(The theorem improves the best lower bound valid in *every* sufficiently large genus from asymptotic constant 2/9 to 1. The every-genus ladder it climbs is Katz-Sabourau's 19/120 and then Liu-Petri's 2/9, the latter also by a random construction. Constant 1 was already reached by Petri-Walker along a subsequence of genera, following Erdos-Sachs, so the new contribution is achieving it uniformly rather than the constant itself. The asymptotic problem stays open, and the remaining gap is wide: Brooks and Buser-Sarnak give 4/3, while the elementary area bound is (S)2log(4g-2), asymptotically 2log g. So this closes much of the liminf gap and determines no optimal constant.)
Hyperbolic geometry
We show that for every sufficiently large genus g, there exists a closed hyperbolic surface Sg with systole sys(Sg)≥logg−12loglogg. In particular, g→∞liminfloggmax{sys(S):S∈Mg}≥1, improving the previously known bound 2/9. This note is a continuation of our previous work on the diameter of finite covers arXiv:2608.12887, using the same framework of constant-twist pants decomposition to study systoles.
The proof was developed by GPT-5.6 Sol through an extended discussion with the author.
Posed by —·Open —·Model GPT-5.6 Sol (OpenAI)·Solved 2026-08-27
PreprintSignificance 22Submitted by VibeGene on 28 Aug 2026
(For two qubits, the paper considers =12|^+⟩⟨^+|+12|01⟩⟨01|, where |^+⟩=(|00⟩+|11⟩)/√2. This rank-2 state has no global supporting affine functional for Entanglement of Formation. Setting _t=(1-t)+t|10⟩⟨10|, Wootters’ formula gives C(_t)=12-√(2t)+O(t). Consequently, _t0^+[E_F()-E_F(_t)]/t=+∞, so E_F is not Lipschitz lower semicontinuous at . By the paper’s criterion, no global supporting affine functional exists there. Thus the claimed universal existence fails even for two qubits, although it remains true for nondegenerate finite-dimensional states.)
Entanglement theory
The paper disproves the assumption that finite-dimensionality and the convex-roof structure of Entanglement of Formation guarantee a global supporting affine functional at every bipartite state. It gives an explicit degenerate two-qubit state ρ for which no Hermitian Λρ satisfies both EF(ρ)=TrΛρρ and EF(σ)≥TrΛρσ for every state σ. The construction uses the equivalence between existence of such a functional and Lipschitz lower semicontinuity of EF, together with Wootters’ formula.
Posed by A.S. Holevo and M.E. Shirokov·Open —·Model Claude Fable 5 (Anthropic)·Solved 2026-08-27
Site-confirmedSignificance 10Submitted by VibeGene on 28 Aug 2026
(The paper proves that not every Heyting algebra can occur as the lattice of subterminal objects of an elementary topos. Specifically, the free Heyting algebra F_2 on two generators cannot occur. Using Bellissima’s representation F_2 O_(K_2), the authors construct an upward-closed subset A⊆ K_2 with A F_2. They show that if some elementary topos E satisfied _ E(1) F_2, then higher-order internal logic would make A definable as a global proposition, forcing A to correspond to an element of F_2, a contradiction. Thus no elementary topos has subterminal lattice isomorphic to F_2, disproving the claim that every Heyting algebra can arise this way. The paper does not classify which Heyting algebras are realizable.)
Intuitionistic propositional logic
We answer the question whether all Heyting algebras can appear as the lattice of subterminal objects of an elementary topos in the negative. Concretely, we have shown that the free Heyting algebra on two generators cannot be such a Heyting algebra. The mathematical results in this document were obtained with the help of ChatGPT 5.6 Sol, although the document itself was written entirely by us and we take full responsibility for its contents.
Posed by —·Open —·Model ChatGPT 5.6 Sol (OpenAI)·Solved 2026-08-27
PreprintSignificance 18Submitted by VibeGene on 28 Aug 2026
(Answered in full: for every k4, a target with exactly two nonadjacent zero digits has F_k(t)=(k+23)3^k-4, independently of the distance between the zeros. Exact at every width, no error term, no hypothesis on k (Theorem 1.1). This is an evaluation, not an extremal result, and the paper is explicit about the difference: the plateau value is not maximal. At k=12 it reads 35·3^8=229,635 while F_12(110101101010)=293,499 at five zero digits, so no global maximizer of F_k is classified. The paper's other results are finite-layer and do not settle the extremal question: balancing monotonicity of [x^≤ C]H_t holds only for C5 (Theorem 1.3), and the all-mass statement is Conjecture 8.1, which the paper states outright does not follow from Theorem 1.3. The chamber where zero digits are adjacent is not addressed.)
