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Claim. On 30 April 2026 Liam Price posted in the site's thread that GPT-5.5 Pro claims a disproof of the first question of Problem 983 and a bound for the second. The post links a write-up as a read-only shared document. The claim is that the Erdős–Straus upper bound $f(\pi(n)+1,n)\le2\pi(n^{1/2})+1$ [Er70b] is attained for infinitely many $n$. Nat Sothanaphan's thread exposition of 19 May 2026 gives the argument. By Pomerance's theorem (The prime number graph, Math. Comp. 33 (1979), 399–408), infinitely many $N$ satisfy $p_{N-i}p_{N+i}<p_N^2$ for $1\le i<N$. For such $N$ take $n=p_N^2-1$, so that $\pi(n^{1/2})=N-1$. Let $A$ consist of the $2N-2$ semiprimes along the path $p_{2N-1},p_1,p_{2N-2},p_2,\ldots,p_{N+1},p_{N-1},p_N$, the two end primes $p_{2N-1}$ and $p_N$, and every prime up to $n$ other than $p_1,\ldots,p_{2N-1}$. Then $A$ has $\pi(n)+1$ elements, all at most $n$, and no set of fewer than $2N-1$ primes covers more elements of $A$ than it has primes. So $f(\pi(n)+1,n)=2\pi(n^{1/2})+1$ for these $n$, the difference $2\pi(n^{1/2})-f(\pi(n)+1,n)$ equals $-1$ infinitely often, and the answer to the first question is no.

Covers. The first question. The post also announces a bound for the second question, but the thread does not state it, and this page does not cover it.

Standing. The site's label is OPEN, and the claim is not on the site's proof-claims tab. On 30 April 2026 Sothanaphan replied that a standard AI check flagged a few minor issues, and Sothanaphan recommended citing Pomerance for the key lemma. On 19 May 2026 Sothanaphan re-derived the argument in the thread, crediting GPT-5.5 Thinking for discussion. Neither post is an acceptance, and there is no refereed version. Claimed.

Depends on. No page of this wiki.