrictions simultaneously.

    Earlier statement and context sources

    Link to this problem
  52. Historical rank328Sinai's Positive Metric Entropy Conjecture for the Chirikov Standard MapFor a set of parameter values of nonzero total length, does the Chirikov standard map have positive metric entropy—a measure of average orbit unpredictability—with respect to ordinary area?Probability, ergodic theory & dynamicsSource-reported proved

    Historical rank #328 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that there is a positive-Lebesgue-measure set Lambda of parameters k for which the Chirikov standard map f_k(x,y) = (x + y + k sin(2 pi x), y + k sin(2 pi x)) mod Z^2 has positive metric entropy with respect to Lebesgue measure; equivalently, it has a positive-measure orbit set with a nonzero Lyapunov exponent.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 146: The ranked Sinai target requests positive metric entropy for a positive-Lebesgue-measure set of parameters in the original standard sine-map family. The source gives positive entropy with respect to normalized area for every sufficiently large positive parameter, a full tail and therefore a positive-measure parameter set. Standard fixed normalization of the sine parameter preserves this implication.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would establish measure-theoretic chaotic behavior throughout a positive-measure parameter set. Refutation would mean that the parameter set with positive entropy has measure zero, even if some such parameters exist.

    Earlier statement and context sources

    Link to this problem
  53. Historical rank334Unrestricted Linear-Circuit Lower Bound for the Discrete Fourier TransformThe discrete Fourier transform converts n inputs into frequency components. Must computing it require at least a constant times n log n steps, even when each step may combine two earlier quantities using arbitrary complex coefficients?Theoretical computer scienceSource-reported disproved

    Historical rank #334 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For each integer n>=2, let C_DFT(n) be the minimum number of steps needed to compute every output of the unnormalized complex DFT y_j=sum_(l=0)^(n-1) exp(-2*pi*i*j*l/n)x_l, 0<=j<n. Initially available functions are input coordinates and scalar constants; each step adds one affine function lambda*f+mu*g from two previously available functions with arbitrary complex scalars. Assertion: there exist c>0 and N such that C_DFT(n)>=c*n*log(n) for all n>=N. No coefficient bound, conditioning or memory restriction is imposed beyond this two-input scalar-step model.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 130: The ranked assertion is an eventual Omega(n log n) lower bound for unrestricted complex two-input linear computation. The companion circuits have o(n log n) size on unbounded lengths, already contradicting an eventual all-length lower bound. Its counted additions, subtractions and scalar multiplications can each be represented by a constant number of the target affine steps. The uniform all-length power saving strengthens the counterexample; arbitrary coefficients and supplied roots are permitted by this target.

    Status reviewed

    Resolution sources

    Earlier ranking context

    This would impose an asymptotic operation limit in a model allowing unrestricted coefficients and reuse of earlier values. Refutation would require arbitrarily large lengths beating every proposed positive constant in that bound.

    Earlier statement and context sources

    Link to this problem
  54. Historical rank340Komlos Conjecture on Signing Unit Vectors with Bounded Infinity NormCombinatorics & discrete geometrySource-reported solved

    Historical rank #340 in the Sep 20, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that there is a universal constant c > 0 such that for every collection of vectors v_1, ..., v_n in the Euclidean unit ball B_2^d (any d, n), there exist signs epsilon_1, ..., epsilon_n in {-1,1} with || sum_i epsilon_i v_i ||_infinity <= c.

    Status update

    Source-reported solved

    User reports the linked proof passes audit. The coordinator verifies the exact theorem and the incidence-matrix implication, without claiming a new independent proof audit, peer review or formal verification. Exact match: the listing requires some universal constant c for all d,n and arbitrary real vectors in the Euclidean unit ball. Theorem 1.2 supplies c=36 with precisely the same sign and infinity-norm conditions. The argument on PDF page 3 passes from rational to real inputs, so there is no rational-only limitation. No optimal constant or efficient algorithm is required by our listing.

    Status reviewed

    Resolution sources

    Link to this problem
  55. Historical rank343Koebe circle-domain uniformization conjectureCan every connected open region of the Riemann sphere—the complex plane together with infinity—be mapped by an angle-preserving bijection to a region whose missing connected pieces are all round closed disks or single points?Analysis & PDESource-reported proved

    Historical rank #343 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every nonempty connected open domain Omega of the Riemann sphere, there exist a circle domain D in the Riemann sphere and a conformal homeomorphism Omega -> D. A circle domain means a nonempty connected open set each of whose complementary components is a closed round disk or a point. Allow arbitrary connectivity, including uncountably many complementary components and point components; the whole sphere (empty complement) and a sphere minus one point are included as trivial circle-domain cases. Conformal homeomorphism means a bijective conformal map, not a possibly noninjective holomorphic covering. No uniqueness, Mobius rigidity, quantitative distortion, explicit construction or boundary-extension requirement is appended. The universal statement is retained, not replaced by only an uncountable-connectivity or otherwise difficult subclass.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 071: Every domain in the Riemann sphere is asserted to be conformally equivalent to a circle domain, with no restriction on connectivity. This matches the frozen existence-only assertion, including bijective conformal equivalence rather than a covering map. The target imposes no rigidity or boundary-extension requirement, so the separate removability-to-rigidity theorem is not needed for its resolution.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would supply circle-domain models even when there are uncountably many complementary components. It would establish existence of such models without asserting their uniqueness or a method for constructing them.

    Earlier statement and context sources

    Link to this problem
  56. Historical rank351Kobayashi Conjecture: Hyperbolic Compact Kähler Manifolds Have Ample Canonical BundleIf a compact connected Kähler manifold admits no nonconstant holomorphic map from the complex plane, must its canonical line bundle be ample—meaning some positive power has enough sections to embed the manifold in projective space?Geometry & topologySource-reported proved

    Historical rank #351 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every compact connected Kähler complex manifold X that is Kobayashi hyperbolic, equivalently every holomorphic map C -> X is constant, the canonical line bundle K_X is ample.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 051: The source states that every compact connected Kähler manifold with no nonconstant entire curve has ample canonical bundle, with projectivity as a consequence. This is the frozen hyperbolicity formulation and conclusion. Its explicit positive-dimensional scope contains the substantive target; a zero-dimensional connected manifold is a point, where the canonical line bundle is already ample.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would force positivity of the canonical bundle from this restriction on entire curves. A counterexample would separate Kobayashi hyperbolicity from the stated algebraic positivity conclusion.

    Earlier statement and context sources

    Link to this problem
  57. Historical rank357Strong Hyperplane Conjecture for Isotropic ConstantsIn every dimension, does a simplex—the higher-dimensional version of a triangle—maximize the isotropic constant among all convex bodies? This constant compares the covariance of a uniformly chosen point with the body's volume in a way unchanged by affine coordinate transformations.Combinatorics & discrete geometrySource-reported proved

    Historical rank #357 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every integer n>=1 and every compact convex body K in R^n with nonempty interior, let X be uniformly distributed on K, let Cov(X) be its centered covariance matrix, and define L_K=(det(Cov(X))/Vol_n(K)^2)^(1/(2n)). Then L_K <= L_Delta_n, where Delta_n is any nondegenerate n-simplex and L_Delta_n=(n!)^(1/n)/((n+1)^((n+1)/(2n))*sqrt(n+2)). Bodies need not be centrally symmetric or smooth. The assertion asks for this sharp bound in each dimension, not uniqueness of equality or merely a dimension-free bound.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 101: The exact ranked target is L_K <= L_Delta_n for every full-dimensional compact convex body in each dimension, with neither symmetry nor smoothness required. The catalogue explicitly asserts that simplices uniquely maximize the isotropic constant among all convex bodies in each dimension. Maximization gives the required inequality, and uniqueness of equality is additional information beyond the ranked target. The separate log-concave entropy theorem is not needed for this implication.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would determine the sharp upper bound separately in every dimension. A counterexample would be a convex body exceeding the simplex value; neither outcome requires classifying all equality cases.

    Earlier statement and context sources

    Link to this problem
  58. Historical rank358CAT(0) Models for All Word-Hyperbolic GroupsDoes every finitely generated word-hyperbolic group act properly and cocompactly by isometries on some proper geodesic CAT(0) space? The space may be chosen freely; it need not be a cube complex or have a prescribed dimension.Geometry & topologySource-reported disproved

    Historical rank #358 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Also known as

    Is every hyperbolic group a CAT(0) group?; Geometric CAT(0) actions of hyperbolic groups

    Exact statement covered by this update

    Does every finitely generated word-hyperbolic group G admit a proper cocompact action by isometries on some proper geodesic CAT(0) metric space X? Proper for X means that closed bounded subsets are compact; proper for the discrete group action means that for each compact K in X, only finitely many g in G have gK intersecting K. A negative answer is one word-hyperbolic group admitting no such action on any such X. Do not restrict X to a cube complex, a fixed dimension, a Rips complex, or a CAT(-1) space; do not replace G by a finite-index subgroup.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 246: For hyperbolic groups whose boundary is S^2, the source gives an action of the original group itself, with finite kernel, on H^3. Hyperbolic three-space is a proper geodesic CAT(0) space, so this supplies the exact requested kind of proper cocompact action for that boundary class. Target 364 quantifies over all word-hyperbolic groups with no boundary restriction, so it remains broader. Family 257: The target is existence of a proper cocompact isometric action on an arbitrary proper geodesic CAT(0) space for every word-hyperbolic group. The constructed hyperbolic group excludes every proper complete CAT(0) model, with no dimension or cubical restriction. Proper metric spaces are complete, so the extra word complete does not leave a class of target models untreated.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A positive answer would give every word-hyperbolic group a CAT(0) geometric model. A negative answer requires a group with no such model at all, rather than failure of one particular construction.

    Earlier statement and context sources

    Link to this problem
  59. Historical rank361Alperins Weight Conjecture for Modular Representations of Finite GroupsIn every p-block of a finite group, is the number of distinct simple modular representations equal to the number of conjugacy classes of weights? The weights encode local subgroup representations, so the conjecture predicts a local description of a global count.Algebra, representation & category theorySource-reported proved

    Historical rank #361 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that the number of isomorphism classes of simple modules of a p-block B of a finite group G equals the number of G-conjugacy classes of B-weights.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 202: For every finite group and prime, both source and target compare the simple modules of a p-block with its conjugacy classes of block weights. Irreducible Brauer characters label the simple modular modules counted in the target. The source is explicitly blockwise rather than only the sum over all blocks, and gives the full numerical equality.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would equate these two blockwise counts throughout finite-group representation theory. A counterexample would identify a block where the simple-module count cannot be recovered from the prescribed weight count.

    Earlier statement and context sources

    Link to this problem
  60. Historical rank363Beck-Fiala ConjectureCombinatorics & discrete geometrySource-reported solved

    Historical rank #363 in the Sep 20, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that there is a universal constant C such that every hypergraph H in which each vertex lies in at most t edges has discrepancy at most C sqrt(t).

    Status update

    Source-reported solved

    User reports the linked proof passes audit. The coordinator verifies the exact theorem and the incidence-matrix implication, without claiming a new independent proof audit, peer review or formal verification. Exact implication: for a finite hypergraph of maximum vertex degree t>0, use one row per edge and one column a_v per vertex in its 0/1 incidence matrix. Each column has Euclidean norm at most sqrt(t). Apply Theorem 1.2 to a_v/sqrt(t). Multiplying back gives a vertex signing with absolute signed sum at most 36 sqrt(t) in every edge. For t=0 all edge sums are zero. This proves the listed universal C sqrt(t) target with C=36, not merely the older linear-in-t bound.

    Status reviewed

    Resolution sources

    Link to this problem
  61. Historical rank364Log-Brunn-Minkowski Inequality for Symmetric Convex BodiesFor two convex bodies symmetric about the origin, form their logarithmic blend by intersecting half-spaces set by weighted geometric means of their supporting distances. Must its volume be at least the corresponding weighted geometric mean of their volumes, in every dimension?Combinatorics & discrete geometrySource-reported proved

    Historical rank #364 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove the conjecture of Boroczky, Lutwak, Yang and Zhang that for origin-symmetric convex bodies K and L in R^n and any lambda in [0,1], the volume of the log-convex combination satisfies |(1-lambda) K +_0 lambda L| >= |K|^(1-lambda) |L|^lambda, strengthening the Brunn-Minkowski inequality in the symmetric case.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 091: The frozen target is the logarithmic Brunn-Minkowski volume inequality for arbitrary origin-symmetric convex bodies, every dimension and each interpolation parameter in [0,1]. The manuscript asserts that exact general symmetric conjecture with no unconditionality or smoothness restriction. Its L_p and even-log-concave-measure consequences add scope but are not required for the ranked Lebesgue-volume statement.

    Status reviewed

    Resolution sources

    Earlier ranking context

    This would impose a stronger volume constraint on symmetric convex shapes than ordinary convex interpolation provides. A counterexample would show precisely where the proposed multiplicative bound fails.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  62. Historical rank366Hyperkähler SYZ (Lagrangian Fibration) ConjectureIn hyperkähler geometry, does a nontrivial nef line bundle—a weakly positive one—with zero Beauville–Bogomolov–Fujiki square always have a positive tensor power whose global sections have no common zero? The manifold is compact and irreducible holomorphic symplectic.Geometry & topologySource-reported proved

    Historical rank #366 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every compact irreducible holomorphic symplectic complex manifold X and every nontrivial nef holomorphic line bundle L with q(c1(L))=0 for the Beauville–Bogomolov–Fujiki form, some positive tensor power of L is generated by global sections.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 041: The strong-SYZ manuscript proves semiampleness of every holomorphic line bundle with nonzero nef isotropic first Chern class on every compact irreducible holomorphic symplectic Kähler manifold, exactly the target conclusion of a globally generated positive tensor power. In this irreducible holomorphic symplectic setting, a nontrivial line bundle has nonzero first Chern class; no dimension or deformation-type restriction is added. The separate projective-space-base result is not needed.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make these positivity and isotropy conditions sufficient for semiampleness. The resulting sections would define a map everywhere on the manifold after taking an appropriate tensor power.