Combinatorial number theory; binary digit sums and cyclic carries
Put N=2k−1 and write wt for the binary Hamming weight. Let Fk(t) count the ordered triples (a,b,c)∈{0,…,N−1}3 with a+b+c≡t(modN) and wt(a)+wt(b)+wt(c)<k, the three-summand analogue of the two-summand count of the Tu-Deng conjecture at the same modulus and weight budget. Evaluate Fk(t) exactly on the targets t whose k-bit cyclic word has exactly two zero digits, no two adjacent: is the value the same for every such t at a given k, and what is it?
Posed by —·Open —·Model GPT-5.6 Sol (high reasoning), Claude Opus 5 (high reasoning) (OpenAI, Anthropic)·Solved 2026-08-26
Site-confirmedSignificance 4Submitted by SilentIbis765 on 27 Aug 2026
(For fixed δ∈(0,1) and all sufficiently large Δ depending only on δ, Glauber dynamics for proper q-colorings mixes rapidly on every graph of girth at least 5 whenever q≥(1+δ)Δ: spectral gap Ω_δ(1/n) and t_mix(ε)=O_δ(n^2log q+nlog(1/ε)). An analogous theorem holds for the anti-ferromagnetic Potts model at q≥(1+δ)(1-β)Δ. What it does and does not improve. On girth it is a large gain: previous results near the (1+δ)Δ threshold needed girth at least eleven (Hayes-Vigoda, extended to constant degrees by Jain-Mizgerd-Vigoda). On the mixing rate it is weaker - those give optimal O(nlog n), this gives O(n^2log q). It does not touch the folklore conjecture that mixing is rapid on every graph for q≥Δ+2; Remark 7 names spanning 4-cycles as the obstruction.)
Glauber dynamics
It is proved that, for every δ∈(0,1), the Glauber dynamics for the uniform distribution on proper q-colorings is rapidly mixing when q≥(1+δ)Δ and the underlying graph has girth at least 5 and maximum degree Δ=Ωδ(1). This result also extends to general multi-spin systems satisfying a local spectral contraction condition, including the anti-ferromagnetic Potts model with q≥(1+δ)(1−β)Δ.
These results are achieved by a new spectral local-to-global principle on graphs with girth at least five for general multi-spin systems, and a novel Fourier analysis for Glauber dynamics on a star. The main ideas behind all the proofs were developed through several rounds of interaction with GPT-5.6 Sol Ultra.
Posed by Mark Jerrum, 1995·Open 31y·Model GPT-5.6 Sol Ultra (OpenAI)·Solved 2026-08-26
PreprintSignificance 30Submitted by VibeGene on 27 Aug 2026
(Let S=k[x_1,x_2,x_3,x_4] over any characteristic-zero field. For each d∈\5,7\ and every r1, the paper proves that r general degree-d forms satisfy Fröberg’s predicted Hilbert series _S/(F_1,…,F_r)(t) = [(1-t^d)^r/(1-t)^4]_+. The genuinely new ranges are 6≤ r11 for quintics and 6≤ r21 for septics. These are reduced to finitely many exact rank tests of Macaulay multiplication matrices; explicit maximal minors are nonzero mod 2, so the required ranks hold in characteristic zero. Thus the conjecture is completely settled for the equal-degree four-variable slices d=5 and d=7, but not in general.)
Commutative algebra
Let k be a field of characteristic zero and let S=k[x1,x2,x3,x4]. We prove Fröberg's predicted Hilbert series for ideals generated by r general forms of equal degree d for every r≥1 in each of the two cases d=5 and d=7. Relative to the classical cases r≤5 and the equal-degree theorem through degree d+2 of Boij--Dannetun--Lundqvist, the generator-count ranges requiring new input are 6≤r≤11 for quintics and 6≤r≤21 for septics. The proof reduces each slice to finitely many endpoint ranks of Macaulay multiplication matrices. For quintics, ten exact endpoint computations based on twenty-one sparse forms suffice. For septics, a nested family of 120 integral forms supplies fifteen endpoint computations. In every endpoint certificate for these new ranges, an explicitly recorded maximal minor is nonzero modulo 2, hence is a nonzero integer. The case r=5 is the classical strong Lefschetz instance; for quintics we also record a matching modular rank and Koszul bound. Zariski openness then gives the result over every characteristic-zero field. The unrestricted Fröberg conjecture remains outside the scope of the paper.
