(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 OpenAI GPT-5 (Codex) (OpenAI)·Solved 2026-08-22
PreprintClaim issueSignificance 8Submitted by WildHeron785 on 24 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
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
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
(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.)
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
(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
Question 6.1 of Chalmoukis, Tsikalas and Yakubovich asks how far the strong Kreiss constant of a matrix can exceed its ordinary Kreiss constant. Answered: for every K>1 there are matrices whose Cayley transforms satisfy K(Ch(An,h))≤K while Ks(Ch(An,h))≥21CnαK with αK=(K−1)/(C+K−1). 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 (OpenAI)·Solved 2026-08-19
(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 2026
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
(Independent of ZFC, which is why this entry is the first to carry that result rather than proved or disproved. Both directions are formalized: Hechler's 1972 construction gives a model where the answer is no, and adding ^+ random reals over a model of CH gives one where it is yes. The credit is shared and mostly human. Newelski, Pawlikowski and Seredynski settled the problem's second question in 1987, and it is formalized here without the boundedness hypothesis. Hechler supplied one direction in 1972. Sungchul Lee derived a positive answer from a real-valued measurable cardinal, assisted by GPT-5.5 Pro, and Nat Sothanaphan observed that the two halves together give independence. What Glazer and Sol added is the removal of the large cardinal. erdosproblems.com still lists #501 as open at the time of writing.)
Erdős #501 · Set theory / forcing
For every x∈R let Ax⊂R be a bounded set of Lebesgue outer measure <1. Must there be an infinite independent set, that is an infinite X⊆R with x∈/Ay for all distinct x,y∈X?
Erdős and Hajnal proved that arbitrarily large finite independent sets exist. Hechler showed in 1972 that the answer is no under the continuum hypothesis, so any positive answer had to come from a model where CH fails, and Sungchul Lee later derived one from a real-valued measurable cardinal.
The answer is that neither side is provable. Dropping Lee's large cardinal by transferring his argument to the extension of a model of CH by random reals gives a model where the answer is yes; Hechler's construction gives one where it is no. The question is independent of ZFC.
Posed by Paul Erdős, 1961·Open 65y·Model Sol, Claude (OpenAI, Anthropic)·Solved 2026-08-19
Every finite simple connected graph G with∣V(G)∣=2d+1,diam(G)=d≥3satisfiesW(G)≤W(C2d+1)=2(2d+1)d(d+1).The claimed equality cases are exactly C2d+1 for every d≥3, the double star D2,3 when d=3, and the nine-vertex tree T1,2,2=S(2,3,3) when d=4.
Posed by E. DeLaViña and B. Waller, 2008·Open 18y·Model GPT-5.6 Sol; Claude Fable 5 (OpenAI, Anthropic)·Solved 2026-08-19
PreprintSignificance 10Submitted by SilentIbis759 on 20 Aug 2026
(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.
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
(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
The Yau–Tian–Donaldson conjecture predicts that a polarized manifold carries a canonical Kähler metric in its polarization class exactly when it is K-polystable. Settled for Kähler–Einstein metrics on Fano manifolds, it remained open for constant scalar curvature. False: there is a polarized smooth projective fivefold that is K-polystable but admits no extremal Kähler metric in c1(A), so K-polystability does not imply existence.
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 big-line-big-clique conjecture of Kára, Pór and Wood asserts that for all k,ℓ there is an n such that every finite point set of at least n points contains ℓ collinear points or k points that pairwise see each other. True for ℓ=4, k=6, the first case left open: every finite point set of size at least 1011055931 has four collinear points or six pairwise visible points.
Posed by Jan Kára, Attila Pór and David R. Wood, 2005·Open 21y·Model GPT-5.6 Sol Pro (OpenAI)·Solved 2026-08-19
(Four results, and the first is partly a refutation. Jakimiuk conjectured c_p = _p - 1 is optimal for every p ≥ 3; the paper proves that for p ≥ 4 and gives a counterexample for every 2 < p < 4, so the conjecture is false as posed and the corrected range is p ≥ 4. The witness is the two-coordinate vector S_2 = (ε_1+ε_2)/√2. The Baranski-Murawski-Nayar-Oleszkiewicz flat-point conjecture is proved outright, in the stronger form that x ↦ ‖x+S_n‖_p/‖x+S_n‖_4 is strictly decreasing on [1,∞) for every real p ≥ 5; that range is the one they conjectured, so nothing is left over. Jakimiuk's second conjecture, dimension-free quadratic stability at p = 3, is proved with an explicit constant, though the optimal constant there is only bracketed and stays open. The paper also records the exact fixed-q moment and Laplace-transform envelopes, from which coefficient-sensitive tail bounds follow.)
Khintchine inequalities
Let ε1,…,εn be independent Rademacher signs, let ∑ai2=1, write S=∑aiεi and q=∑ai4, and let μp=E∣G∣p for a standard Gaussian G. Two 2025 conjectures say that q alone governs how far S falls short of Gaussian.
Jakimiuk proved E∣S∣p≤μp−cpq for p≥3 and conjectured the optimal constant is cp=μp−1 throughout that range; separately he conjectured a dimension-free quadratic stability bound at the critical exponent p=3.
Baranski, Murawski, Nayar and Oleszkiewicz reduced the finite-dimensional Lp/L4 Khintchine constant for p≥5 to supx≥1∥x+ε1+⋯+εN∥p/∥x+ε1+⋯+εN∥4 and conjectured the supremum is attained at x=1 - that is, the flat coefficient vector is the extremizer.
