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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Boris Alexeev, Moe Putterman, Mehtaab Sawhney, Mark Sellke and Gregory Valiant, Short proofs in combinatorics, probability and number theory II, arXiv:2604.06609v1 (8 April 2026), Section 6, answer Problem 1141 in the negative. For a fixed integer a≥1a\geq1 call nn aa-good when n−ak2n-ak^2 is prime for every integer k≥1k\geq1 with (k,n)=1(k,n)=1 and ak2<nak^2<n. Theorem 6.1 states that for each fixed a≥1a\geq1 only finitely many nn are aa-good. The case a=1a=1 is the problem's property, so only finitely many nn have n−k2n-k^2 prime for every kk coprime to nn with k2<nk^2<n, and the answer to the question is no. The authors attribute the proof to an internal OpenAI model; their comment on the use of AI adds that ChatGPT-5.4 Pro solved Problem 1141 in all five independent attempts they made and, asked as a follow-up, generalized the proof to n−ak2n-ak^2 for every aa. The paper has a [[../library/discrete_geometry/alexeev_2026_short_proofs_combinatorics_probability_number_theory/_index|library card]].

The argument is a short deduction from Theorem 1.3 of Pollack's paper on prime character nonresidues, carded as [[../library/primes/pollack_2017_bounds_first_several_prime_character_nonresidues/_index|Pollack 2017]]: for A>0A>0, ε>0\varepsilon>0 and every large enough modulus mm, a quadratic character χ\chi modulo mm takes the value 11 at at least (log⁡m)A(\log m)^A primes p≤m1/4+εp\leq m^{1/4+\varepsilon} (restated as Theorem 6.3 of the paper for A≥1A\geq1). Suppose nn is aa-good and large, and write an=u2dan=u^2d with dd squarefree. When d>1d>1, Pollack's theorem applied to the quadratic character of Q(d)\mathbb Q(\sqrt d) viewed modulo 4an4an gives an odd prime p∤anp\nmid an with p≪an3/8p\ll_a n^{3/8} for which ax2≡n(modp)ax^2\equiv n\pmod p has two roots r1,r2r_1,r_2. Every k<n/ak<\sqrt{n/a} coprime to nn in one of these two classes makes pp divide the prime n−ak2n-ak^2, which forces n−ak2=pn-ak^2=p, an equation with at most one solution kk. A Möbius count over the prime factors of nn shows that the number of such kk is $\frac{2\sqrt{n/a}}{p}\cdot\frac{\varphi(n)}{n}+O(2^{\omega(n)})\gg_a n^{1/8}/\log\log n$, which exceeds 11 for large nn, a contradiction. When d=1d=1 the congruence is solvable for every odd prime not dividing anan, the least such prime is Oa(log⁡n)O_a(\log n), and the same count gives ≫an/(log⁡nlog⁡log⁡n)\gg_a\sqrt n/(\log n\log\log n) admissible kk. Remark 6.2 notes that the bound is ineffective because Pollack's theorem rests on Siegel's theorem, and that computation suggests 17221722 is the largest 11-good nn.

Reviewed. The site's curator, Thomas Bloom, marks Problem 1141 disproved and credits the resolution to the internal OpenAI model of this paper in the site's commentary (last edited 9 April 2026). The claim has no refereed evidence.

Formalizations. Four Lean files declare themselves formalizations of this result, following the paper, so they are links on this page and not claims of their own. The first, posted in the site's thread on 11 April 2026 by Yuta Oriike and made with GPT-5.4 Pro, proves the formal-conjectures statement and the aa-general variant with Pollack's Theorem 1.3 and Mertens' third theorem taken as axioms; its #print axioms lists those two beside the standard three. The link pins the revision of 13 April 2026 that updated the reference to Pollack's theorem, to which the thread post was edited to point; the file was first uploaded on 11 April 2026. The second, Boris Alexeev's lean-proofs file, names the model and the five authors as informal authors and GPT-5.4 Pro and Yuta Oriike as formal authors, and its header marks the proof unconditional; the pinned commit of 25 August 2026 imported a proof of Pollack's theorem, which the file takes from ErdosProblems.Erdos1141.PollackTheorem. The same commit adds a companion, Erdos1141b.lean, which names the same authors and proves the same statements without Pollack's theorem, from a split prime below (8an)31/64(8an)^{31/64} that a weak Burgess estimate supplies. The third, in the erdos-lean repository that the formal_proof attribute of the formal-conjectures statement names, inlines the lean-proofs file with its dependencies into one self-contained file. The second and third contain no sorry, axiom or native_decide token. The corpus has built none of these files, so the claim lists no formalized evidence.

Tang's earlier bound of N1/2+o(1)N^{1/2+o(1)} on the number of good n≤Nn\le N, described on the problem page, is not used by the proof.

Depends on. Nothing beyond the cited papers.