Posed by Ralf Fröberg, 1985·Open 41y·Model GPT-5.6 Sol; Claude Fable 5; Grok 4.6 (OpenAI; Anthropic; xAI)·Solved 2026-08-25
Site-confirmedSignificance 14Submitted by VibeGene on 28 Aug 2026
(Let T be a complete first-order theory in discrete or continuous logic, let M\, and let \\∈\_x() be Borel-definable over M. The paper proves that the following three conditions are equivalent:(i) \ is a frequency interpretation measure (fim) over M;(ii) \ is definable over M and its canonical random extension r_\ is generically stable over M^\Ω;(iii) \ is self-averaging over M. The new work proves the reverse implications (iii)\⇒(ii)\⇒(i) and extends the characterization to continuous logic. The authors therefore make the equivalent conditions into a definitive definition of generic stability for Keisler measures. The paper also proves a further characterization in terms of an order-property condition and derives consequences for closure under Morley products.)
Model theory
Given a first-order theory T (in discrete or continuous logic) and a Borel-definable global Keisler measure mu in T, we show that the following conditions are equivalent: (i)mu is a frequency interpretation measure; (ii)mu is definable and its canonical “random extension” rmu is generically stable in the randomization theory TR; (iii)mu is “self-averaging”.
This result establishes a robust notion of generic stability for Keisler measures, which resolves a long-term research objective from previous work. The implications (i)Rightarrow(ii)Rightarrow(iii) were previously established by the authors (for T discrete). The primary focus of this paper is the reverse implications (iii)Rightarrow(ii)Rightarrow(i), which we obtain through the use of AI models.
Posed by Gabriel Conant, Kyle Gannon, 2020·Open 6y·Model ChatGPT 5.5; Kimi K3; Claude Fable 5; ChatGPT 5.6 Sol (OpenAI; Moonshot AI; Anthropic)·Solved 2026-08-25
PreprintSignificance 20Submitted by VibeGene on 26 Aug 2026
(Claims an explicit compact complex threefold X, fibred over P^1 by complex 2-tori via period functions on the (3,4,∞) orbifold, degenerating to a del Pezzo-of-degree-six fibre (identified opposite sides of its hexagon) at one point and to bielliptic multiple fibres of multiplicities 3 and 4 at the other two. Argues X is simply connected with H_*(X;Z)=H_*(S^6;Z), hence diffeomorphic to S^6, with algebraic dimension exactly 1. This directly contradicts [CDP20, Cor. 2.3], a published (and once-corrected) theorem; the paper states this and argues where the two accounts diverge, rather than overlooking it. Posted hours before this entry, with no independent check, no formalisation, and no refutation yet in any venue found. Filed as a candidate specifically because none of that has happened, not because a problem with the argument has been found.)
Complex geometry; differential topology
Hopf's problem, posed in 1948: does the six-sphere S6 admit an integrable complex structure? S6 is one of only two spheres carrying an almost complex structure at all (the other is S2), from the octonions' multiplication, but almost complex structures need not be integrable, and whether that one - or any other - integrates has stood open for 78 years through a history of disputed attempts, including a widely discussed 2016 argument by Atiyah that did not hold up. This paper claims yes: it builds an explicit compact complex threefold X, fibred over P1 by complex 2-tori degenerating at three points, and argues X is simply connected with the integral homology of S6, hence diffeomorphic to it.
Posed by Heinz Hopf, 1948·Open 78y·Model Claude (Anthropic)·Solved 2026-08-24
(The paper constructs a real meromorphic function F on C satisfying F^-1(\0,1,∞\), with each of the three fibers F^-1(0), F^-1(1), and F^-1(∞) infinite, such that for every a∈\0,1,∞\, the a-point divisor in each of the upper and lower half-planes fails the Blaschke condition. Consequently, F is not of bounded type in either half-plane. This gives a negative answer to Nevanlinna’s century-old question asking whether an entire-plane meromorphic function that omits three distinct values in a half-plane must be of bounded type there. By postcomposition with Möbius transformations, the three exceptional values \0,1,∞\ may be replaced by any prescribed triple of distinct values in . By affine change of variables, the construction applies to any Euclidean half-plane.)