    Earlier statement and context sources

    Related problems

    • #363 · Boundedness of Compact Hyperkahler Manifolds: Finiteness of Deformation Types in Each DimensionRelated problemBoundedness of Compact Hyperkahler Manifolds: Finiteness of Deformation Types in Each Dimension

      With the recorded fibration theorem, a proof of Hyperkähler SYZ would yield deformation finiteness for the second-Betti-number-at-least-five scope.

      Scope qualification: That conditional partial result does not cover the unrestricted finiteness conjecture. No unconditional implication between the complete targets is established here.

    Link to this problem
  63. Historical rank368Benjamini-Schramm conjecture p_c<p_u for non-amenable Cayley graphsOn a graph built from finitely many generators of an infinite group, is there a percolation range with multiple infinite clusters exactly when the group is nonamenable, meaning it has no translation-invariant averaging rule?Probability, ergodic theory & dynamicsSource-reported proved

    Historical rank #368 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove the Benjamini-Schramm conjecture that for any Cayley graph G of a finitely generated infinite group, p_c(G)<p_u(G) if and only if the group is non-amenable, where p_c is the percolation threshold and p_u the threshold for a unique infinite cluster.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 214: Every Cayley graph in the nonamenable direction of the ranked conjecture is a locally finite transitive, hence quasi-transitive, graph covered by the theorem. The displayed stronger inequality p_c<p_(2 to 2)<=p_u gives the required strict inequality. Together with the established uniqueness direction for amenable Cayley graphs, this settles the ranked biconditional; the new result itself applies to the larger nonamenable quasi-transitive class.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would characterize nonamenability through a gap between the emergence of infinite clusters and their uniqueness. A counterexample would separate this group property from the proposed percolation criterion.

    Earlier statement and context sources

    Link to this problem
  64. Historical rank369Dixmier unitarisability problem for discrete groupsSuppose every uniformly bounded linear action of a discrete group on a complex Hilbert space can be made length-preserving by a bounded invertible change of coordinates. Must the group admit a translation-invariant average on all bounded functions—that is, be amenable?Algebra, representation & category theorySource-reported proved

    Historical rank #369 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Every unitarisable discrete group G is amenable. Here unitarisable means: for every complex Hilbert space H and every group representation pi:G->GL(H) by bounded invertible linear operators with sup_{g in G} ||pi(g)||<infinity in operator norm, there exists a bounded invertible S on H such that S pi(g) S^{-1} is unitary for all g. There is no countability or Hilbert-space separability restriction. Equivalently, every nonamenable discrete group admits at least one uniformly bounded Hilbert-space representation not similar to a unitary one.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 251: The ranked formulation quantifies over all discrete groups and all complex Hilbert spaces, without countability or separability restrictions. The first manuscript's abstract explicitly gives amenability if and only if every uniformly bounded Hilbert-space representation is similar to a unitary representation for a discrete group. This matches the needed unitarizable-implies-amenable direction in full. Its path contains Countable, but the frozen abstract and family summary both explicitly state all discrete groups; the second, strong-Ulam-stability manuscript has a separate countable-group restriction.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would connect this universal representation property to invariant averaging on the group, without countability restrictions. A counterexample would separate the similarity-to-unitary property from amenability.

    Earlier statement and context sources

    Link to this problem
  65. Historical rank374Fujita Freeness ConjectureOn a smooth complex projective variety of dimension n, does adding n+1 copies of any ample divisor to the canonical divisor always give a linear system with no base point—so that every point is avoided by some member?Geometry & topologySource-reported proved

    Historical rank #374 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every smooth complex projective variety X of dimension n and every ample divisor L, prove that K_X + (n+1)L is basepoint-free.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 038: For every smooth complex projective n-fold and ample line bundle L, the manuscript proves global generation of K_X+mL for every integer m >= n+1. Choosing m=n+1 gives exactly the ranked basepoint-freeness assertion, with identical field, smoothness, dimensional and ampleness hypotheses and no extra conditions.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would ensure that at each point some global section of this adjoint divisor is nonzero. It would give a dimension-dependent guarantee for defining the associated map everywhere.

    Earlier statement and context sources

    Link to this problem
  66. Historical rank375Nagata Conjecture on Plane Curve Linear SystemsFor r≥10 very general points in the complex projective plane, how much degree must a curve spend to vanish to orders m₁,...,mᵣ at them? Must its degree always exceed (m₁+⋯+mᵣ)/√r?Geometry & topologySource-reported proved

    Historical rank #375 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that for any r >= 10 very general points in the complex projective plane P^2 and nonnegative integer multiplicities m_1 through m_r, every plane curve of degree d having multiplicity at least m_i at the i-th point satisfies d > (1 / sqrt{r}) sum_{i=1}^r m_i.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 039: The source proves sum of multiplicities < d*sqrt(r) for every nonzero effective plane curve at r >= 10 very general complex points, including arbitrary nonhomogeneous multiplicities and reducible/nonreduced curves. Dividing by sqrt(r) is the strict inequality in the frozen Nagata target. The higher-dimensional and positive-characteristic Seshadri companions are not needed for this exact match.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would turn prescribed local multiplicities into a strict global lower bound on curve degree. A counterexample would exhibit a curve meeting the conditions with degree at or below that threshold.

    Earlier statement and context sources

    Related problems

    Link to this problem
  67. Historical rank378Toms–Winter: Strict Comparison Implies Z-StabilityIn a separable simple nuclear C*-algebra that is not an algebra of compact operators, does strict comparison of positive elements force tensoring with the Jiang–Su algebra to leave it unchanged up to isomorphism? This asks whether an order-comparison property forces structural regularity, even without an identity element.Algebra, representation & category theorySource-reported proved

    Historical rank #378 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every separable, simple, nuclear, non-elementary C*-algebra A, unital or nonunital, assert that strict comparison of positive elements implies A is isomorphic to A tensor_min Z, where Z is the Jiang–Su algebra. Non-elementary excludes algebras of compact operators. In the Cuntz semigroup Cu(A) of positive elements of A tensor K modulo Cuntz equivalence, strict comparison means: x≤y whenever phi(x)≤phi(y) for every ordered-semigroup functional phi:Cu(A)→[0,infinity] preserving zero and increasing sequential suprema, with strict inequality for every such phi for which 0<phi(y)<infinity. This is the convention following Conjecture 5.1 of Castillejos–Evington–Tikuisis–White. No UCT, unitality, finite trace-boundary, uniform Gamma or finite nuclear dimension hypothesis is added.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 291: The exact target includes separable simple nuclear non-elementary algebras without unitality, UCT or bounded-trace restrictions, and uses extended Cuntz-semigroup functionals. The Cuntz-comparison manuscript explicitly states that strict comparison in the extended-functional sense implies Jiang-Su absorption for that full class, including nonunital algebras and unbounded traces. This manuscript covers the target; the unital tracial paper and the stably finite amenable-action theorem alone would not cover all of it. Family 302: Minimal homeomorphisms of infinite compact metrizable spaces give simple separable unital nuclear non-elementary crossed products. For this subclass, strict comparison forces radius of comparison zero; the stated equality then forces mean dimension zero, and the stated equivalence gives Jiang-Su stability. This provides the Toms-Winter implication for these integer-action crossed products only. It does not cover arbitrary simple nuclear algebras, nonunital cases, or general group actions.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would connect an order property of positive elements to this tensorial structural regularity. A counterexample would separate strict comparison from Jiang–Su absorption within the same class.

    Earlier statement and context sources

    Link to this problem
  68. Historical rank379Margulis–Platonov normal-subgroup conjectureFor a simply connected, absolutely almost simple algebraic group over a number field, can every noncentral normal subgroup of its rational points be recovered from local information? The conjecture uses the rank-zero groups at finite places, taking preimages of open normal subgroups of their product.Algebra, representation & category theorySource-reported proved

    Historical rank #379 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every absolutely almost simple simply connected algebraic group G over a number field K, let A be the finite set of nonarchimedean places v with rank over K_v of G equal to zero, and let delta:G(K)→∏_{v∈A}G(K_v) be the diagonal map. Every noncentral normal subgroup N of G(K) is delta^−1(W) for some open normal subgroup W of that product, with the empty product interpreted as the trivial group.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 018: The number-field manuscript has the same absolutely almost simple, simply connected group hypothesis, the same noncentral abstract normal subgroups of G(K), and the inverse-image description by open normal subgroups of the anisotropic nonarchimedean local product. It therefore covers the complete frozen number-field target; the global-function-field companion extends beyond it.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would describe all such normal subgroups using just these local groups. When there are no rank-zero finite places, it would force every noncentral normal subgroup to equal the entire rational-point group.

    Earlier statement and context sources

    Link to this problem
  69. Historical rank392Measure-zero singular set of stationary integral varifoldsA stationary integral varifold is a generalized surface with integer multiplicities that is stationary for area. Must its nonsmooth part always have zero measure in the surface's own dimension, in every positive dimension and codimension in Euclidean space?Analysis & PDESource-reported proved

    Historical rank #392 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For all positive integers m and n, every stationary integral m-varifold V in an open subset U of Euclidean R^{m+n} has H^m(Sing(V))=0, where Sing(V) is the complement in spt(V) intersect U of the points having a neighborhood on which V is an integer multiple of a smooth embedded minimal m-submanifold.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 346: The target asks that the singular set of every stationary integral m-varifold in Euclidean open sets of arbitrary positive dimension and codimension have zero m-dimensional Hausdorff measure. The first manuscript bounds its Hausdorff dimension by m-1, which directly implies this measure-zero conclusion. The companion also states the exact almost-everywhere regularity formulation: near almost every support point the varifold is a constant positive integer multiple of a smooth embedded minimal submanifold. No stability, orientability, multiplicity-two or codimension-one restriction is imposed; the sphere extension is extra.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would guarantee that such objects are locally integer multiples of smooth embedded minimal submanifolds outside a set of zero m-dimensional measure. A counterexample would exhibit a singular part of positive measure.

    Earlier statement and context sources

    Link to this problem
  70. Historical rank401Serres Positivity Conjecture for Intersection Multiplicities over Regular Local RingsWhen two finitely generated modules over a regular local ring have complementary dimensions and a tensor product of finite length, must Serre's intersection multiplicity—the alternating sum measuring their algebraic intersection—be strictly positive?Algebra, representation & category theorySource-reported proved

    Historical rank #401 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every regular local ring R and finitely generated R-modules M and N with finite length M tensor_R N and dim(M)+dim(N)=dim(R), prove that chi^R(M,N)=sum_i (-1)^i length(Tor_i^R(M,N)) is strictly positive.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 193: The catalogue asserts strict positivity for nonzero finitely generated modules over every regular local ring, with finite-length tensor product and complementary dimensions. These are the ranked intersection-multiplicity hypotheses, and ramified mixed characteristic is expressly included. No field, unramifiedness, or extra dimension restriction remains.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would guarantee positivity of intersection multiplicity throughout this algebraic setting. A counterexample would show that complementary dimensions and a finite intersection do not suffice for a positive value.

    Earlier statement and context sources

    Link to this problem
  71. Historical rank403Rokhlin problem on multiple mixingFor a probability-preserving transformation, if each pair of events approaches statistical independence as its time separation tends to infinity, must every finite collection do the same as all successive time gaps tend to infinity?Probability, ergodic theory & dynamicsSource-reported proved

    Historical rank #403 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether every mixing measure-preserving transformation of a probability space is mixing of all orders: for a measure-preserving transformation T on a probability space, does mixing (that is, mu(A intersect T^{-n}B) -> mu(A)mu(B) as |n| -> infinity for all measurable A,B) imply k-fold mixing for every k >= 3, where k-fold mixing requires the correlation mu(A_1 intersect T^{-n_1}A_2 intersect ... intersect T^{-(n_1+...+n_{k-1})}A_k) to converge to the product of the measures as the gaps n_i -> infinity?

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 145: Theorem 1.1 explicitly proves every finite order on arbitrary probability spaces, with no standardness or countable-generation restriction, and uses precisely the successive positive gaps and cumulative times displayed in the frozen target. The target defines ordinary mixing using |n|, the usual two-sided invertible convention, so the theorem covers that reading directly. The supplied introduction further states that Corollary 8.1 extends the conclusion to mixing endomorphisms on arbitrary probability spaces. Thus invertibility is not a remaining scope obstruction even if the intended version includes noninvertible probability-preserving maps. This conclusion uses the published statement of the extension, without inspecting its proof.

    Status reviewed

    Resolution sources

    Earlier ranking context

    This would determine whether pairwise loss of dependence guarantees all finite-order mixing. A counterexample would separate correlations among several widely separated events from those detectable using only two events.

    Earlier statement and context sources

    Link to this problem
  72. Historical rank419Lück's Determinant Conjecture for Discrete GroupsFor square matrices over the integer group ring of any discrete group, is their determinant-like logarithmic spectral average always nonnegative? The average comes from the matrix's adjoint times itself, is normalized by matrix size, and omits the zero part of the spectrum.Geometry & topologySource-reported disproved

    Historical rank #419 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every discrete group G, n≥1 and A∈M_n(Z[G]), let μ be the spectral measure of A* A for the canonical group trace tensored with normalized matrix trace. Determine whether ∫_(0,∞) log(t) dμ(t)≥0 always holds, with the atom at zero omitted and a priori value −∞ allowed.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 197: The integral group-ring matrix is a square matrix in the ranked class and is invertible over the rational group ring. Its strictly positive Fuglede–Kadison determinant below one gives a negative logarithmic determinant. Passing to the spectral logarithmic integral of A* A and normalized matrix trace multiplies this logarithm by a positive factor, preserving its negative sign; the stated finite logarithmic integral avoids a convention-dependent divergence.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would impose a universal lower bound on this logarithmic spectral quantity, even when the operator has a kernel. Failure would include either a negative value or divergence to minus infinity.