Posed by Jacek Jakimiuk; Adam Barański, Daniel Murawski, Piotr Nayar, Krzysztof Oleszkiewicz, 2025·Open 1y·Model ChatGPT 5.6 Sol (OpenAI)·Solved 2026-08-18
PreprintSignificance 8Submitted by VibeGene on 19 Aug 2026
(The theorem is the vector-valued endpoint bound ‖Rf‖_L^1,∞ ≤ 2‖f‖_L^1 for R = (R_1,…,R_n), so the constant 2 also serves each component R_j uniformly in the dimension; the best previously known component bound grew like clog n. The mechanism is a decomposition theorem stated as Theorem 1.2: for every nonnegative f ∈ L^1 L^2 and every λ > 0, write f = + (-Δ)^α/2u with ≤ λ and u in the fractional Sobolev space H^α, obtained from an obstacle problem for the fractional Laplacian together with a Lewy-Stampacchia type estimate on an unbounded domain. That replaces the Calderon-Zygmund decomposition, whose cube geometry is where the dimensional loss enters.)
Harmonic analysis
The Riesz transforms R1,…,Rn on Rn are the Fourier multipliers −iξj/∣ξ∣, the natural higher-dimensional Hilbert transforms. Stein proved in 1983 that their Lp bounds can be taken independent of the dimension for every 1<p<∞. At the 1986 ICM he asked whether the same holds at the endpoint p=1: is there an absolute constant C, independent of n, with∣{x:∣Rf(x)∣>λ}∣≤λC∥f∥L1(Rn)for every λ>0? The Calderon-Zygmund route gives a constant that grows with the dimension, and the best known was Janakiraman's clogn.
This paper answers yes, with C=2, for the vector transform R=(R1,…,Rn) - so the same constant serves every single component Rj uniformly in n.
Posed by Elias M. Stein, 1986·Open 40y·Model Claude Opus 5.0, GPT-5.6 Sol (Anthropic, OpenAI)·Solved 2026-08-18
PreprintSignificance 38Submitted by VibeGene on 19 Aug 2026
(A record, not a resolution, and a small one by design. The bound moves from 2.371339 to 2.371177, about 1.6 × 10^-4, and the authors describe it as a small step. Whether = 2 is untouched, and nothing here suggests the laser method can reach it. The interesting claim is methodological rather than numerical. The bottleneck in this line of work is a hard optimization problem, and the paper reports progress by reformulating that problem and then improving the optimizer, with AlphaEvolve doing the final refinement. That is a different kind of contribution from a new mathematical identity, and it is why the entry is filed as computation.)
Algebraic complexity
The matrix multiplication exponent ω is the infimum of all t for which two n×n matrices can be multiplied in O(nt) arithmetic operations. Strassen showed in 1969 that ω<3, and sixty years of work has driven the upper bound down without anyone knowing the true value. Whether ω=2 is one of the central open questions of algebraic complexity.
The current bounds come from the laser method as refined by combination loss analysis. This paper attacks the optimization problem at the core of that refinement, reformulating it so it can be solved in a larger setting, designing a new optimization algorithm for it, and then refining that algorithm with AlphaEvolve.
The result is ω<2.371177, improving the previous best of 2.371339.
Posed by Volker Strassen, 1969·Open 57y·Model AlphaEvolve (Google DeepMind)·Solved 2026-08-17
(The claim is the exact conjectured decay: __a(u) ≤ C_a/√(log u) for every u > 1 and every n, with C_a dimension-free - concretely _a^2(log_a/_a-1)^1/2 where _a = (1+a)/(1-a). What is new is one step in a three-paper chain rather than a proof from scratch, and the paper is explicit about it. Chen's reverse-heat and Boolean-bridge framework and Xiang-Zhang's localized terminal-discrepancy method are taken as given; the addition is a power coupling that splits each reverse edge ratio into two geometric powers, producing a switched exponential weight that restores the exact reverse jump rate of the perturbed coordinate. Because the frozen exponent then has a fixed numerator, no growing stopping buffer is needed and the loglog u factor disappears. That last loglog is what stood between the previous work and Talagrand's statement.)
Analysis of Boolean functions
On the Boolean hypercube G={−1,1}n with uniform measure λ, let Tμf(x)=∫Gf(x⊙y)dμ(y) be convolution by a finite positive measure μ, and setψμ(u)=sup{uλ({Tμf≥u}):f≥0,∥f∥1=1},which measures how much better than Markov's inequality convolution makes the tail. In 1989 Talagrand conjectured that for the biased-coin product measure μa=(21+aδ1+21−aδ−1)⊗n with 0<a<1,ψμa(u)≤loguCa(u>1),with Ca depending on a alone and not on the dimension n. He offered a \$1000 prize for a proof. The Gaussian analogue was settled by Eldan and Lee; the hypercube case, the original, stayed open.
This paper claims the conjectured bound.
Posed by Michel Talagrand, 1989·Open 37y·Model Odin Automatic AI Research Agent·Solved 2026-08-16
PreprintSignificance 37Submitted by VibeGene on 19 Aug 2026
For every real ξ>0 the sequence of integer parts [ξ7n], n=0,1,2,…, contains infinitely many composite numbers. Second, there is no infinite right truncatable prime in base~7.
Posed by Forman and Shapiro (1967), Dubickas and Novikas (2005), 2005·Open 21y·Model Fable 5, Opus 4.8 (Anthropic)·Solved 2026-08-15
Lean-checked, statement unauditedSignificance 10Submitted by LucidKestrel185 on 16 Aug 20261comment
Posed by The Hessian conjecture (de Bondt, van den Essen line)·Open —·Model GPT-5.6 Sol, GPT-5.6 Luna, Claude Fable 5, DeepSeek V4 Pro (OpenAI, Anthropic, DeepSeek)·Solved 2026-08-14