Complex analysis
We construct a real meromorphic function F on C such that F−1({0,1,∞})⊂R, while F is not of bounded type in either half-plane. More strongly, for every a∈C∖{0,1,∞}, the a-point divisor in either half-plane fails the Blaschke condition. Thus the construction provides an independent negative answer to a question going back to Nevanlinna’s 1925 work that had remained open for over a century.
Postcomposition gives the analogous counterexample for any prescribed triple of distinct values in the Riemann sphere. The core construction and proof were generated during an autonomous run of GPT-5.6 Sol Ultra.
Posed by Rolf Nevanlinna, 1925·Open 101y·Model GPT-5.6 Sol Ultra (OpenAI)·Solved 2026-08-24
PreprintSignificance 45Submitted by VibeGene on 27 Aug 2026
(Two tiers, and only the first is the record. Rank ≥ 31 is unconditional: 31 explicit points, independence asserted via the leaderboard's stated general practice of exact 2-descent (not reproduced here - see the verification note). Rank exactly 31 is conditional on GRH and BSD, per the submitters' commentary, in the same style as the sibling record's Bober-bound argument; no numeric derivation has been published for this curve specifically. The entry is a partial result because the open question - whether ranks are unbounded at all - remains unanswered by any single record.)
Elliptic curves
How large can the Mordell-Weil rank of an elliptic curve over Q be? Whether ranks are unbounded is open, and progress is measured by explicit records, tabulated by Dujella: rank ≥28 from 2006, raised to ≥29 by Elkies and Klagsbrun in 2024, and to ≥30 three days before this one by the same team (see the related entry). Now ≥31, witnessed by an explicit curve y2+xy+y=x3+x2+a4x+a6 with a4 of 67 digits and a6 of 99, carrying thirty-one independent rational points.
Posed by Classical; rank records tabulated by Andrej Dujella·Open —·Model Claude (Anthropic)·Solved 2026-08-23
(The manuscript claims C_a,b is transcendental for every a1 and b1-a, settling the positive integer-valued affine subclass of Erdős Problem 270. Two pieces of context matter. Problem 270 as Erdős and Graham posed it, for every f(n)→∞, was already answered no by Crmarić and Kovač in 2025: for any α>0 some such f makes the series sum to α. What survives is the non-decreasing case, and the affine family sits inside it. Separately, the checkable parts here were already known - the short irrationality proof for C_1,0 is Crmarić and Kovač's, posted by Kovač on the Erdős Problems forum in July 2026 and credited in the repository, and base-case transcendence follows a 2023 MathOverflow argument. The new content is the extension to the whole affine family, which is the part with neither formalization nor review.)
Erdős #270 · Transcendence theory
For integers a≥1 and b≥1−a, the series Ca,b=∑n=1∞n!/((a+1)n+b)! is transcendental. Equivalently, the series in Erdős Problem 270 is transcendental whenever f(n)=an+b is a positive integer-valued affine function.
Posed by Paul Erdős and Ronald Graham, 1980·Open 46y·Model GPT-5.6 Sol (Codex) (OpenAI)·Solved 2026-08-22
AnnouncedSignificance 12Submitted by CobaltMongoose239 on 23 Aug 2026
(Proves Haglund's Conjecture 4 for k=1: every non-real first-quadrant zero of _1+t_2 is simple with strictly decreasing imaginary part, no branch escapes forward, and every finite-multiplicity real collision stays real afterwards. The cases k2 remain open. Two readings worth separating: Conjecture 4 asserts the monotone descent alone, so the no-escape and stays-real statements are this paper's own additions rather than Haglund's text, and they are the stronger part of the theorem. The descent itself, part (i), is the part that rests on the unavailable interval-arithmetic certificate.)
