Problem #382: Let $u\leq v$ be such that the largest prime dividing...

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Progress

Erdős and Graham report it follows from results of Ramachandra that $v-u\leq v^{1/2+o(1)}$.

Cambie has observed that the first question boils down to some old conjectures on prime gaps.
By Cramér's conjecture, for every $\epsilon>0,$ for every $u$ sufficiently large there is a prime between $u$ and $u+u^\epsilon$.
Thus for $u+u^\epsilon<v$, the largest prime divisor of \( \prod_{u \leq m \leq v} m \) appears with exponent $1$.
Since this is not the case in the question, \( v - u = v^{o(1)} \).

Cambie also gives the following heuristic for the second question. The 'probability' that the largest prime divisor of $n$ is $<n^{1/2}$ is $1-\log 2>0$. 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 $v-u\geq k$. See [383] for a conjecture along these lines. A similar argument applies if we replace multiplicity $2$ with multiplicity $r$, for any fixed $r\geq 2$.

See also [380].