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Claim. Theorem 1 of Will Sawin's preprint, Sets of unit fractions without two members whose average is a unit fraction (arXiv:2607.15419v1, 16 July 2026), the result page of the source card: for a positive integer $N$ let $A_N$ be the set of $a\in\{1,\ldots,N\}$ such that $a+b\nmid2ab$ for every $b\in\{1,\ldots,N\}$ with $b\ne a$ and $\Omega(b)\le\Omega(a)$, where $\Omega$ counts prime factors with multiplicity. Then distinct members $a,b$ of $A_N$ satisfy $a+b\nmid2ab$, and there is an absolute constant $c>0$ with $|A_N|>cN$ for all large $N$. In the notation of Problem 327, $f_2(N)>cN$ for large $N$, so the second question's answer is negative: a set in which no two distinct members have $a+b\mid2ab$ need not have size $o(N)$. The first part is immediate, since one of any two integers has at most as many prime factors as the other; the second is proved by restricting to integers without very small prime factors and with controlled numbers of prime factors of each size, and bounding the integers excluded from $A_N$ through a mean-value theorem of de la Bretèche and Tenenbaum, along the lines of Stef's 1992 argument for integers without two close divisors. The paper makes no attempt to compute $c$.

Submission note. Posted to erdosproblems.com as a proof claim by Will Sawin (account WillSawin) on 18 July 2026, giving "ChatGPT 5.5 for data analysis and reference search and ChatGPT 5.6 was used for proofreading. The proof strategy and writeup are mine." as the AI used:

My proof gives a negative answer to the second part of the question: More precisely, there is a constant $c>0$ such that for large $N$ there is a set $A \subseteq {1,\dots, N}$ such that if $a,b\in A$ have $a\neq b$ then $a+b \nmid 2ab$ and $\lvert A\rvert \geq cN$. This is proved by an explicit construction: We take $A$ to consist of all $a \in \{1,\dots, N\}$ such that if $b\in \{1,\dots ,N\}$ has $a\neq b$ and $\Omega(b) \leq \Omega(a)$ then $a+b\nmid 2ab$. The reason this works is that for a typical $a \in {1,\dots,N}$, the number of $b \in \{1,\dots, N\}$ with $\Omega(b)\leq \Omega(a)$ and $a+b \mid 2ab$ is $O(1)$. Once this is checked, it is not difficult to find a positive density set of $a$ on which the average number of $b$ satisfying this condition is less than $1/2$, Checking this requires expressing the problem in terms of the anatomy of integers, and using estimates from analytic number theory.

Covers. The second question only. The paper says that its method gives, for the first question ($a+b\nmid ab$), a lower bound weaker than the odd numbers' $\lceil N/2\rceil$; through the doubling correspondence recorded on the problem page, the construction yields even sets of positive density for the first condition without exceeding the odd numbers. The constant $c$ is not computed.

Provenance. The preprint is the claim's first posting (arXiv v1 of 16 July 2026, the only version on the arXiv listing no journal reference). The author filed it on the site's proof-claim tab on 18 July 2026, naming the AI systems ChatGPT 5.5 for data analysis and reference search and ChatGPT 5.6 for proofreading, with the proof strategy and the write-up their own; the paper's own account (p. 2) credits ChatGPT with noticing, in Cambie's largest example for $N=500$, the pattern that led to the construction. The tab entry was filed as a full claim and corrected by the author in a comment of 29 July 2026 to the second part of the problem only.

Standing. Claimed. The site's label is OPEN and its commentary does not mention the result (2026-10-07). The comments under the claim are not an acceptance: the site's curator added the size clause to the claim text on 18 July 2026 and said they looked forward to the paper, and a reader's comment of 23 July 2026 reports a missing bracket in Lemma 6 and otherwise no issue. No journal acceptance, independent review or citing work was found on 2026-09-17. The source card records the statement as checked and the proof as not independently verified; neither awards standing. A later full claim, Della Pietra's manuscript of 29 July 2026, says it reproves this conclusion independently.