Analytic number theory and entire-function zero dynamics
Haglund's Conjecture 4 reads: for k≥1, the imaginary part of each non-real zero of Ξk(z)+tΦk+1(z) decreases monotonically as t goes from 0 to 1, where the Φn are the incomplete-gamma summands of Riemann's series for Ξ and Ξk=∑n≤kΦn. This work proves the case k=1, the pencil Φ1+tΦ2: every non-real zero in the closed first quadrant is simple and the imaginary part of its analytic branch strictly decreases. It adds two statements Conjecture 4 does not itself assert - no non-real branch escapes to infinity on a bounded forward parameter interval, and at a real collision of any finite multiplicity the full local Weierstrass-Puiseux multiset stays real to the right. The cases k≥2 remain open, and nothing is claimed about the zeros of Ξ or the Riemann hypothesis.
Posed by James Haglund, 2009·Open 17y·Model ChatGPT and Codex (OpenAI)·Solved 2026-08-22
(Disproved on the Hopf threefold X=(C^3\0\)/⟨ z↦ e^-1z⟩: Xia and Zhang construct a smooth Hermitian form and smooth functions _j with +dd^c_j>0 whose Monge-Ampère masses tend to infinity, so the universal bounded mass property fails already in complex dimension three. The construction uses the Hopf threefold's elliptic fibration, an exact mass identity reducing excess Monge-Ampère mass to a fibrewise Dirichlet energy, and heat-kernel regularizations of Green functions that make that energy diverge while preserving positivity. What the negative answer removes is load-bearing rather than incidental: finiteness of this mass is the starting point for the theory of volumes of Bott-Chern classes, and it enters as a standing hypothesis in recent Hermitian pluripotential theory.)
Complex geometry
A compact complex manifold X of dimension n has the bounded mass property if, for one (equivalently every) Hermitian form ω, the Monge-Ampère masses ∫X(ω+ddcφ)n are uniformly bounded over all smooth φ with ω+ddcφ>0. On a compact Kähler manifold Stokes' theorem makes that mass independent of φ outright; for a merely Hermitian ω, which is not closed, it genuinely depends on φ, and controlling it is a recurring theme of Hermitian pluripotential theory. The property is known to hold in dimension n≤2 and on manifolds of Fujiki class. Boucksom, Guedj and Lu left open whether it holds on every compact complex manifold, raising the question explicitly for Hopf manifolds of dimension at least three. This paper answers it in the negative on the Hopf threefold X=(C3∖{0})/⟨z↦e−1z⟩.
Posed by Sébastien Boucksom, Vincent Guedj, Chinh H. Lu, 2025·Open 1y·Model Rethlas (GPT-5.6 Sol) (OpenAI)·Solved 2026-08-21
PreprintSignificance 16Submitted by VibeGene on 25 Aug 2026
(Answers Problem 3 and generalizes it: the classification √(m) ∈ S ⇔ m = 2 covers every square root, and a further theorem replaces parity by divisibility by any p ≥ 2. Note the scope of the machine-checking, which is narrower than the paper: the author states that the case m = 3 is what is verified in Lean, and the repository flags the thickness computation of section 4.1 and all of section 8 as not formalized.)
Distribution mod 1; Mahler Z-numbers
Dubickas splits (1,+∞) into the set Z of those α for which some nonzero real ξ makes every integral part ⌊ξαn⌋ even, and its complement S; at α=3/2 the question of which side one lies on is Mahler's. His Problem 3 asks which side 3 is on. Answered: 3∈Z, with the explicit witness ξ=1.34160899796112665163…, and more generally m∈S if and only if m=2. The mechanism is Cantor-set arithmetic rather than Diophantine approximation: since m2 is an integer, the two-scale problem collapses to a base-m covering induction on restricted-digit expansions.
Posed by Artūras Dubickas, 2006·Open 20y·Model Fable 5, Opus 5 (Anthropic)·Solved 2026-08-21
Lean-checked, statement unauditedSignificance 20Submitted by LucidKestrel185 on 22 Aug 2026
(Constructed an explicit class of smooth random, time-dependent incompressible velocity fields on T^3, obtained by alternating smooth shear flows with iid random phases on finite time blocks. For every fixed sufficiently small resistivity, the magnetic field has an almost-sure exponential growth rate at least 1/2, together with a time-uniform lower bound whose random prefactor has a resistivity-uniform inverse-moment estimate. This is one variant case of Arnold's 1994 fast-dynamo problem, not the problem itself: Arnold asks for a field that is smooth, autonomous and deterministic all at once, and this one keeps the smoothness while giving up the other two. The sibling entry on this site relaxes the opposite hypothesis, keeping an autonomous deterministic field at Lipschitz regularity. Neither settles Arnold's problem as posed, which remains open.)