    Earlier status note

    Open · subcases solved

    October 6, 2026: Failure of determinant approximation is different from Lück’s determinant lower bound. The reported determinants 2 and at most 5^(1/3) do not contradict a lower bound of 1. The complete proof was not audited in this review.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Link to this problem
  73. Historical rank420Absolute continuity of Bernoulli convolutions outside reciprocal Pisot parametersChoose independent, equally likely signs in the geometric series ±1 ± λ ± λ² ± ⋯. For every 1/2<λ<1 whose reciprocal is not a Pisot number, must the resulting random sum have a probability density?Probability, ergodic theory & dynamicsSource-reported disproved

    Historical rank #420 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every lambda in (1/2,1) whose reciprocal is not a Pisot number, is the law nu_lambda of the random series sum_{n>=0} epsilon_n lambda^n, with independent signs epsilon_n equally likely +1 or -1, absolutely continuous with respect to Lebesgue measure?

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 153: The source proves singular unbiased Bernoulli convolutions at reciprocals of quartic Salem numbers in (1,2). Such reciprocals lie in (1/2,1), and Salem numbers have unit-circle conjugates, so they are not Pisot numbers. These are counterexamples within the exact ranked parameter class. The classification being an infinite approximation criterion rather than a finite algorithm does not weaken this negative answer.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would exclude singular distributions for all parameters in this class. A counterexample would identify a parameter outside reciprocal Pisot values whose distribution has a nonzero singular component.

    Earlier statement and context sources

    Link to this problem
  74. Historical rank424Irrationality of Catalan's ConstantIs the infinite alternating sum 1−1/3²+1/5²−1/7²+⋯, called Catalan’s constant, impossible to express exactly as a ratio of any two integers?Number theory & arithmetic geometrySource-reported proved

    Historical rank #424 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that Catalan's constant G = beta(2) = sum over n >= 0 of (-1)^n / (2n+1)^2 = 0.9159655941... is an irrational number.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 005: The alternating reciprocal odd-square series in the abstract is exactly the frozen definition G=beta(2). The claimed irrationality therefore affirms the full target, with no additional parameter, genericity, or conditional hypothesis; no transcendence assertion is inferred.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof of irrationality would exclude every rational representation of this specific constant. A negative answer would identify a rational value equal to the entire infinite sum.

    Earlier status note

    Open · disputed claim

    The irrationality of Catalan’s constant remains open. Sun’s September 3, 2026 proof claim is challenged by Wachs’s September 16 critique, which identifies an omitted prime-2 contribution and discrepancies in the arithmetic estimate. The critique is a substantive objection to this proof claim; it does not establish rationality of Catalan’s constant.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Formerly in the Top 100 most famous open problems.

    Link to this problem
  75. Historical rank425Separable Quotient Problem for Banach SpacesCan every infinite-dimensional complete normed space have a closed subspace collapsed to zero so that the resulting quotient is still infinite-dimensional but has a countable dense subset?Analysis & PDESource-reported relative independence · assuming Con(ZFC + a measurable cardinal)

    Historical rank #425 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that every infinite-dimensional Banach space has a quotient by a closed subspace that is both infinite-dimensional and separable.

    Status update

    Source-reported relative independence · assuming Con(ZFC + a measurable cardinal)

    Resolved (independent) on the basis of the OpenAI mathematics release of 6 October 2026. Relative independence assumes Con(ZFC + a measurable cardinal); this is neither an unconditional proof nor an unconditional disproof of the assertion. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 323: The paper defines SQ_F by bounded linear surjection onto a separable infinite-dimensional Banach space and explicitly identifies the equivalent quotient-by-closed-subspace formulation. Theorem 1.1 applies to both real and complex fields and says that real-valued measurability of the continuum implies SQ_F while CH implies its failure. It then explicitly states independence of the unrestricted assertion relative to consistency of ZFC with a measurable cardinal. This determines the ranked question metamathematically under that named consistency assumption; it is neither an unconditional ZFC proof of the universal quotient assertion nor an unconditional ZFC construction of a counterexample. Corollary 1.2 lowers the consistency assumption to Con(ZFC) only for the separate norm-density-aleph-one restriction, which must not replace the full-target assumption. The publisher formalization covers the conditional CH-negative direction only. The positive consistency direction and full independence theorem are not included in that selected Lean statement.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would guarantee a separable infinite-dimensional quotient even for spaces too large to have a countable dense subset themselves. A counterexample would identify a Banach space with no such quotient.

    Earlier status note

    Open · subcases solved

    October 6, 2026: Araújo, Barbosa and Seoane-Sepúlveda cover infinite-dimensional Banach spaces lacking the Lotz property. Their sufficient condition does not cover every infinite-dimensional Banach space. The complete proof was not audited in this review.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Link to this problem
  76. Historical rank430Long-Time Existence of the Calabi Flow on Compact Kaehler ManifoldsCalabi flow changes a Kähler metric to reduce variation in scalar curvature. If a class on a compact Kähler manifold contains a constant-scalar-curvature metric, must the flow from every smooth Kähler metric in that class run forever and converge smoothly to one?Analysis & PDESource-reported disproved

    Historical rank #430 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For a compact Kaehler manifold in a Kaehler class containing a constant-scalar-curvature Kaehler metric, prove that the Calabi flow from every smooth initial Kaehler metric in the class exists for all time and converges smoothly to a cscK metric.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 352: The ranked statement assumes that the Kahler class contains a cscK metric and asserts all-time existence and smooth convergence for every smooth initial metric. The supplied example is a smooth metric on CP^10 in the Fubini-Study class, which contains a cscK metric, and its Calabi flow becomes singular in finite time. Thus it violates the existence clause within the exact permitted class, and hence refutes the full conjunction.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would provide a globally convergent evolution within every such class, from every smooth starting metric. Failure could involve finite-time breakdown or lack of the asserted smooth convergence.

    Earlier statement and context sources

    Link to this problem
  77. Historical rank432Combinatorial Invariance of Kazhdan-Lusztig PolynomialsFor Bruhat intervals in Coxeter systems, is the associated Kazhdan–Lusztig polynomial determined entirely by the interval's directed graph? Must matching directed graphs give the same polynomial even when they come from different Coxeter systems?Algebra, representation & category theorySource-reported proved

    Historical rank #432 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For Coxeter systems (W_1,S_1) and (W_2,S_2), prove that any directed-graph isomorphism between Bruhat intervals [u,v] in W_1 and [w,z] in W_2 implies equality of the Kazhdan-Lusztig polynomials P_{u,v}(q)=P_{w,z}(q), or construct a counterexample.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 168: The abstract gives equality of equal-parameter Kazhdan-Lusztig polynomials for isomorphic Bruhat intervals in arbitrary Coxeter systems, including across different systems. The frozen target asks only for a directed-graph isomorphism of Bruhat intervals; such an isomorphism preserves the Bruhat order recovered by directed reachability. Thus the abstract-order statement covers the target's directed-graph hypothesis.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make the directed interval structure sufficient to determine these polynomials. A counterexample would show that additional information from the surrounding Coxeter systems is needed.

    Earlier statement and context sources

    Link to this problem
  78. Historical rank438Zariski Cancellation Problem (characteristic zero, dimension ≥ 3)If the product of an affine variety with a line is algebraically identical to ordinary (n+1)-dimensional affine space, must the variety already be n-dimensional affine space? Ask this for every n≥3 over every characteristic-zero field.Algebra, representation & category theorySource-reported disproved

    Historical rank #438 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every characteristic-zero field k and n >= 3, prove or disprove that B[X] isomorphic to k[X_1,...,X_{n+1}] implies B isomorphic to k[X_1,...,X_n] for affine k-algebras B.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 047: For k=C and n=4 the integral affine fourfold has X times A1 isomorphic to A5 but X not isomorphic to A4. Passing to coordinate rings gives precisely a counterexample B to the universally quantified implication for characteristic-zero fields and n >= 3. A single n=4 counterexample disproves that target; it does not answer the separate n=3 case. Stable-coordinate and affine-fibration consequences are not additional indexed targets.

    Status reviewed

    Resolution sources

    Earlier ranking context

    This asks whether adding an extra coordinate can conceal a difference from affine space. A positive answer would allow that coordinate to be cancelled from every isomorphism in the stated setting.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  79. Historical rank440Stein's Conjecture on the Hilbert Transform along Lipschitz Vector FieldsAssign a unit direction at each point of the plane, varying at a bounded rate. Must the Hilbert transform taken along those directions preserve square-integrability with a bounded increase in norm, when restricted to a sufficiently short scale controlled only by that rate?Analysis & PDESource-reported proved

    Historical rank #440 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every Lipschitz unit vector field v:R^2->S^1, prove that there exists epsilon>0, depending only at the natural scale on the Lipschitz norm of v, such that H_{v,epsilon}f(x)=p.v. integral from -epsilon to epsilon of f(x-yv(x)) dy/y defines a bounded operator on L^2(R^2).

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 083: The target requires an L^2-bounded short-scale principal-value Hilbert transform for every planar Lipschitz unit vector field, with scale controlled by its Lipschitz norm. The family states a uniform strong L^2 bound at an absolute multiple of the reciprocal Lipschitz seminorm, uniform in inner truncation, for fields depending on both coordinates, and explicitly gives the bounded principal-value operator. These are the target hypotheses and conclusion.

    Status reviewed

    Resolution sources

    Earlier ranking context

    The bound would control how much this direction-dependent singular integral can enlarge an L2 input. Its specified scale ties the control to the rate at which the direction field varies.

    Earlier statement and context sources

    Link to this problem
  80. Historical rank448Smooth Realization of Finite-Entropy Ergodic TransformationsCan every invertible ergodic probability-preserving system with finite entropy be modeled by a smooth map with a smooth inverse on some compact finite-dimensional manifold? The model must be measurably equivalent and preserve a smooth probability volume; the original space is standard and nonatomic.Probability, ergodic theory & dynamicsSource-reported disproved

    Historical rank #448 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that every invertible ergodic measure-preserving transformation of a standard nonatomic probability space with finite Kolmogorov–Sinai entropy is measurably isomorphic to a C-infinity diffeomorphism of some compact finite-dimensional smooth manifold preserving a smooth probability volume form.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 152: The source example is invertible and ergodic on a standard nonatomic probability space and has zero, hence finite, Kolmogorov–Sinai entropy. It excludes measurable conjugacy to any smooth diffeomorphism preserving a strictly positive smooth probability density in every finite dimension. A smooth probability volume form induces such a density, so the example refutes the ranked universal smooth-realization assertion, including its possible manifold choices.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would provide smooth finite-dimensional models for all transformations in this class. A counterexample would reveal measure-theoretic dynamics that no such smooth volume-preserving system can realize.

    Earlier statement and context sources

    Link to this problem
  81. Historical rank450Deligne–Drinfeld Freeness Conjecture for grt_1Is the rational Grothendieck–Teichmüller Lie algebra built from exactly one generator in every odd weight starting at three, with no extra relations beyond the Lie-algebra rules? This asks whether its known free motivic subalgebra is already the whole algebra.Number theory & arithmetic geometrySource-reported proved

    Historical rank #450 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Is the weight-graded Grothendieck–Teichmüller Lie algebra grt_1 over Q freely generated by one homogeneous generator sigma_(2k+1) in each odd weight 2k+1≥3, equivalently is the known free motivic Lie subalgebra Q-Lie(sigma_3,sigma_5,...) all of grt_1 in every weight? For the completed convention, require the corresponding isomorphism after weight completion.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 008: The abstract specifies the rational weight-graded Grothendieck-Teichmuller Lie algebra, one free generator in every odd weight >=3, and the corresponding weight-completed isomorphism. These are exactly the graded and completed alternatives in the frozen grt_1 target.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would identify the entire algebra with the free Lie algebra on these generators in every weight. In the completed setting, the same identification must hold after weight completion.

    Earlier statement and context sources

    Link to this problem
  82. Historical rank453Shub entropy conjectureCan growth in a map's action on topological features force at least as much dynamical complexity? For every continuously differentiable map of a compact smooth manifold to itself, Shub predicts topological entropy at least the logarithm of the spectral radius on real homology.Probability, ergodic theory & dynamicsSource-reported disproved

    Historical rank #453 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every C1 self-map f of a compact smooth manifold M, prove that the topological entropy h(f) is at least the logarithm of the spectral radius of the induced map on total real homology.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 151: The ranked assertion quantifies over every C1 self-map of a compact smooth manifold, not just invertible maps. The source constructs a noninvertible C1 map on a compact manifold with zero topological entropy and eigenvalue 2 on second real homology. Total real homology then has spectral radius at least 2, so the required entropy lower bound fails. The noninvertibility is allowed by this exact target.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would turn growth detected by homology into a universal lower bound on dynamical complexity at C1 regularity. A counterexample would show that homological growth can exceed this proposed entropy control.

    Earlier statement and context sources

    Link to this problem
  83. Historical rank458LeBrun-Salamon Conjecture: positive quaternion-Kahler manifolds are symmetric spacesMust every compact quaternion-Kähler manifold with positive scalar curvature be a symmetric space, meaning that reflection through each point extends to a global isometry?Geometry & topologySource-reported proved

    Historical rank #458 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that every compact quaternion-Kahler manifold with positive scalar curvature and holonomy contained in Sp(n)Sp(1) is a symmetric space.

