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PROVED This has been solved in the affirmative.
We call an interval $[u,v]$ 'bad' if the greatest prime factor of $\prod_{u\leq m\leq v}m$ occurs with an exponent greater than $1$. Let $B(x)$ count the number of $n\leq x$ which are contained in at least one bad interval. Is it true that\[B(x)\sim \#\{ n\leq x: P(n)^2\mid n\},\]where $P(n)$ is the largest prime factor of $n$?
Erdős and Graham only knew that $B(x) > x^{1-o(1)}$. Similarly, we call an interval $[u,v]$ 'very bad' if $\prod_{u\leq m\leq v}m$ is powerful. The number of integers $n\leq x$ contained in at least one very bad interval should be $\ll x^{1/2}$. In fact, it should be asymptotic to the number of powerful numbers $\leq x$.

We have\[\#\{ n\leq x: P(n)^2\mid n\}=\frac{x}{\exp((c+o(1))\sqrt{\log x\log\log x})}\]for some constant $c>0$.

Tao notes in the comments that if $[u,v]$ is bad then it cannot contain any primes, and hence certainly $v<2u$, and in general $v-u$ must be small (for example, assuming Cramer's conjecture, $v-u\ll (\log u)^2$).

Tao [Ta26c] has proved this asymptotic, and in fact\[B(x) = \left(1+O((\log x)^{-1+o(1)})\right) \#\{ n\leq x: P(n)^2\mid n\}.\]Furthermore, the number of $n\leq x$ contained in at least one very bad interval, but are not themselves powerful, is $O(n^{2/5+o(1)})$.

See also [382].

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Additional thanks to: Terence Tao

When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:

T. F. Bloom, Erdős Problem #380, https://www.erdosproblems.com/380, accessed 2026-04-23