Dynamo theory
Arnold's fast-dynamo problem asks for a smooth divergence-free velocity field on T3, chosen independently of the magnetic diffusivity, that drives exponential growth of the magnetic field at every sufficiently small diffusivity. This constructs a genuinely C∞ field with that behaviour: random and time-dependent, refreshing iid on finite time blocks, for which the almost sure exponential growth rate is at least 1/2 at each fixed small enough resistivity, with a time-uniform lower bound whose random prefactor has a resistivity-uniform inverse-moment bound. The field is neither autonomous nor deterministic, so Arnold's smooth autonomous problem on T3 remains open.
Posed by Arnold's fast-dynamo problem (1994); the random formulation has no single named proposer·Open —·Model ChatGPT 5.6 Sol Ultra (OpenAI)·Solved 2026-08-20
PreprintSignificance 30Submitted by VibeGene on 22 Aug 2026
Conjecture 13 of King, Gosset, Kothari and Babbush asserts that for the set Bε(ρ) of Pauli observables with expectation value at least ε in magnitude, the fractional chromatic number of the induced anticommutation graph is O(ε−2); it would give a triply efficient Pauli shadow tomography algorithm. False: there are states and observables for which no finite constant bounds χfε2.
Posed by Robbie King, David Gosset, Robin Kothari and Ryan Babbush, 2025·Open 1y·Model GPT Sol 5.6 (OpenAI)·Solved 2026-08-20
(Two tiers, and only the first is the record. Rank ≥ 30 is unconditional, being thirty explicit independent points. Rank exactly 30 is conditional: applying Bober's bound (arXiv:1112.1503) with Δ = 4.25 gives an analytic rank of at most 31, and the root number is +1 so the rank is even, hence 30 - but that argument assumes GRH, and equating analytic rank with rank assumes BSD. The entry is a partial result because the open question is whether ranks are unbounded at all, which no single record answers. Superseded three days later by this project's own rank ≥ 31 record (see the related entry); left unedited otherwise as a record of what was known at the time.)
Elliptic curves
How large can the Mordell-Weil rank of an elliptic curve over Q be? Whether ranks are unbounded is open, and progress is measured by explicit records, tabulated by Dujella: rank ≥28 from 2006, raised to ≥29 by Elkies and Klagsbrun in 2024. Now ≥30, witnessed by an explicit curve y2+xy=x3+a4x+a6 with a4 of 63 digits and a6 of 94, carrying thirty independent rational points.
Posed by Classical; rank records tabulated by Andrej Dujella·Open —·Model Claude (Anthropic)·Solved 2026-08-20
Marton's inner bound, proposed in 1979, is the best known achievable region for a general discrete memoryless broadcast channel, and whether it always achieves the capacity region had been open ever since. It does not: there is a finite two-receiver discrete memoryless broadcast channel whose two-letter Marton value strictly exceeds twice its one-letter value, so the complete one-letter Marton region is strictly contained in the capacity region.
Posed by Katalin Marton, 1979·Open 47y·Model GPT-5.6 Sol, Claude Fable 5, Claude Opus 5 (OpenAI, Anthropic)·Solved 2026-08-20
(The paper's appendix draws a distinction worth keeping: a counterexample may reduce to a finite certificate, checkable once the object is written down, or it may itself be a theorem quantified over all degenerations. This is the second kind. The method field records construction, because the resolution exhibits an explicit fivefold, but the difficulty lay elsewhere - candidate manifolds of this shape have been available since 2008, and what was missing was the proof that the mechanism works.)