    Status update

    Source-reported proved

    OpenAI’s 6 October 2026 release reports a proof of the LeBrun–Salamon conjecture in the standard quaternionic-Kähler setting: compact connected manifolds of real dimension 4m with m≥2, holonomy contained in Sp(m)Sp(1), and positive scalar curvature. Corollary 1.4 concludes that the metric is a compact symmetric Wolf metric up to scale, without an extra simply-connectedness assumption. The claim does not extend to arbitrary positively scalar-curved oriented four-manifolds. No independent proof audit, confirmation of refereeing, or local Lean verification is claimed.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would restrict the stated geometric class to symmetric spaces. A counterexample would exhibit a nonsymmetric manifold satisfying the same positivity and holonomy conditions.

    Earlier status note

    Open

    October 6, 2026: The new LeBrun-Salamon claim covers quaternionic dimension at most 14, or real dimension at most 56. It does not settle all dimensions. The complete proof was not audited in this review.

    Recorded . This note predates the resolution reported above.

    Earlier statement and context sources

    Link to this problem
  84. Historical rank461Kaplansky's Direct Finiteness ConjectureIn the group ring of every group over every field, does a product ab equal to one always force the reversed product ba to equal one?Algebra, representation & category theorySource-reported disproved

    Historical rank #461 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that for every group G, every field K, and all a,b in the group ring K[G], the equation ab=1 implies ba=1; equivalently, every group ring over a field is directly finite (and, by the known equivalence, stably finite).

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 197: The ranked assertion covers every group and field and asks whether ab=1 forces ba=1 in the scalar group ring. The characteristic-two companion explicitly gives finite sums with ab=1 and ba not equal to 1; the first manuscript even makes the group torsion-free over F2. The odd-characteristic companion supplies another counterexample but only at one specified odd prime, not all odd primes.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would guarantee that one-sided inverses in all these group rings are two-sided. A counterexample would identify a group and field for which multiplication admits this specific asymmetry.

    Earlier statement and context sources

    Link to this problem
  85. Historical rank465David–Semmes Riesz-transform problem in intermediate dimensionsFor an n-dimensional Ahlfors–David regular measure, meaning its mass has uniform n-dimensional growth, does uniform L² control of the truncated Riesz transforms force uniform n-rectifiability—a quantitative form of being surface-like? The target is 2≤n≤d−2 in ℝᵈ.Analysis & PDESource-reported proved

    Historical rank #465 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every pair of integers 2<=n<=d-2 and every n-Ahlfors–David regular Radon measure mu on R^d, does uniform L^2(mu) boundedness over epsilon>0 of R_{mu,epsilon}f(x)=integral_{|x-y|>epsilon}(x-y)|x-y|^(-n-1)f(y)dmu(y) imply that mu is uniformly n-rectifiable?

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 081: The frozen target is exactly the remaining intermediate-dimensional David-Semmes question: n-AD regular Radon measures on R^d, 2<=n<=d-2, uniform L^2 boundedness of all positive hard truncations of the n-dimensional vector Riesz transform, and uniform n-rectifiability. The abstract explicitly covers these dimensions, hard truncations and scalar-to-vector L^2 formulation, with big pieces of Lipschitz images as conclusion.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make this analytic boundedness condition sufficient for quantitative geometric regularity of the measure. A counterexample would exhibit bounded transforms without the stated uniformly rectifiable structure.

    Earlier statement and context sources

    Link to this problem
  86. Historical rank467Small Cohen-Macaulay ConjectureDoes every complete Noetherian local ring have a nonzero finitely generated module on which its parameters act one by one without killing a nonzero element, even after quotienting out the earlier parameters?Algebra, representation & category theorySource-reported disproved

    Historical rank #467 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that every complete Noetherian local ring R admits a nonzero finitely generated R-module on which a system of parameters of R is a regular sequence, equivalently a small maximal Cohen-Macaulay module.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 195: A three-dimensional complete Noetherian normal local domain containing C belongs to the target class of complete Noetherian local rings. Absence of every nonzero finitely generated maximal Cohen–Macaulay module directly contradicts the universal existence assertion. One domain counterexample suffices; it need not treat every characteristic.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would guarantee a finite module where the ring's parameters act successively without the specified zero-divisor obstructions. Failure would exhibit a ring for which no such small maximal Cohen-Macaulay module exists.

    Earlier statement and context sources

    Link to this problem
  87. Historical rank468Ergodicity of Irrational Triangular BilliardsIn every triangle whose angles are irrational multiples of π, is ideal billiard motion ergodic? That means there is no measurable collection of positions and directions, of neither zero nor full measure, that stays isolated under the motion.Probability, ergodic theory & dynamicsSource-reported proved

    Historical rank #468 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that every triangular billiard whose angles are irrational is ergodic for billiard flow.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 150: The frozen target asks ergodicity for triangular billiards whose angles are irrational. The family proves weak mixing, hence ergodicity, for every nondegenerate triangle with even one angle irrational relative to pi. Normalized area times uniform direction is the billiard-flow Liouville measure. The weaker companion states ergodicity explicitly and excludes only the usual null set of vertex-hitting trajectories.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make the stated condition on the angles sufficient for ergodicity. A counterexample would show that a triangle can meet that condition while retaining a measurable obstruction to ergodic behavior.

    Earlier statement and context sources

    Link to this problem
  88. Historical rank472CAT(0) Conjecture for All Artin GroupsDoes every Artin group given by a finite Coxeter matrix admit a proper, cocompact action by isometries on a CAT(0) geodesic metric space? The space may be chosen freely: it need not be the standard Salvetti complex or a cube complex.Geometry & topologySource-reported disproved

    Historical rank #472 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Decide whether every Artin group associated with a finite Coxeter matrix acts properly and cocompactly by isometries on a CAT(0) metric space, in the CAT(0)-group convention of Haettel Conjecture 1.1. The space is geodesic, and is not required to be the standard Salvetti complex or a CAT(0) cube complex.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 254: The finite 116-generator Artin matrix in Theorem 1.1 has off-diagonal labels 2, 3, and infinity, so it belongs to the finite-Coxeter-matrix class in the target; finite matrix does not mean that every label or the Coxeter group is finite. The source defines geometric action as proper and cocompact isometric action on a nonempty proper CAT(0) space and identifies the expected universal assertion with Haettel Conjecture 1.1, the same convention frozen in the target. Crucially, its definition paragraph explicitly states that ambient properness follows from the other action hypotheses for a length space. CAT(0) spaces in the target are geodesic and therefore length spaces, so allowing a merely geodesic ambient description does not avoid the obstruction under that convention. The theorem excludes every dimension and every choice of complex or metric, not only the Salvetti or cubical models. It therefore supplies a full counterexample.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would give every such Artin group a geometric model of nonpositive curvature. The question concerns the existence of that action across the whole class, including groups beyond the established special cases.

    Earlier statement and context sources

    Link to this problem
  89. Historical rank475Katok entropy rigidity conjectureA geodesic flow follows free motion along locally shortest paths. On a closed negatively curved manifold, do its Liouville and topological entropies—two measures of dynamical complexity—agree exactly when the geometry is locally symmetric?Probability, ergodic theory & dynamicsSource-reported proved

    Historical rank #475 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For a closed negatively curved Riemannian manifold, the Liouville entropy of the geodesic flow equals its topological entropy if and only if the metric is locally symmetric.

    Status update

    Source-reported proved

    OpenAI’s 6 October 2026 release reports entropy rigidity for closed connected smooth negatively curved manifolds of dimension at least three; the surface case is attributed to Katok’s earlier theorem. The named conjecture uses the connected-manifold convention stated by De Simoi and coauthors and cited in Humbert’s account. Equality of normalized Liouville entropy and topological entropy is equivalent to local symmetry, at arbitrary scale. A disconnected-manifold extension is outside this claim. No independent proof audit, confirmation of refereeing, or local Lean verification is claimed.

    Status reviewed

    Resolution sources

    Earlier ranking context

    The proposed equivalence would let equality of two measurements of geodesic-flow complexity characterize local symmetry. Its failure would show that this entropy comparison does not distinguish the metrics in the claimed way.

    Earlier statement and context sources

    Link to this problem
  90. Historical rank480Furstenberg No-Bigeodesics (Planar FPP, Standard Minimum Moment)Give square-grid edges independent nonnegative travel times from a law with no point masses. With probability one, is there no doubly infinite, non-self-intersecting route whose every finite segment is fastest between its endpoints, when the minimum of four independent times has finite second moment?Probability, ergodic theory & dynamicsSource-reported proved

    Historical rank #480 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every non-atomic probability law G on [0,infinity) such that E[min(t_1,t_2,t_3,t_4)^2]<infinity for four independent G-distributed variables, assign independent G-distributed passage times t_e to the unoriented nearest-neighbor edges of Z^2. Is it almost surely the case that there is no doubly infinite self-avoiding nearest-neighbor path (v_i)_{i in Z} for which, for every i<j, the segment from v_i to v_j minimizes the sum of edge passage times over all finite nearest-neighbor paths with those endpoints? The probability is for the product edge law. This is the standard minimum-moment version selected from the legacy canonical finite-moment scope using Damron–Hanson A1' and Ahlberg–Hoffman A1, not a claim of equivalence to Damron–Hanson's separately stated moment-free conjecture. No first moment of a single edge, exponential moment, absolute continuity of a density, curvature assumption, or deterministic asymptotic direction is imposed.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 212: Both statements concern nearest-neighbor undirected planar FPP with iid nonnegative nonatomic weights and exactly the finite second moment of the minimum of four independent weights. The manuscript rules out all doubly infinite geodesics simultaneously without curvature, density, exponential-moment, or prescribed-direction assumptions. Thus the published scope matches the carefully restricted ranked minimum-moment formulation.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would rule out all these globally extending optimal paths under the stated minimum-moment condition. A disproof would need a permitted law giving positive probability to such a path, without requiring a prescribed direction.

    Earlier statement and context sources

    Link to this problem
  91. Historical rank483Gaboriau's Fixed Price Problem for Countable GroupsDo all essentially free probability-measure-preserving actions of the same infinite countable group have the same cost, on standard probability spaces? Extended costs are allowed.AnalysisSource-reported disproved

    Historical rank #483 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Does every infinite countable group Gamma have fixed price: do all essentially free probability-measure-preserving actions of Gamma on standard probability spaces have the same cost? The cost of an action is the infimum, over symmetric irreflexive Borel graphings generating its orbit relation on a conull set, of half the integral of the vertex degree; extended values are allowed. A counterexample consists of one such group with two such actions of different cost. Finite groups are a settled subcase of the customary all-countable-groups formulation.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 259: The target is refuted by one infinite countable group with two essentially free probability-measure-preserving actions of different costs. The manuscript constructs exactly such a pair for a finitely generated group; finite generation ensures countability. The different-cost assertion, rather than just estimates on a sequence of extensions, directly supplies the required counterexample.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A positive answer would make cost independent of the chosen free action. A counterexample would consist of one group with two such actions of different cost.

    Earlier statement and context sources

    Link to this problem
  92. Historical rank484Persistence Conjecture for Weakly Reversible Mass-Action SystemsIn a chemical reaction network where each reaction can be followed by a path of reactions back to its starting complex, must every species stay bounded away from zero in the long run? Assume mass-action kinetics, positive initial concentrations, and a bounded stoichiometric compatibility class.Applied & computational mathematicsSource-reported proved

    Historical rank #484 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every weakly reversible chemical reaction network with mass-action kinetics, prove that every positive trajectory whose stoichiometric compatibility class is bounded is persistent: no species concentration tends to zero along the omega-limit set; or construct a counterexample within that class.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 149: The ranked persistence target is restricted to positive trajectories in bounded stoichiometric classes. The source proves a common compact absorbing set inside every positive class, with eventual strictly positive lower concentration bounds, and also explicitly states persistence without assuming a bounded class. It thus covers every target trajectory and excludes any boundary omega-limit point, with stronger classwise uniform bounds.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would rule out asymptotic species extinction under these structural and boundedness conditions. A counterexample would show that a positive trajectory can approach the boundary despite satisfying both conditions.

    Earlier statement and context sources

    Related problems

    • #292 · Global Attractor Conjecture for Complex-Balanced Mass-Action SystemsSource-reported proved · historical rankRelated problemGlobal Attractor Conjecture for Complex-Balanced Mass-Action Systems

      The global-attractor question asks for convergence under complex balance without assuming the whole compatibility class is bounded. This persistence question instead ranges over weakly reversible systems with bounded classes.

      Scope qualification: The overlap supports a bounded-class persistence-to-convergence argument. No implication between the two complete targets is established here.

    Link to this problem
  93. Historical rank488Köthe Conjecture on Nil One-Sided IdealsAlgebra, representation & category theorySource-reported disproved

    Historical rank #488 in the Oct 6, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or refute that a ring with no nonzero nil two-sided ideal has no nonzero nil one-sided ideal. Equivalent formulations include: the sum of two nil right ideals in any ring is nil; and for every nil ring R, the 2x2 matrix ring over R is nil.

    Status update

    Source-reported disproved

    The ranking source’s coordinator reports: Disproved by a nil-ideal matrix counterexample. At tadamcz/koethe commit 2463511ec0111a8decac6d5cfdf761a8cf783314, the inspected CI Comparator job builds the solution, compares the conjecture negation against the challenge, and records acceptance by both Lean and NanoDa using only propext, Quot.sound and Classical.choice. The formal construction supplies a nonnilpotent 2x2 matrix over a nil two-sided ideal. The standard column-ideal argument, reviewed here and given in the specialist manuscripts, refutes the original nil-one-sided-ideal statement. The original-formulation bridge is mathematical rather than part of this Lean development. Adamczewski-Böhmler-Marczinzik (7 September) give a human-readable account; Greenfeld-King-Vendramin (14 September) give a further point-module construction. No fresh local kernel run or journal acceptance is asserted. The paper’s contribution statement credits GPT-6 Astra with finding the initial construction in an experiment run by Tom Adamczewski.

