Erdős Problem #382

declared status 'open'. Formalized: no. Let $u\leq v$ be such that the largest prime dividing $\prod_{u\leq m\leq v}m$ appears with exponent at least $2$. Is it true that $v-u=v^{o(1)}$? Can $v-u$ be arbitrarily large? Current best: Erd\H{o}s and Graham report it follows from results of Ramachandra that $v-u\leq v^{1/2+o(1)}$. For any fixed $k$, there is therefore a positive 'probability' that there are $k$ consecutive integers around $q^2$ (for a prime $q$) all of whose prime divisors are bounded above by $q$, when $v-u\geq k$. A similar argument applies if we replace multiplicity $2$ with multiplicity $r$, for any fixed $r\geq 2$. Prize: no. OEIS: A388850. Tags: number theory.

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