Kähler geometry
Posed by Shing-Tung Yau, Gang Tian and Simon Donaldson, 1993·Open 33y·Model Fable 5, GPT-5.6-sol, Danus (Anthropic, OpenAI, FrenzyMath(AI4M@PKU))·Solved 2026-08-19
(The claim is R(c) = 40c+41 for every c ≥ 2, reduced to three finite facts: the base value R(2) = 121, and the unsatisfiability of a 321-position and a 521-position spoke template. The reduction is Lean-checked and holds for every D ≥ 1; the two unsatisfiability results carry DRAT proofs. This completes the partial entry for the same conjecture, which proved it for roughly two thirds of integers via a scaling lemma; that lemma is now one of three legs, covering the branch where d is divisible by 3. The supporting results are worth more than the headline for anyone deciding whether to believe it: the paper also shows every band relaxation is satisfiable, which is why previous attempts stalled, and that the affine method alone is exactly sharp and can never finish.)
Rado numbers / partition regularity
For a constant c, the 4-colour Rado number R(c) is the least N such that every colouring of {1,…,N} in four colours contains a monochromatic solution to x+y+c=z. Myers (Rutgers thesis, 2015, Conjecture 4.9) and Ahmed, Boza, Emamy-Khansary, Marin, Revuelta and Sanz (Math. Comp. 85, 2016, §5.5) conjecturedR(c)=40c+41for all sufficiently large c, with the small values R(0)=45 and R(1)=83 as exceptions. Previous methods reached individual values but not the general case.
This claims the conjecture for every c≥2, by reducing it to three finite facts: the single base value R(2)=121 and the unsatisfiability of two "spoke" templates. The reduction is formalised in Lean 4 and holds for every D≥1; the two templates are settled by SAT with DRAT certificates.
(Only the smooth case falls. Hamburger's real-analytic theorem is untouched, and the counterexample is explicitly a C^∞ object, so the conjecture's classical analytic form remains true. The gap between the two is the whole content of the result.)
Differential geometry
Carathéodory's conjecture, Problem 8.1 of Ghomi's list and traceable to 1922, asks whether every closed convex surface in R3 has at least two umbilic points. Hamburger settled the real-analytic case in 1940-41 and it stands. The C∞ case is false: an explicit support function gives a smoothly embedded two-sphere bounding a convex body with exactly one umbilic point. The same family disproves the smooth Loewner conjecture, whose member at k=1 has an isolated trace-free Hessian zero of winding number three.
Posed by Constantin Carathéodory, 1922·Open 104y·Model Claude, Codex (Anthropic, OpenAI)·Solved 2026-08-19
Lean-checked, statement unauditedSignificance 55Submitted by VelvetFalcon287 on 21 Aug 20261comment
For the Kasami APN function F(x)=x4k−2k+1 on GF(2n) with gcd(k,n)=1, the conjecture asserts that for Δ={F(b)+F(b+1)+1} and all distinct nonzero v1,v2, the number of triples in Δ3 with v1x+v2y+(v1+v2)z=0 is exactly 22n−3. Proved for kmodn∈{1,2,n−2,n−1} and verified exhaustively for n≤13; the general case remains open.
Posed by Proposed anonymously at the NSUCRYPTO cryptographic olympiad, 2019·Open 7y·Model Claude Fable 5, Aristotle (Anthropic, Harmonic)·Solved 2026-08-19
Question 6.1 of Chalmoukis, Tsikalas and Yakubovich asks how far the Power boundedness constant P(T) of a matrix can exceed its ordinary Kreiss constant K(T). Answered more strongly: for every K>1 there are matrices whose Cayley transforms satisfy K(Ch(An,h))≤K while the strong Kreiss constant satisfies Ks(Ch(An,h))≥21CnαK with αK=(K−1)/(C+K−1). Since P(T)≥Ks(T), this solves the question. Moreover, since the Kreiss matrix theorem gives Ks(T)≤P(T)≤edK(T) in dimension d, the exponent α<1 is optimal up to an arbitrarily small power loss.
Posed by Nikolaos Chalmoukis, Georgios Tsikalas and Dmitry Yakubovich, 2025·Open 1y·Model ChatGPT 5.6 Pro, Claude Fable (OpenAI)·Solved 2026-08-19
Koivisto asked at Dagstuhl in 2013 whether the linear extensions of an arbitrary n-element poset can be counted exactly in time O∗(cn) for some c<2. Yes: a deterministic exact algorithm runs in O∗(1.89n), breaking the 2n barrier for the general problem.
Posed by Mikko Koivisto, at Dagstuhl, 2013·Open 13y·Model Claude Opus 5, ChatGPT 5.6 Sol (Anthropic, OpenAI)·Solved 2026-08-19