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Claim. For every integer $\ell\geq 2$,

$$\limsup_{n\to\infty}\frac{Q_2(n(n+1)\cdots(n+\ell))}{n^2}=\infty,$$

where $Q_2(m)$ is the powerful part of $m$; this answers the second question of Problem 935 yes. Tony Feng and twenty-three coauthors, Semi-Autonomous Mathematics Discovery with Gemini: A Case Study on the Erdős Problems, arXiv:2601.22401, carded as Feng et al. 2026, give the proof in Section 3.2 of version 1 (29 January 2026) and in Section 4.2 of version 3 (5 February 2026).

Covers. The second question: for every $\ell\geq 2$ the ratio of $Q_2(n(n+1)\cdots(n+\ell))$ to $n^2$ is unbounded. The first and third questions are not addressed.

The argument. Since $n(n+1)(n+2)$ divides $n(n+1)\cdots(n+\ell)$, it is enough to treat $\ell=2$. Let $x_k+y_k\sqrt8=(3+\sqrt8)^k$, so that $x_k^2-8y_k^2=1$, and put $n_k=8y_k^2$. Then $n_k=2^3y_k^2$ and $n_k+1=x_k^2$ are both powerful. The paper's Lemma 5 shows that for every prime $p\equiv 5\pmod 8$ some $k$ has $p^2\mid n_k+2$. No odd prime divides two of $n_k$, $n_k+1$, $n_k+2$, so for that $k$ the powerful part of the product is at least $n_k(n_k+1)Q_2(n_k+2)$, and the ratio to $n_k^2$ exceeds $p^2$. As $p$ runs over the infinitely many primes $p\equiv 5\pmod 8$, the ratio is unbounded.

Claimant. The paper attributes the solution to Aletheia, a research agent built on Gemini Deep Think. Version 1 lists the result as a partial AI solution. Version 3 lists it as an independent rediscovery: its Addendum 4.1 records that Wouter van Doorn gave the same construction in a comment of 20 November 2025 on the site's thread for Problem 367, an almost identical question, and the authors report that the agent did not access that page.

Standing. The claim is claimed. The site labels the problem OPEN, so its commentary crediting van Doorn's construction, also given by Aletheia, with the second part is not acceptance; the preprint has no journal publication.

Depends on. No page of this wiki.