    Status reviewed

    Resolution sources

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  94. Historical rank491Gottschalk's surjunctivity conjectureA cellular automaton updates configurations by applying the same local rule at every group-indexed position. With finitely many possible symbols, must every such automaton that never merges distinct configurations also be able to produce every configuration, for every group?Probability, ergodic theory & dynamicsSource-reported disproved

    Historical rank #491 in the Sep 22, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that every group is surjunctive; equivalently, for every group G and every finite alphabet A, every injective cellular automaton A^G -> A^G is surjective.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 197: The ranked surjunctivity assertion concerns all finite alphabets and all configurations A^G. The characteristic-two companion states an injective nonsurjective cellular automaton on K^G for a finite field K, so its alphabet is finite and its configuration space has the required unrestricted form. This is a full counterexample rather than failure only on periodic or finitely supported configurations.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would connect injectivity and surjectivity for all the specified cellular automata. A counterexample would identify a group-based system that preserves distinctions between inputs while excluding some configurations from its outputs.

    Earlier statement and context sources

    Link to this problem
  95. Historical rank497Quillen's Conjecture on Posets of Elementary Abelian p-SubgroupsFor a prime p, build a geometric complex from chains of nontrivial elementary abelian p-subgroups of a finite group. If that complex can be continuously shrunk to a point, must the group contain a nontrivial normal subgroup of prime-power order pᵏ?Algebra, representation & category theorySource-reported proved

    Historical rank #497 in the Sep 22, 2026 edition. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that for every finite group G and prime p, if the order complex of the poset A_p(G) of nontrivial elementary abelian p-subgroups is contractible, then G has a nontrivial normal p-subgroup, i.e. O_p(G) != 1.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 310: The exact target says that contractibility of A_p(G) implies O_p(G) is nontrivial, for every finite G and prime p. The manuscript proves the stronger contrapositive: when O_p(G)=1, augmented reduced rational homology of A_p(G) is nonzero. A contractible order complex has zero reduced rational homology, so this directly forces noncontractibility and proves the target with all group and prime quantifiers intact.

    Status reviewed

    Resolution sources

    Earlier ranking context

    A proof would make the topology of this subgroup complex detect a nontrivial normal p-subgroup. A counterexample would show that contractibility can occur even when the group's p-core is trivial.

    Earlier statement and context sources

    Research on ProofAtlas

    Link to this problem
  96. Source rank509Trace Reconstruction with Constant Deletion ProbabilityTheoretical computer scienceSource-reported disproved

    Historical source rank #509 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether an unknown string x in {0,1}^n can be reconstructed with high probability from a number of independent traces that is polynomial in n, where each trace is obtained by independently deleting each bit of x with a fixed constant probability delta; the best known bounds are an upper bound of exp(tilde-O(n^{1/5})) and a lower bound of tilde-Omega(n^{3/2}) traces (Chase).

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 122: The ranked question asks whether polynomially many independent deletion traces suffice to reconstruct every binary string with high probability at a fixed nontrivial deletion probability. The lower-bound manuscript asserts n^{Omega(log log n)} samples for every fixed q in (0,1), even with unrestricted computation and any fixed positive success probability. That excludes polynomial sample complexity, hence answers the ranked question negatively. The quasipolynomial upper bounds and low-deletion polynomial regime do not weaken this fixed-q lower bound or determine the exact optimal sample order.

    Status reviewed

    Resolution sources

    Link to this problem
  97. Source rank524Grothendieck's Homotopy Hypothesis for Globular Weak Infinity-GroupoidsAlgebra, representation & category theorySource-reported proved

    Historical source rank #524 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every coherator C in Ara's Grothendieck–Maltsiniotis globular formalism, prove that the fundamental infinity-groupoid functor induces an equivalence Hot → Ho(infinity-Gpd_C), where both categories are localized at their homotopy-group weak equivalences, as in Ara Conjecture 8.14.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 312: The coordinator retrieved the missing statement in Section 5. Corollary 5.2, PDF pages 24-25, explicitly gives the geometric-realization/fundamental-infinity-groupoid adjunction for every Grothendieck coherator C and says it is a Quillen equivalence. Together with the previously matched Ara coherator and homotopy-group weak-equivalence definitions, this supplies the requested fundamental-infinity-groupoid functor equivalence, not merely an unspecified abstract equivalence. Only the statement and definitions were used; the comparison proof was not audited. Statement reviewer context: The coherator issue is resolved at the definition level: Definition 2.1 explicitly identifies the free contractible globular-theory convention with Ara Sections 2.3 and 2.9-2.11, with the height bound on generating admissible fillers, not on all composite operations. The introduction defines weak equivalences by components and all based homotopy groups, matching the target localization. Theorem 1.1 proves elementary expansions are weak equivalences for every such coherator and cellular model. The introduction then claims a Quillen equivalence with spaces in Section 5. However, the supplied first-ten-page extract does not state that the particular fundamental infinity-groupoid functor of Ara Conjecture 8.14 induces that equivalence, nor give the comparison functors in Section 5. An abstract equivalence of homotopy theories cannot alone identify the specified functor. Preserve open status pending the Section 5 comparison statement, while recording that the coherator convention and weak-equivalence class match.

    Status reviewed

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  98. Source rank525Deterministic polynomial-time factorization of dense polynomials in F_q[x]Theoretical computer scienceSource-reported proved

    Historical source rank #525 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Construct a deterministic algorithm that, given a densely represented polynomial f in F_q[x] of degree n, outputs its factorization over F_q in time polynomial in n and log(q), or prove that no such algorithm exists in the stated computational model.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 142: Theorem 1.1 supplies uniform deterministic complete factorization over F_p, including multiplicities, in polynomial bit complexity in degree and log p, without randomness, an integer-factorization or primitive-root oracle, or GRH. The standard prime-field reduction transfers this to F_q with q=p^m in the explicitly represented finite-field model stated in Section 2. For a squarefree f, form A=F_q[x]/(f) as an F_p algebra and compute the kernel B of a->a^p-a by F_p linear algebra. If f has r irreducible factors, B is F_p^r. For each basis element b of B, compute its minimal polynomial over F_p and factor that polynomial using Theorem 1.1; it splits into linear factors. The gcds of b-a with current factors of f, for those roots a in F_p, separate all components as the basis is exhausted. Every space and polynomial involved has dimension or degree at most m deg(f), and only polynomially many operations/calls are required. Since m log p=log q, the overhead remains polynomial in degree and log q. Usual deterministic squarefree decomposition, with finite-field pth-root extraction when needed, restores multiplicities. This is a scope transfer using the released algorithm as a black box, not a review of its proof.

    Status reviewed

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  99. Source rank527Lipschitz Isomorphism Problem for Separable Real Banach SpacesAnalysisSource-reported disproved

    Historical source rank #527 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every pair of separable real Banach spaces X and Y, if there exists a bijection f:X->Y and finite constants 0<a<=b such that a||x-x'||<=||f(x)-f(x')||<=b||x-x'|| for all x,x' in X, must there exist a bounded real-linear bijection T:X->Y with bounded inverse? A proof for all such X,Y or a counterexample resolves the question. No requirement is made that the given map f be linear or that T preserve the norm. An existence proof of a counterexample suffices; no algorithmic construction is required.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 324: The first abstract constructs separable real Banach spaces that are globally bi-Lipschitz equivalent but not linearly isomorphic, which is precisely a counterexample to the indexed universal implication. The companion gives a more explicit mechanism: Z has no linear copy of c0 but is bi-Lipschitz equivalent to Z direct-sum-infinity c0. A linear isomorphism would carry the second space's c0 summand to a linear copy in Z, contradicting the stated property. These are global equivalences, not coarse embeddings or merely sphere maps.

    Status reviewed

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  100. Source rank532Zariski-Lipman ConjectureGeometry & topologySource-reported disproved

    Historical source rank #532 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that every complex algebraic variety or complex space whose tangent sheaf is locally free is smooth.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 048: The source constructs a singular normal affine complex surface with free rank-two tangent sheaf. A free tangent sheaf is locally free, and an affine complex surface lies within the ranked class of complex algebraic varieties. Its singularity directly negates the asserted smoothness conclusion, so the universal target is disproved by this example.

    Status reviewed

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  101. Source rank533Fixed Point Property for Nonexpansive Maps on Reflexive Banach SpacesAnalysis & PDESource-reported proved

    Historical source rank #533 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Also known as

    Does reflexivity imply the fixed point property?; Reflexive Banach space FPP problem

    Exact statement covered by this update

    For every reflexive Banach space X with its given norm, every nonempty norm-closed bounded convex subset C of X, and every map T:C->C satisfying ||T(x)-T(y)||<=||x-y|| for all x,y in C, must there exist x in C with T(x)=x? Convexity uses real coefficients, for real or complex Banach spaces; T need not be linear or affine. A negative resolution gives such an X,C,T with no fixed point. No change to the given norm and no stronger Cesàro-mean nonexpansiveness assumption is allowed.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 328: The source gives a fixed point for every nonexpansive selfmap of every nonempty closed bounded convex subset of a real reflexive Banach space, retaining the given norm. These are the target's quantifiers, with no uniform convexity, affine-map, renorming or Cesaro-mean hypothesis. A complex reflexive Banach space viewed as real remains reflexive with the same norm, real convex subsets and nonexpansive maps, so the real statement also covers the target's complex-field convention.

    Status reviewed

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  102. Source rank551Ryser Conjecture on r-Partite Hypergraph Vertex CoverCombinatorics & discrete geometrySource-reported disproved

    Historical source rank #551 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every r-partite hypergraph H, prove that its vertex-cover number tau(H) is at most (r-1) times its matching number nu(H).

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 162: An intersecting hypergraph has matching number one. For r=q+1 the source constructs an r-partite, r-uniform example with covering number r, while the ranked inequality would require tau<=r-1. Existence for sufficiently large prime q is enough to refute the all-r universal assertion; neither small-r cases nor all prime orders need be settled. Both constituent manuscripts give this kind of counterexample.

    Status reviewed

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  103. Source rank552Saxl Conjecture on the Staircase Kronecker SquareAlgebra, representation & category theorySource-reported proved

    Historical source rank #552 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Also known as

    Saxl conjecture; Staircase Kronecker-square conjecture; Staircase tensor-square conjecture

    Exact statement covered by this update

    For every integer k >= 1, with N = k(k+1)/2 and staircase partition rho_k = (k,k-1,...,1), does every irreducible complex representation S^lambda of S_N occur in S^{rho_k} tensor S^{rho_k}, equivalently is g(rho_k,rho_k,lambda) > 0 for every partition lambda of N?

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 205: The Saxl companion states the tensor-square inclusion for every staircase partition and every irreducible complex representation of its symmetric group. This exactly covers every k and partition lambda in the frozen target. The separate universal-square result excludes sizes 2, 4 and 9 but those exclusions are not imposed on the all-staircase Saxl statement; the two manuscripts must not be conflated.

    Status reviewed

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  104. Source rank553Kaplansky's Quasitrace Problem for C*-AlgebrasAnalysisSource-reported disproved

    Historical source rank #553 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every complex C*-algebra A, unital or nonunital, is every bounded 2-quasitrace tau:A->C additive on all of A, hence a positive linear trace? Here a 1-quasitrace satisfies tau(x*x)=tau(xx*)>=0 for every x, is complex-linear on each commutative C*-subalgebra, and satisfies tau(a+ib)=tau(a)+i tau(b) for self-adjoint a,b. A 2-quasitrace additionally extends to a 1-quasitrace on M_2(A) whose value on diag(a,0) is tau(a). Bounded means sup{|tau(a)|: ||a||<=1}<infinity. Resolve by proving this assertion or exhibiting an A and bounded 2-quasitrace that is not additive. No exactness, nuclearity, simplicity or separability hypothesis is imposed; unbounded extended-valued variants are not added to this contract.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 294: The ranked target allows any complex C*-algebra and asks whether every bounded 2-quasitrace is additive. The abstract supplies a separable unital algebra admitting normalized 2-quasitraces but no tracial state. A normalized 2-quasitrace on a unital algebra is bounded, and additivity would make it a tracial state, contradicting the stated property. Thus the example lies within the exact bounded contract; it is not merely an unbounded extended-valued quasitrace.

    Status reviewed

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  105. Source rank558Wall's D(2) ProblemGeometry & topologySource-reported disproved

    Historical source rank #558 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every finite connected CW complex X of dimension at most 3 such that H_3(\widetilde X; Z)=0 and H^3(X; M)=0 for every local coefficient system M of Z[pi_1(X)]-modules, there exists a finite CW complex Y of dimension at most 2 that is homotopy equivalent to X. Here \widetilde X is the universal cover, coefficients range over all modules with the usual local-system convention, and homotopy equivalence is ordinary, not necessarily simple.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 321: The target permits finite connected complexes of dimension at most three and demands a finite complex of dimension at most two with the same ordinary homotopy type. The source supplies a finite three-dimensional complex with the required vanishing of universal-cover homology above degree two and third cohomology for every local module, but no such finite two-dimensional homotopy model. Its infinite fundamental group is allowed by the unrestricted target; neither simple homotopy equivalence nor finite fundamental group is required.

    Status reviewed

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  106. Source rank559The Finitistic Dimension Conjecture for Finite-Dimensional AlgebrasAlgebra, representation & category theorySource-reported disproved

    Historical source rank #559 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that for every finite-dimensional algebra A over a field, the little finitistic dimension findim(A) = sup {projdim(M) | M in mod A, projdim(M) < infty} is finite.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 198: The target asks finiteness of the supremum of finite projective dimensions among finitely generated modules for every finite-dimensional algebra over a field. A finite-dimensional complex algebra with finite-dimensional modules of unbounded finite projective dimensions has infinite little finitistic dimension, exactly contradicting that universal target. This is not merely the big finitistic-dimension variant.

