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Problem 729

../

claims/: The 1 claim page of Problem 729, one per claimant's result; the problem's standing derives from them.


Statement. Let C>0C>0 be a constant. Are there infinitely many integers a,b,na,b,n with a+b>n+Clog⁡na+b> n+C\log n such that the denominator of

n!a!b!\frac{n!}{a!b!}

contains only primes ≪C1\ll_C 1?

Status. PROVED (LEAN). The site labels the problem PROVED (LEAN) (page last edited 11 January 2026) and credits Barreto and Leeham, using ChatGPT and Aristotle, for an affirmative proof that modifies the argument used for Problem 728; the result is recorded on the claim page Barreto and Price 2026. The Lean qualifier refers to Aristotle's formalization of the GPT-5.2 Pro argument, in Boris Alexeev's repository of formalized Erdős problems, which this corpus has not built; nothing is refereed. The standing in the frontmatter derives from the claim page.

Source. erdosproblems.com/729, accessed 2026-09-04 and, with its discussion thread and the community database, 2026-10-07. Cite as: T. F. Bloom, Erdős Problem #729, https://www.erdosproblems.com/729.

References.

  • [Er68c] P. Erdős, Aufgabe 557. Elemente Math. (1968), 111-113.
  • [EGRS75] Erdős, P. and Graham, R. L. and Ruzsa, I. Z. and Straus, E. G., On the prime factors of (\sp2n\sbn)(\sp{2n}\sb{n}). Math. Comp. (1975), 83-92. Library home: erdos_1975_prime_factors.
  • [So26] Sothanaphan, N., Resolution of Erdős Problem #728: a writeup of Aristotle's Lean proof. arXiv:2601.07421 (2026). Library home: sothanaphan_2026_resolution_erdos_problem_728_writeup_aristotle.

Formalization. Statement in formal-conjectures, tagged solved, which reads the denominator in Q\mathbb{Q} and asks for a bound KK depending on CC beyond which no prime divides it; it names as the problem's formal proof the Aristotle-generated Lean file in Alexeev's repository, linked at pinned commits from the claim page. This corpus has not built it.

Current assessment

The question, as the site states it (page last edited 11 January 2026): for a constant C>0C>0, are there infinitely many a,b,na,b,n with a+b>n+Clog⁡na+b>n+C\log n such that the denominator of n!/(a! b!)n!/(a!\,b!) contains only primes bounded in terms of CC? Erdős [Er68c] proved that a! b!∣n!a!\,b!\mid n! forces a+b≤n+O(log⁡n)a+b\le n+O(\log n), and the proof needs only the prime 22: by Legendre's formula the exponent of 22 in n!n! is nn minus the binary digit sum of nn, so the divisibility gives a+b≤n+O(log⁡n)a+b\le n+O(\log n) at once. The problem, a remark of [EGRS75], asks whether the bound persists when the small primes are ignored. The answer to the stated question is yes: for every CC there are infinitely many triples with a+b>n+Clog⁡na+b>n+C\log n whose denominator involves only primes below a threshold depending on CC. The bound itself survives for each fixed set of ignored primes, with a constant depending on the set (Legendre's formula at the least prime outside the set gives it), and fails only once the bound on the ignored primes may depend on CC.

Proof. The accepted claim page Barreto and Price 2026 records the result the site credits: an informal argument of GPT-5.2 Pro, adapting the proof of Problem 728 on the claim page Barreto 2026, formalized by Harmonic's Aristotle from the TeX alone and posted by Kevin Barreto to the thread, the informal proof on 2026-01-08 and the Lean proof on 2026-01-10. With n=2mn=2m, b=mb=m, a=m+ka=m+k and k=⌊clog⁡m⌋k=\lfloor c\log m\rfloor, the denominator, the numerator of k!(m+kk)/(2mm)k!\binom{m+k}{k}/\binom{2m}{m} in lowest terms, is controlled prime by prime: Kummer's theorem gives the carries in νp(2mm)\nu_p\binom{2m}{m}, and a Chernoff-and-union-bound count over the primes p≤2kp\le2k finds mm for which νp(2mm)≥νp(m+kk)+νp(k!)\nu_p\binom{2m}{m}\ge\nu_p\binom{m+k}{k}+\nu_p(k!) at every prime p≥pmin⁡(c)p\ge p_{\min}(c). Sothanaphan's writeup [So26] of the Problem 728 proof derives the same statement, in its second appendix, from a general valuation theorem extracted from that method. Pomerance's refereed note on the middle binomial coefficient (claim page Pomerance 2026 of Problem 728) proves that (m+1)⋯(m+k)∣(2mm)(m+1)\cdots(m+k)\mid\binom{2m}{m} for almost all mm and every k≤ηlog⁡mk\le\eta\log m with η<1/log⁡4\eta<1/\log4, the stronger integrality with no denominator at all, but only for constants below 1/log⁡41/\log4, so it is a partial result for this problem and carries no claim page here. Problem 401 is a later, more precise question in the same spirit, settled the next day by the same route.

Search scope, 2026-10-07: the site's problem page, its discussion thread (30 comments), the community database, the formal-conjectures file and Alexeev's repository. The site lists no proof claim for the problem, and the thread's literature searches, including an inquiry to Pomerance, found no earlier solution.

Linked library material

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