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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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Claim. For the triples 1≤i<j≤n/21\le i<j\le n/2 of Problem 699, some prime p≥ip\ge i divides both (ni)\binom ni and (nj)\binom nj whenever j≤3i/2j\le 3i/2, and whenever n=2jn=2j. The write-up is the Overleaf document linked above, submitted to the site's proof-claims tab on 2026-07-18; it comes with exact-integer code for the finite checks the argument needs.

Submission note. Posted to erdosproblems.com as a proof claim by Liam Price (account Leeham) on 18 July 2026, giving "GPT 5.6 Sol Pro" as the AI used, which the site marks as accepted as correct:

GPT-5.6 Sol Pro proves the conjecture when j≤3i/2j\le 3i/2 or n=2jn=2j, including reproducible exact-integer code for the finite verification.

Argument. The curator's reply on the claim lays out the first case. Let vv be the part of (ni)\binom ni supported on the primes p≥ip\ge i. A counterexample makes vv coprime to (nj)\binom nj. From the identity

(ni)(n−ij−i)=(nj)(ji),\binom ni\binom{n-i}{j-i}=\binom nj\binom ji,

a divisor of (ni)\binom ni coprime to (nj)\binom nj divides (ji)\binom ji, so v∣(ji)v\mid\binom ji, and Vandermonde's identity gives (ji)2<(2j2(j−i))\binom ji^2<\binom{2j}{2(j-i)}. The theorem of Ecklund, Eggleton, Erdős and Selfridge that v>(ni)1/2v>\binom ni^{1/2} outside finitely many explicit exceptions then forces (ni)<(2j2(j−i))\binom ni<\binom{2j}{2(j-i)} at any counterexample. When j≤3i/2j\le3i/2, 2(j−i)≤i<j≤n/22(j-i)\le i<j\le n/2 gives (2j2(j−i))≤(2ji)≤(ni)\binom{2j}{2(j-i)}\le\binom{2j}i\le\binom ni, a contradiction. The central case n=2jn=2j needs a longer elementary argument. The same inequality shows that for fixed i<ji<j only finitely many nn can be counterexamples, which the site states as an easy fact. Proposition 2.4 of van Doorn and Rocca's manuscript on the van Doorn–Rocca claim page proves the case j≤3i/2j\le3i/2 from the same two inputs.

Covers. Every triple with j≤3i/2j\le 3i/2 and every triple with n=2jn=2j. The triples with 3i/2<j<n/23i/2<j<n/2 remain, and the problem's label is unchanged.

Depends on. No page of this wiki.

Claimant and system. Liam Price submitted the claim and attributes the proof to GPT 5.6 Sol Pro; the site credits the result to GPT 5.6, prompted by Price.

Standing. Pending. The site's proof-claims tab marks the claim as accepted by the site, and the curator's remarks on the problem page (edited 19 July 2026) credit GPT 5.6, prompted by Price, with the cases j≤3i/2j\le 3i/2 and n=2jn=2j; Thomas Bloom's reply on the claim (2026-07-19) and Bloom's exposition on the problem page write out the case j≤3i/2j\le3i/2 only. The site labels the problem FALSIFIABLE, an open label, and the problem has no parts, so this credit is not acceptance. No refereed publication exists, the claim lists no formalization, and this corpus has not checked the write-up.