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Source. T. Feng, T. Trinh, G. Bingham et al., Semi-Autonomous Mathematics Discovery with Gemini: A Case Study on the Erdős Problems, arXiv:2601.22401v3 (5 February 2026); Section 4.2, the problem on p. 21, Remark 4.2 on pp. 21--22, the solution on pp. 22--23 with Lemma 5 on p. 22 and its proof on pp. 22--23, Addendum 4.1 on p. 23. The result is unnumbered. The artifact is identified on the source card.

Read depth. Claims checked: the assertion, Lemma 5 and the proof (pp. 22--23) were read in full on the print. Nothing here is independently reviewed. A preprint.

Statement

For $n=\prod_pp^{k_p}$ let $Q_2(n)=\prod_{k_p\ge2}p^{k_p}$ be its powerful part. The paper proves (pp. 22--23) that for every integer $\ell\ge2$

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

• Lemma 5 (p. 22). With $x_k+y_k\sqrt8=(3+\sqrt8)^k$ and $n_k=8y_k^2$, for every prime $p\equiv5\pmod 8$ there is a positive integer $k$ with $n_k+2\equiv0\pmod{p^2}$.

Proof pointer

Since $n(n+1)(n+2)$ divides $n(n+1)\cdots(n+\ell)$, the case $\ell=2$ suffices. Along $n_k=8y_k^2$ both $n_k$ and $n_k+1=x_k^2$ are powerful, so the ratio exceeds $Q_2(n_k+2)$. Lemma 5 uses that $p\equiv5\pmod8$ is inert in $\mathbb Z[\sqrt2]$ and the Frobenius map to find an odd power of $3+\sqrt8$ congruent to $-1$ modulo $p^2$; Dirichlet's theorem gives infinitely many such $p$, so $Q_2(n_k+2)\ge p^2$ is unbounded (pp. 22--23).

Dependencies

Dirichlet's theorem on primes in arithmetic progressions.

Bears on

• Problem 935: answers the second of its three questions affirmatively; the first and third are not addressed (Remark 4.2, p. 21). Remark 4.2 calls the argument well known to experts.
• Problem 367: provenance only. Addendum 4.1 (p. 23) records that the question solved is almost identical to one in Problem 367 and that the construction is the same as in van Doorn's comment of 20 November 2025 on that problem's page, and on that basis reclassifies the case as an independent rediscovery. No result about Problem 367 is credited here.