    Status reviewed

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  107. Source rank562Zarankiewicz Crossing Number Conjecture for Complete Bipartite GraphsCombinatorics & discrete geometrySource-reported proved

    Historical source rank #562 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove for all positive integers m,n that cr(K_{m,n})=floor(m/2)floor((m-1)/2)floor(n/2)floor((n-1)/2).

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 165: The second manuscript states the ordinary crossing-number equality for K_{m,n} for every pair of positive integers m,n, with exactly the four floor factors in the frozen target. The domain, quantifiers and crossing-number model agree. This does not concern the different extremal-edge Zarankiewicz problem for K_{t,t}-free graphs.

    Status reviewed

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  108. Source rank565Automorphism Problem for the Turing DegreesLogic, foundations & set theorySource-reported proved

    Historical source rank #565 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether the partially ordered set of Turing degrees is rigid: prove that every order automorphism of the Turing degrees is the identity, or construct a nonidentity order automorphism.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 241: The target asks whether every order automorphism of the full partial order of Turing degrees is the identity. The manuscript states exactly this full-order rigidity theorem, with no restriction to a cone, definability class, degree segment or particular automorphism. It therefore supplies a positive exact-target resolution under the user's release-based policy, without an independent proof audit.

    Status reviewed

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  109. Source rank566Kapovich-Kleiner Conjecture for Hyperbolic Groups with Sierpiński Carpet BoundaryGeometry & topologySource-reported proved

    Historical source rank #566 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Decide whether every Gromov-hyperbolic group whose boundary at infinity is homeomorphic to the standard Sierpiński carpet admits a proper cocompact isometric action on a convex subset of real hyperbolic three-space with nonempty totally geodesic boundary.

    Status update

    Source-reported proved

    The Cannon theorem reported in OpenAI’s 6 October 2026 release implies the Kapovich–Kleiner carpet claim through the original paper’s Theorem 5 and the paragraph following Conjecture 6 (p. 649). This gives the stated proper cocompact action on a convex subset of hyperbolic three-space with nonempty totally geodesic boundary. The implication has no torsion-free restriction; proper actions allow finite kernels. Corollary 13 alone treats the torsion-free case. No independent proof audit, confirmation of refereeing, or local Lean verification is claimed.

    Status reviewed

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  110. Source rank568The PPT-Squared Conjecture on Compositions of PPT Quantum ChannelsQuantum information & computationSource-reported disproved

    Historical source rank #568 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Decide whether, in every finite dimension, the composition of any two quantum channels that are positive partial transpose (PPT) is an entanglement-breaking channel; equivalently, whether two PPT links in a repeater chain can never be used to swap entanglement to the terminal systems.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 272: The target universally asserts that composing any two finite-dimensional PPT quantum channels is entanglement breaking. The separate trace-preserving PPT channel on M_21 has a square that is not entanglement breaking; choosing the same channel for both maps is already a counterexample to the universal two-map assertion. Thus this conclusion does not depend on any secret-key protocol equivalence.

    Status reviewed

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  111. Source rank581Kadison-Kastler Spatial Isomorphism Conjecture for von Neumann AlgebrasAnalysis & PDESource-reported proved

    Historical source rank #581 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every epsilon > 0, prove that there is delta > 0 such that any two von Neumann algebras M,N in B(H) with Kadison-Kastler Hausdorff distance between unit balls less than delta are unitarily conjugate by a unitary u with ||u - I|| < epsilon; or produce a counterexample.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 289: The target is already the strong form: for each epsilon there must be a universal delta forcing conjugacy by u with ||u-I||<epsilon when unit balls are delta-close. The principal abstract states exactly this, uniformly over all von Neumann algebras, representations and Hilbert spaces in the common unital setting. The one-sided near-inclusion and norm-separable C*-algebra counterexamples concern different hypotheses and do not refute the ranked two-sided von Neumann statement.

    Status reviewed

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  112. Source rank583Kadison's Single-Generator Problem for Separable von Neumann AlgebrasAnalysis & PDESource-reported proved

    Historical source rank #583 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether every separable von Neumann algebra is generated, as a von Neumann algebra, by one operator; equivalently whether every separable von Neumann algebra is generated by two self-adjoint elements or two unitaries, with the unresolved core being separable II_1 factors.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 296: The ranked target is single generation of separable von Neumann algebras, with the unresolved core in II_1 factors. The manuscript proves generation by one operator, equivalently two self-adjoint operators, for every II_1 factor with separable predual and explicitly states that the established direct-integral reduction gives all von Neumann algebras with separable predual. This matches the target separability convention and reaches beyond the relative-generation strengthening.

    Status reviewed

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  113. Source rank588C1 Interior Regularity of Infinity-Harmonic Functions in Higher DimensionsAnalysis & PDESource-reported proved

    Historical source rank #588 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Also known as

    Continuous differentiability of infinity-harmonic functions; C1 regularity for the infinity-Laplace equation; Regularity of absolutely minimizing Lipschitz extensions

    Exact statement covered by this update

    For every integer n>=3, every open connected set Omega in R^n, and every real continuous viscosity solution u of Delta_infinity u = sum_{i,j=1}^n (partial_i u)(partial_j u)(partial_ij u)=0 in Omega, must u belong to C^1(Omega)? A universal proof or an infinity-harmonic counterexample in one such dimension resolves the question. This is interior continuity of the gradient; no boundary regularity, fixed Holder exponent, or restriction to convex-ring potentials is required.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 377: The source asserts uniform interior C1,alpha_d estimates in every dimension d>=3 and explicitly states local C1 regularity on arbitrary open subsets. A continuous viscosity infinity-harmonic function in the ranked target is bounded on each compactly contained ball, so the local boundedness required by the estimate is automatic; rescaling such balls yields continuous gradients throughout Omega. Positive Holder exponents and quantitative estimates are stronger than the target, which requests only interior C1 and no boundary assertion.

    Status reviewed

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  114. Source rank589Hindman's four-expression partition-regularity conjectureCombinatorics & discrete geometrySource-reported proved

    Historical source rank #589 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every integer r>=1 and every coloring chi:N_{>0}-> {1,...,r}, do there exist positive integers x,y with chi(x)=chi(y)=chi(x+y)=chi(xy)? No distinctness of x and y or of the four displayed values is additionally required; zero, rational and finite-field domains are not the target.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 164: Take k=2 in the source's all-finite-colorings theorem on positive integers. The resulting two-element set {x,y} has nonempty subset sums x,y,x+y and nonempty subset products x,y,xy all in one common color, exactly the four required expressions. The source even supplies distinct x and y, which the frozen target does not require. No infinite-set assertion is being inferred.

    Status reviewed

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  115. Source rank591Kadison-Ringrose Vanishing Conjecture for Continuous Hochschild CohomologyAnalysisSource-reported proved

    Historical source rank #591 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every von Neumann algebra M and every integer n>=1, is H^n_cont(M,M)=0? Cohomology uses all bounded n-linear maps M^n to M with the usual Hochschild differential, and coefficients in the natural M-bimodule M. Equivalently every bounded cocycle is a bounded coboundary. No separability restriction is imposed. A universal proof or a counterexample resolves the target; degree one is already known and the unresolved scope lies in higher degrees for type II_1 algebras.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 295: The target uses all bounded multilinear Hochschild cocycles with coefficients in the von Neumann algebra itself, for every positive degree and without separability. The abstract asserts bounded primitives in every degree at least two on every complex von Neumann algebra and explicitly combines this with the established degree-one inner-derivation theorem. It therefore supplies the full continuous-cohomology vanishing target, not only normal or completely bounded cohomology.

    Status reviewed

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  116. Source rank595Banach's Simple Lebesgue Spectrum ProblemProbability, ergodic theory & dynamicsSource-reported proved

    Historical source rank #595 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Construct, or prove impossible, an invertible ergodic probability-measure-preserving transformation T whose Koopman operator on L^2_0 has Lebesgue maximal spectral type with multiplicity one; equivalently, find f whose two-sided orbit {f∘T^k:k∈Z} is a complete orthogonal system in L^2_0.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 144: A smooth standard-volume-preserving diffeomorphism of the three-torus is invertible and probability preserving. Simple Lebesgue spectrum on the entire complex mean-zero L2 space gives exactly the ranked multiplicity-one spectral condition, and its absence of nonzero invariant mean-zero functions implies ergodicity. The smooth realization is stronger than the target requirement of an abstract probability-preserving transformation.

    Status reviewed

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  117. Source rank612Exact Overlaps Conjecture for self-similar sets on the real lineProbability, ergodic theory & dynamicsSource-reported proved

    Historical source rank #612 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Let Phi = {phi_i}_{i in I} be a finite iterated-function system of contracting similarities of the real line, let X be its self-similar attractor, and let s be the similarity dimension. Prove that if dim_H(X) < min{1,s}, then Phi has an exact overlap: there exist distinct finite words u and v with phi_u = phi_v.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 148: The measure theorem implies the ranked set statement. If there is no exact overlap, distinct words give distinct composed maps, so random-walk entropy is the ordinary symbol entropy. Choose weights p_i=|r_i|^s, where s is the similarity dimension; then entropy divided by the Lyapunov exponent is s. The theorem gives a measure supported on the attractor with dimension min(1,s), preventing the ranked dimension drop. Unequal and negative contractions are explicitly included.

    Status reviewed

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  118. Source rank622Aaronson-Kuperberg Unitary Synthesis ProblemQuantum information & computationSource-reported proved

    Historical source rank #622 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether there is an efficient quantum query algorithm R such that, for every dimension d and every unitary U in U(d), some Boolean function f:{0,1}^m->{0,1} lets R^f approximate U within diamond-distance error 1/2 while the query complexity t and oracle input length m obey t m <= poly(log d); or prove that no such algorithm exists.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 283: The manuscript gives one uniform polynomial-size circuit for each qubit count, with target-dependent Boolean oracle, polynomial query length and calls, and full diamond error at most 1/2. These are the target's oracle and error conventions; efficient construction of the oracle is not required there. Arbitrary d can be encoded in ceil(log2 d) qubits by extending U to the next power-of-two dimension, preserving polynomial log-d resources and the restricted channel error.

    Status reviewed

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  119. Source rank626Graham Conjecture on Euclidean Ramsey SetsCombinatorics & discrete geometrySource-reported disproved

    Historical source rank #626 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether a finite set X in Euclidean space is Euclidean Ramsey if and only if X is spherical, where Euclidean Ramsey means that for every finite number of colors some dimension n has a monochromatic congruent copy of X in every coloring of R^n.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 172: The primary definition is the exact original-scale Ramsey property: for every finite number of colors a dimension must work for arbitrary colorings, with congruence preserving all distances and no measurability requirement. The paper’s stated geometric consequences give nine points on a circle failing its necessary-and-sufficient tensor criterion. The publisher scope further explicitly states a twelve-point unit-circle configuration with a fifty-color obstruction in every positive dimension, as well as a nine-point non-Ramsey circle set. Such a finite circle configuration is spherical, while one fixed finite coloring obstruction in every dimension negates the Ramsey quantifiers. It therefore refutes Graham’s spherical-implies-Ramsey direction. This conclusion uses the spherical counterexamples themselves, not the separate non-subtransitive Ramsey quadrilaterals that refute Leader–Russell–Walters.

    Status reviewed

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  120. Source rank629Brennan's Conjecture on Integrability of Derivatives of Conformal MapsAnalysis & PDESource-reported proved

    Historical source rank #629 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that for every conformal map f from the unit disk D onto a simply connected planar domain (equivalently, for the inverse Riemann map F of a simply connected domain Omega), the integral of |f'|^p over D is finite for every p in (-2, 2/3) (equivalently, the integral of |F'|^p over Omega is finite for every p in (4/3, 4)).

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 072: The ranked target explicitly includes the inverse Riemann map formulation: integrability of its derivative to every exponent strictly between 4/3 and 4 on the simply connected domain. The first manuscript states precisely that assertion for every conformal bijection onto the disk. No endpoint claim is needed. The different Kraetzer spectrum counterexample in the companion manuscript is not a disproof of Brennan.

    Status reviewed

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  121. Source rank635Blaschke Conjecture for Compact Rank-One Symmetric SpacesGeometry & topologySource-reported proved

    Historical source rank #635 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Every closed connected Riemannian manifold M of positive dimension satisfying inj(M)=diam(M) is isometric, after multiplying its metric by a positive constant, to a compact rank-one symmetric space with its standard metric: a sphere, a real, complex or quaternionic projective space, or the Cayley plane.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 344: Both source and target quantify over all closed connected smooth Riemannian manifolds of positive dimension with global injectivity radius equal to diameter. The source concludes the metric, up to scaling, is a standard compact rank-one symmetric-space metric, exactly the target's isometric classification including spheres and real, complex, quaternionic and Cayley projective cases. It is a full metric conclusion, not merely a diffeomorphism, homotopy or cohomology classification.

    Status reviewed

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  122. Source rank637Balanced L3 Boundedness of the Triangular Hilbert TransformAnalysis & PDESource-reported proved

    Historical source rank #637 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Also known as

    Triangular Hilbert transform conjecture; Symmetric L3 estimate for the triangular Hilbert transform

    Exact statement covered by this update

    For Schwartz functions f,g,h:R^2->C, define Lambda_3(f,g,h) as the limit as epsilon decreases to zero of the integral over {(x,y,z) in R^3: |x+y+z|>epsilon} of f(x,y)g(y,z)h(z,x)/(x+y+z) dx dy dz. Is there a universal finite constant C such that |Lambda_3(f,g,h)| <= C ||f||_(L^3(R^2)) ||g||_(L^3(R^2)) ||h||_(L^3(R^2)) for every such triple? A negative resolution must show that no such C exists. This is the flat Euclidean triangular Hilbert transform, without extra structure on an input, averaging over directions, curvature or a truncation-dependent loss.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 082: The target asks for the flat scalar triangular Hilbert form at the balanced L^3 point for arbitrary complex Schwartz inputs. The continuous maximal theorem gives L^3 x L^3 to L^(3/2) control over both hard truncation endpoints, with joint norm and almost-everywhere principal values. Pairing with the third L^3 input yields the target form bound, as the abstract explicitly states. Annular r-variation is additional strength; the separate real dyadic theorem is not used as a substitute for the Euclidean complex result.

