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Problem #935: For any integer $n=\prod p^{k_p}$ let $Q_2(n)$ be the...

For any integer $n=\prod p^{k_p}$ let $Q_2(n)$ be the powerful part of $n$, so that\[Q_2(n) = \prod_{\substack{p\\ k_p\geq 2}}p^{k_p}.\]Is it true...

Problem Statement

For any integer $n=\prod p^{k_p}$ let $Q_2(n)$ be the powerful part of $n$, so that\[Q_2(n) = \prod_{\substack{p\\ k_p\geq 2}}p^{k_p}.\]Is it true that, for every $\epsilon>0$ and $\ell\geq 1$, if $n$ is sufficiently large then\[Q_2(n(n+1)\cdots(n+\ell))<n^{2+\epsilon}?\]If $\ell\geq 2$ then is\[\limsup_{n\to \infty}\frac{Q_2(n(n+1)\cdots(n+\ell))}{n^2}\]infinite?

If $\ell\geq 2$ then is\[\lim_{n\to \infty}\frac{Q_2(n(n+1)\cdots(n+\ell))}{n^{\ell+1}}=0?\]
Categories: Number Theory Powerful

Progress

Erdős [Er76d] writes that if this is true then it 'seems very difficult to prove'.

A result of Mahler implies, for every $\ell\geq 1$,\[\limsup_{n\to \infty}\frac{Q_2(n(n+1)\cdots(n+\ell))}{n^2}\geq 1.\]All these questions can be asked replacing $Q_2$ by $Q_r$ for $r>2$, only keeping those prime powers with exponent $\geq r$.

Source: erdosproblems.com/935 | Last verified: January 19, 2026

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