    Status reviewed

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  123. Source rank643The Auslander-Reiten Conjecture on Vanishing of Self-ExtensionsAlgebra, representation & category theorySource-reported disproved

    Historical source rank #643 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that for any Artin algebra Lambda and finitely generated Lambda-module M, if Ext^i_Lambda(M, M oplus Lambda) = 0 for all i > 0, then M is projective.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 199: A finite-dimensional algebra over the stated rational-function field is an Artin algebra, and the supplied finite-dimensional module is finitely generated. The abstract gives vanishing of Ext^i(Z,Z) and Ext^i(Z,Lambda) for every positive i while Z is nonprojective, which is precisely vanishing against Z direct-sum Lambda in the target. Characteristic two is allowed by its universal field scope.

    Status reviewed

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  124. Source rank645Eilenberg–Ganea ConjectureAlgebra, representation & category theorySource-reported disproved

    Historical source rank #645 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that every group G of integral cohomological dimension cd(G)=2 has geometric dimension gd(G)=2, equivalently that G admits a two-dimensional Eilenberg–Mac Lane complex K(G,1).

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 249: The source constructs a group with integral cohomological dimension exactly 2 and geometric dimension exactly 3. This directly falsifies the target assertion that every integral-cd-2 group admits a two-dimensional K(G,1). The extra finite generation and residual finiteness strengthen the counterexample, and no finite-cell restriction is being used to replace geometric dimension.

    Status reviewed

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  125. Source rank657Petty's Conjectured Projection InequalityCombinatorics & discrete geometrySource-reported proved

    Historical source rank #657 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every integer n≥3 and compact convex body K⊂R^n with nonempty interior, determine whether V_n(ΠK)/V_n(K)^(n−1)≥V_n(ΠB)/V_n(B)^(n−1), with equality if and only if K is an ellipsoid, where B is the Euclidean unit ball and h_(ΠK)(u)=V_(n−1)(K|u⊥) for unit u.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 088: Theorem 1.1 treats every compact convex body with nonempty interior in every dimension n>=4, with no symmetry or boundary-regularity hypothesis, and gives equality exactly for translates of invertible linear images of the Euclidean ball. Its constant kappa_(n-1)^n*kappa_n^(2-n) is exactly the target ball ratio because Pi(B)=kappa_(n-1)B. The introduction then explicitly states that Chen, Feng, Li, Xi and Xu, Theorem 1.1, prove the unrestricted n=3 inequality including ellipsoid equality, and explicitly says the combination establishes the conjecture for every n>=3. This new frozen release assertion supplies the previously missing dimension-three statement and provenance within the user-authorized source policy. Full coverage follows by the union of n=3 and n>=4; it is not inferred from the catalogue word remaining alone. The Brannen simplex maximum concerns the opposite bound and plays no role.

    Status reviewed

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  126. Source rank664One-Dimensional Lieb-Thirring Sharp-Constant ConjectureMathematical physicsSource-reported proved

    Historical source rank #664 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove the conjectured sharp value of the one-dimensional Lieb-Thirring constant L_gamma,1 for Schrodinger operators for every gamma with 1/2 < gamma < 3/2; equivalently show that the optimal constant equals the one-bound-state or Keller value throughout that open interval.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 262: The finite-matrix theorem gives the scalar one-bound-state sharp constant for the full sum of negative eigenvalue moments at every 1/2<gamma<3/2, independently of matrix size. Matrix size one is precisely the ranked scalar one-dimensional inequality, also stated directly in the scalar companion. No commutativity or rank hypothesis narrows the scalar conclusion.

    Status reviewed

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  127. Source rank676Finite Lattice Representation ProblemAlgebra, representation & category theorySource-reported disproved

    Historical source rank #676 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether every finite lattice is isomorphic to the congruence lattice of some finite algebra.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 206: The target asks whether every finite lattice is the full congruence lattice of a finite algebra. The negative-solution companion expressly produces a finite nonempty lattice not representable by any finite nonempty algebra of any finite signature. That directly refutes the universal claim, while the companion undecidability and subgroup-interval statements are additional results rather than necessary qualifications.

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  128. Source rank693Guy Conjecture on the Crossing Number of the Complete Graph K_nCombinatorics & discrete geometrySource-reported proved

    Historical source rank #693 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that the crossing number cr(K_n) of the complete graph K_n equals Z(n) = floor(n/2)*floor((n-1)/2)*floor((n-2)/2)*floor((n-3)/2)/4, for all n.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 165: For every positive integer n the first manuscript gives the ordinary crossing number of K_n as the product of the same four floor factors divided by four. This is precisely the frozen Guy/Harary-Hill formula. The source concerns unrestricted ordinary plane drawings, not only rectilinear or specially constrained drawings, and has no sufficiently-large-n restriction.

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  129. Source rank701Le's mu-Constant Conjecture for Surface Singularities in C^3Geometry & topologySource-reported proved

    Historical source rank #701 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every holomorphic one-parameter family f_t:(C^3,0)→(C,0) of isolated hypersurface singularities whose Milnor number at 0 is constant for t near 0, prove that the embedded topological type of the surface germ f_t^{-1}(0) is locally constant in t.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 064: For every holomorphic one-parameter family of isolated hypersurface singularities in C3 with constant Milnor number, the manuscript provides a jointly continuous ambient topological right-trivialization fixing parameter and origin and preserving defining functions. Restricting that trivialization to their zero sets gives local constancy of embedded surface-germ topology, exactly the ranked target and in fact a stronger function-preserving conclusion.

    Status reviewed

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  130. Source rank707Generalized Mukai Conjecture on Picard Number and PseudoindexGeometry & topologySource-reported proved

    Historical source rank #707 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every smooth complex projective Fano manifold X of dimension n≥1, let ρ_X be its Picard number and ι_X=min{−K_X·C:C⊂X is an irreducible rational curve} its pseudoindex. Then ρ_X(ι_X−1)≤n, with equality if and only if X is isomorphic to (P^(ι_X−1))^(ρ_X).

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 063: The source and target both quantify over every positive-dimensional smooth complex projective Fano manifold and use the pseudoindex, the minimum anticanonical degree of a rational curve. Both give rho*(iota-1) <= n, with equality precisely for the product of rho copies of P^(iota-1). Thus the inequality and the complete equality characterization match, with no Picard-rank, dimension or family restriction.

    Status reviewed

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  131. Source rank708Courtade-Kumar Most Informative Boolean Function ConjectureCryptography, coding, information & optimizationSource-reported proved

    Historical source rank #708 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every n>=1, Boolean f:{0,1}^n->{0,1}, and p in [0,1/2], let X be uniform and Y=X xor Z for independent Bernoulli(p) noise coordinates Z. Prove or disprove I(f(X);Y)<=1-h_2(p), with information in bits and binary entropy h_2; unbalanced f are included.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 119: The ranked statement covers every Boolean function on a uniform binary cube, including unbalanced outputs, under independent symmetric bit-flip noise. The first manuscript explicitly resolves Courtade-Kumar as the Boolean specialization of its stronger fixed-information binary-channel contraction theorem. The second independently supplies Hellinger contraction for arbitrary Boolean bias and all noise correlations, and states that this yields the information bound. No balanced-output restriction is introduced; p=0 and p=1/2 are the elementary entropy and independence endpoints.

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    Link to this problem
  132. Source rank734Modified integer round-up property for one-dimensional bin packingApplied & computational mathematicsSource-reported disproved

    Historical source rank #734 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Given item sizes a_1,...,a_n in (0,1], let OPT be the minimum number of unit-capacity packing patterns needed to cover all items and let OPT_f be the optimum of the pattern-based linear programming relaxation. Prove that OPT <= ceil(OPT_f)+1 for every instance.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 118: The ranked modified integer round-up inequality requires OPT<=ceil(OPT_f)+1 for every bin-packing instance. The catalogue constructs arbitrarily large additive gaps between the same configuration-LP value and the integral optimum, already for rational sizes greater than 1/6. Since ceil(OPT_f)+1 is at most OPT_f+2, any asserted gap greater than two is a counterexample. Thus this restricted item-size construction suffices to disprove the universal ranked claim; the separate additive NP-hardness theorem is not required.

    Status reviewed

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    Link to this problem
  133. Source rank737Purely Cosmetic Surgery ConjectureGeometry & topologySource-reported proved

    Historical source rank #737 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that no nontrivial knot in S^3 admits purely cosmetic surgeries: if Dehn surgeries on a knot K along distinct slopes r and r' yield orientation-preservingly homeomorphic 3-manifolds, then K is the unknot or r = r'.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 306: Both the exact target and the abstract concern distinct Dehn surgery slopes on a nontrivial knot in S^3 and orientation-preserving homeomorphism of the resulting closed manifolds. The source covers every nontrivial smooth knot and, according to the family summary, includes the meridional slope. No hyperbolicity, genus or slope restriction appears. The orientation-preserving qualifier is essential and does not assert the chirally cosmetic version.

    Status reviewed

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    Link to this problem
  134. Source rank743Integrality and l-independence of the pairing of a Lefschetz class with an absolute Hodge classNumber theory & arithmetic geometrySource-reported proved

    Historical source rank #743 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Let A be a CM abelian variety over Q-bar and c an absolute Hodge class. Reduce A to A0 over F_p and take a Lefschetz class x on A0 of complementary dimension. Determine whether x·c is in Q and independent of l.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 001: The frozen target asks rationality and l-independence of the complementary Lefschetz/absolute-Hodge pairing after reduction of a CM abelian variety. Lefschetz classes are generated by divisor products, so the stated rational pairing theorem for all abelian varieties with good reduction includes this CM case; it also includes the crystalline realization and residue characteristic 2. The word integrality in the title does not add an integral-valued requirement to the exact target, which asks membership in Q. Family 032: Theorem 1.1 proves rational algebraicity for every Hodge class in every codimension on every complex CM abelian variety. For the target CM variety over Q-bar, choose a complex embedding; its rational absolute Hodge class is in particular a rational Hodge class and therefore has a rational algebraic cycle representative. The usual descent/spreading of an algebraic representative to Q-bar and specialization at the chosen good reduction identify the reduced l-adic class with a specialized rational algebraic cycle. A Lefschetz class is, by the conventional definition, a rational combination of products of divisor classes on the reduced abelian variety, so it too is represented by a rational algebraic cycle. Their complementary-degree pairing is the degree of their intersection, a single rational number, and the cycle-class and specialization compatibilities identify that same number for every admissible l. The divisors defining the Lefschetz class need not lift to characteristic zero. Thus the stronger CM Hodge assertion supplies the target rationality and l-independence by this elementary cycle-class implication. The target asks membership in Q, not integrality in Z despite its inherited title.

    Status reviewed

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  135. Source rank747Kervaire-Laudenbach Conjecture on Adjoining a Generator and a RelatorAlgebra, representation & category theorySource-reported proved

    Historical source rank #747 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every nontrivial group G and every element w of the free product G * Z, determine whether the quotient (G * Z)/<<w>> by the normal closure of w is nontrivial.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 256: The target asks whether (G*Z)/<<w>> is nontrivial for every nontrivial G and arbitrary w. Saying that no such free product is normally generated by one element is precisely that assertion for every w. The companion Kervaire manuscript supplies the full statement, so the mapping does not rely only on the nonsingularity hypothesis in the Howie manuscript.

    Status reviewed

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  136. Source rank750Erdős–Gallai Cycle Decomposition ConjectureCombinatorics & discrete geometrySource-reported proved

    Historical source rank #750 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether there exists an absolute constant C such that every finite simple graph G on n>=1 vertices has an edge partition into at most Cn pieces, each a simple cycle or a single edge.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 181: The source and target both quantify over every finite simple graph on n vertices and permit an edge partition into at most Cn pieces, each a simple cycle or a single edge, for one absolute constant C. No connectedness or even-degree hypothesis is imposed, and the desired linear rather than exact optimal constant is precisely what is supplied.

    Status reviewed

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    Link to this problem
  137. Source rank753Lech's Conjecture on Multiplicities under Flat Local ExtensionsAlgebra, representation & category theorySource-reported proved

    Historical source rank #753 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every flat local homomorphism (R,m)->(S,n) of Noetherian local rings, the Hilbert–Samuel multiplicities at the maximal ideals satisfy e_m(R)<=e_n(S).

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 194: Both statements quantify over flat local homomorphisms of Noetherian local rings and assert that Hilbert–Samuel multiplicity cannot decrease. The abstract expressly allows arbitrary dimension, residue fields and characteristics, so its nonzero local-ring convention does not leave a ranked characteristic or dimension regime uncovered.

    Status reviewed

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  138. Source rank754The Bass (Hattori-Stallings) Trace Conjecture for Group RingsAlgebra, representation & category theorySource-reported proved

    Historical source rank #754 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every group G and every finitely generated projective ZG-module P, prove that the Hattori-Stallings rank of P has zero coefficient on every conjugacy class of G other than the identity class.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 207: The algebraic companion explicitly distinguishes the complex conclusion from the integral target: complex projective traces may have nonidentity finite-order support, whereas the integral assertion has identity-only support. Immediately after Theorem 1.1, which applies to every discrete group, the introduction states that this integral identity-only assertion follows in Section 9 by combining the theorem with the known restriction on integral finite-order coefficients. This is the requested published statement of the passage to ZG, not an inference that complex finite-order support alone equals the target. It matches all finitely generated projective integral group-ring modules and does not impose torsion-freeness, countability, or a geometric hypothesis on G. The finite-order coefficient issue is therefore addressed at the author-stated scope level, sufficient under the authorized release-based policy without auditing the Section 9 argument.

    Status reviewed

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  139. Source rank758Critical Dimension for Singular Minimizing Cones in the One-Phase Bernoulli ProblemAnalysis & PDESource-reported determined

    Historical source rank #758 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine the smallest positive integer k* for which there exists a nonzero nonnegative function u in H^1_loc(R^{k*}), homogeneous of degree one, which is a global minimizer of J(u;D)=integral_D (|grad u|^2+1_{u>0}) dx and is not of the form (x dot e)^+ for a unit vector e. Global minimizer means J(u;D)<=J(v;D) for every bounded Lipschitz domain D and every nonnegative v in H^1(D) with v-u in H^1_0(D). The complete answer must identify k*, exhibit or prove existence of such a minimizer there, and exclude such minimizers in every smaller dimension.

    Status update

    Source-reported determined

    Resolved (determined) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 367: The target asks for the least dimension with a nonflat nonnegative one-homogeneous global minimizer of the one-phase Bernoulli functional, including both existence there and exclusion below. The abstract supplies exactly both clauses: all nonzero homogeneous minimizers through dimension six are flat, and a nonflat homogeneous global minimizer exists in dimension seven. Thus the determined answer is k*=7; the singular-set bounds are additional consequences.

    Status reviewed

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  140. Source rank759Lane-Emden Liouville conjecture in dimensions at least fiveAnalysis & PDESource-reported proved

    Historical source rank #759 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every integer N>=5 and every p,q>0 satisfying 1/(p+1)+1/(q+1)>1-2/N, there do not exist functions u,v in C^2(R^N), both strictly positive everywhere, satisfying -Delta u=v^p and -Delta v=u^q on all of R^N.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 370: Set the source weights A=B=0 and n=N>=5. Its subcritical inequality becomes N/(p+1)+N/(q+1)>N-2, exactly the ranked inequality after division by N. Every everywhere-positive C2 pair in the target is continuous at the origin and classical elsewhere, and the source imposes no growth or symmetry assumption. Its nonexistence theorem therefore covers every target parameter and solution while also handling lower dimensions and nonzero weights.

    Status reviewed

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  141. Source rank764Zero-Error Capacity of Binary Symmetric Channels with Complete FeedbackCryptography, coding, information & optimizationSource-reported solved

    Historical source rank #764 in source edition claude-ranking.2026-09-20.v2. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine the zero-error capacity C_0(p) of the binary symmetric channel with error probability p under complete noiseless feedback.

    Status update

    Source-reported solved

    For the ordinary memoryless BSC, fixed-blocklength zero-error capacity with noiseless feedback is zero for 0<p<1 and one bit per use at deterministic endpoints p=0 or p=1. Under either message and any causal feedback encoding strategy every finite output word has positive probability when 0<p<1. Thus two messages cannot have disjoint decoding supports. A single message gives zero rate; deterministic endpoints transmit one bit exactly. An adversarial bounded-error-fraction or list-decoding problem is a distinct target. The Berlekamp citation does not change the written random-error-probability formulation. No retitling, rank transfer or substitute target is authorized.

    Status reviewed

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  142. Source rank769Kalikow's directional zero-one problem in dimensions at least threeProbability, ergodic theory & dynamicsSource-reported proved

    Historical source rank #769 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every d>=3, every iid uniformly elliptic law of nearest-neighbor transition probabilities on Zd, and every fixed unit vector ell, does the annealed law of the walk X started at the origin satisfy P(lim_n X_n dot ell=+infinity) in {0,1}?

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 220: The ranked Kalikow question assumes iid uniformly elliptic nearest-neighbor transition laws on Zd for d>=3 and asks whether directional escape has annealed probability zero or one. The strict-ellipticity companion proves precisely that zero–one conclusion for every nonzero direction under weaker positivity assumptions, so uniform ellipticity and unit directions are included. The finite-range-dependent theorem also contains the iid uniformly elliptic case; ballisticity is a separate stronger dynamical conclusion rather than a needed hypothesis.

    Status reviewed

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  143. Source rank771Torsion-Freeness Conjecture for All Artin-Tits GroupsGeometry & topologySource-reported proved

    Historical source rank #771 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For every finite Coxeter matrix M with off-diagonal entries in {2,3,...,infinity}, decide whether the associated Artin-Tits group A_M has no nonidentity element of finite order. For a finite entry m_st, the defining relation equates the two alternating words in s,t of length m_st; an infinite entry gives no relation.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 254: The same universal finite-rank Salvetti asphericity result supplies a finite-dimensional K(A_M,1) for every finite Coxeter matrix. Its standard consequence is torsion-freeness: a group with a finite-dimensional classifying space has no nontrivial finite subgroup. Thus this mapping uses an explicit mathematical implication from the stronger asphericity statement, rather than treating the family title as a direct torsion theorem.

    Status reviewed

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  144. Source rank783Barnette's Conjecture on Hamiltonicity of Cubic Planar Bipartite GraphsCombinatorics & discrete geometrySource-reported proved

    Historical source rank #783 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that every cubic 3-connected planar bipartite graph has a Hamiltonian cycle.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 180: The abstract states Hamiltonicity for every finite simple cubic bipartite planar 3-vertex-connected graph, exactly the standard graph class in the frozen Barnette target. It imposes no order bound or extra face-size condition. Only this explicit planar conclusion is used; the broader Pfaffian formulation in the family summary is unnecessary.

    Status reviewed

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  145. Source rank787Maximum Number of Mutually Unbiased Bases in Dimension SixQuantum information & computationSource-reported proved

    Historical source rank #787 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that the maximum number of mutually unbiased bases (MUBs) in complex dimension d = 6 is 3.

    Status update

    Source-reported proved

    OpenAI’s 6 October 2026 release reports that the maximum number of mutually unbiased bases in complex dimension six is three. The exclusion of four arbitrary complex bases depends on complete execution of the documented computer-assisted pipeline under the stated binary64 arithmetic and compiler conditions; no completed-run receipt was independently inspected. ProofAtlas has not rerun or audited that computation. No independent proof audit, confirmation of refereeing, or full-target formal verification is claimed.

    Status reviewed

    Resolution sources

    Link to this problem
  146. Source rank794Burr-Erdos Conjecture on the Ramsey Number of the HypercubeCombinatoricsSource-reported proved

    Historical source rank #794 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    For n>=1 let Q_n have vertex set {0,1}^n, with two vertices adjacent exactly when they differ in one coordinate. Let R(Q_n,Q_n) be the least N such that every red-blue coloring of the edges of K_N contains a monochromatic, not necessarily induced, copy of Q_n. Does an absolute constant C exist such that R(Q_n,Q_n)<=C*2^n for every n? Resolve by proving such a bound or proving that the ratios R(Q_n,Q_n)/2^n are unbounded.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 171: The source constructs the bound R(Q_n,Q_n)<=C 2^n with one absolute constant C for the two-color Ramsey number of the binary n-cube. This matches the target for all n, with ordinary monochromatic copies rather than induced copies. The abstract's linear order is in the cube's 2^n vertices, not in its dimension n.

    Status reviewed

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  147. Source rank824Tingley's Problem on Isometries Between Unit Spheres of Banach SpacesAnalysis & PDESource-reported proved

    Historical source rank #824 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Given real Banach spaces X and Y and a surjective isometry f:S_X -> S_Y between their unit spheres, prove that f extends to a surjective real linear isometry from X onto Y for all Banach spaces, or give a counterexample; for complex spaces the expected extension is real-linear.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 322: The source covers every surjective unit-sphere isometry between real Banach spaces, with no finite-dimensional, separability or geometric hypothesis, and gives a surjective real-linear isometric extension. This is exactly the indexed assertion; uniqueness is additional. Complex spaces are covered as real Banach spaces, since the target expects only a real-linear extension. The zero-space case is trivial and does not create a gap in the nonzero formulation of the summary.

    Status reviewed

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  148. Source rank841Seymour Second Neighborhood ConjectureCombinatorics & discrete geometrySource-reported proved

    Historical source rank #841 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove that every oriented graph D (directed graph with no loops and no 2-cycles) has a vertex x whose second out-neighborhood N^{++}_D(x) is at least as large as its first out-neighborhood N^{+}_D(x).

    Status update

    Source-reported proved

    OpenAI’s 6 October 2026 release reports a proof of Seymour’s second-neighborhood conjecture for nonempty finite oriented graphs, including disconnected graphs and sinks. The second neighborhood counts vertices at directed distance exactly two. This is the standard convention used in the conjecture literature cited by Bai–Li–Park; empty and infinite graphs are outside this claim. No independent proof audit, confirmation of refereeing, or local Lean verification is claimed.

    Status reviewed

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  149. Source rank876Barendregt-Geuvers-Klop conjectureLogic, foundations & set theorySource-reported proved

    Historical source rank #876 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that every weakly normalizing pure type system is strongly normalizing: if every typable term has some normal form, then no typable term admits an infinite reduction sequence.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 245: The ranked weak-normalization-implies-strong-normalization statement for pure type systems is matched by the manuscript's beta-reduction theorem. It quantifies over every legal expression in every valid context, permits reductions inside type annotations, and imposes no functionality hypothesis. Thus it covers the ordinary beta-Barendregt-Geuvers-Klop target rather than just functional rules, closed terms or a restricted evaluation strategy.

    Status reviewed

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  150. Source rank899Nakayama Conjecture on Infinite Dominant DimensionAlgebra, representation & category theorySource-reported disproved

    Historical source rank #899 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove: if A is an Artin algebra whose minimal injective resolution has every term projective (equivalently, A has infinite dominant dimension), then A is self-injective.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 199: The Tachikawa companion defines infinite dominant dimension as projectivity of every term in the regular left module's minimal injective resolution. Corollary 1.2 constructs Gamma_K=End_{A_K}(A_K direct-sum M_K)^op and states that it is a finite-dimensional K-algebra whose every minimal-injective-resolution term is projective but whose self-injective dimension is infinite. Finite-dimensional algebras over fields are Artin algebras; infinite self-injective dimension excludes self-injectivity, which would have injective dimension zero. Hence this is precisely a counterexample to the ranked implication, with the left-module convention made explicit by the opposite algebra. Characteristic two and the transcendental base field do not narrow the universal Artin-algebra target: one permitted counterexample suffices. No inference from the neighboring Tachikawa conjecture alone is needed.

    Status reviewed

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  151. Source rank936Gaussian Moat ProblemNumber theory & arithmetic geometrySource-reported disproved

    Historical source rank #936 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that there is an infinite sequence of distinct Gaussian primes z_1,z_2,... in Z[i] and a constant B such that |z_{n+1}-z_n| <= B for all n.

    Status update

    Source-reported disproved

    Resolved (disproved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 028: The frozen assertion asks whether there exists an infinite sequence of distinct Gaussian primes with some finite uniform step bound. The source denies precisely this and proves the stronger uniform finiteness of all connected components for each bound, including axes and every starting prime. Thus the existence assertion is false; no effective component-size bound is needed by the target.

    Status reviewed

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  152. Source rank988Barker Sequence ConjectureCryptography, coding, information & optimizationSource-reported proved

    Historical source rank #988 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Prove or disprove that no binary Barker sequence of length greater than 13 exists, where a Barker sequence is a finite {+1,-1} sequence whose nontrivial aperiodic autocorrelations are all at most 1 in magnitude.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 179: The source classifies all Barker sequence lengths greater than one as exactly 2,3,4,5,7,11,13, using the circulant Hadamard result together with classical Barker facts. Every listed length is at most 13, which immediately proves the frozen nonexistence target for binary sequences of length greater than 13 with the stated aperiodic autocorrelation condition.

    Status reviewed

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  153. Source rank1018Convergence of the Flint Hills SeriesNumber theory & arithmetic geometrySource-reported proved

    Historical source rank #1018 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether the Flint Hills series sum_{n=1}^infty 1/(n^3 sin(n)^2) converges.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 017: The abstract expressly concludes convergence of exactly sum 1/(n^3 sin^2 n), with angles in radians, from the stronger irrationality-exponent assertion for pi. The frozen target only asks whether this series converges, so the stated consequence fully decides it affirmatively; no claim about normality or an effective irrationality bound is required.

    Status reviewed

    Resolution sources

    Formerly in the Top 100 most famous open problems.

    Link to this problem
  154. Source rank1095FPT Three-Machine Unit Precedence Makespan Scheduling by Poset WidthTheoretical computer scienceSource-reported proved

    Historical source rank #1095 in source edition recovery-ranking.2026-10-06.v1. This entry is outside the current open ranking.

    Exact statement covered by this update

    Determine whether the scheduling problem P3|prec,p_j=1|Cmax has an algorithm running in f(w) n^{O(1)} time, where w is the width of the partial order induced by the precedence constraints.

    Status update

    Source-reported proved

    Resolved (proved) on the basis of the OpenAI mathematics release of 6 October 2026. Exact scope reviewed; proofs not independently audited, refereeing not independently confirmed, and Lean artifacts not executed. Family 124: The ranked problem P3|prec,p_j=1|Cmax asks for running time f(w)*n^{O(1)} parameterized by precedence-poset width. The catalogue asserts one uniform deterministic polynomial-time algorithm for all explicit precedence graphs, nonpreemptive unit jobs, and exactly three identical parallel machines, computing an optimal makespan and deadline feasibility. Its polynomial bound is independent of width, so it gives the requested FPT algorithm with a constant parameter factor and solves a stronger unrestricted version.

    Status reviewed

    Resolution sources

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Earlier status updates

The previous edition’s archived data retains its status updates and sources, including problems outside this list. Its ranks and qualifications belong to that earlier edition. Previous edition and status history · Oct 6, 2026 (JSON).