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How many integers $n\leq X$ are there with the property that $\prod_{p\in S} p \geq n^{1/2-\epsilon}$? Here (to keep notation readable) I've written $p\in S$ if and only if $p||n$ (that is, $p|n$ and $p^2\nmid n$) and $n/p$ is a square modulo $p$.

I'd like it to be $\ll X^{1-\delta}$ for a positive $\delta$, but I don't even have a heuristic for how large it should be! Hopefully I'm not being foolish and this is plausible.

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    $\begingroup$ Maybe, -epsilon should be replaced to +epsilon? $\endgroup$
    – Fedor Petrov
    Commented Oct 4, 2014 at 10:22
  • $\begingroup$ Nah, primes satisfy this condition. Sorry! $\endgroup$
    – alpoge
    Commented Oct 4, 2014 at 10:40
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    $\begingroup$ Shouldn't this be at least a positive proportion? A positive proportion are squarefree, then for each $p$ dividing $n$, half the numbers are squares mod $p$, so we should expect half the primes to satisfy this condition, so $\prod_{p \in S} p \geq n^{1/2}$ about half the time. Subtract the $\epsilon$ can only increase this. $\endgroup$
    – Will Sawin
    Commented Oct 4, 2014 at 11:25
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    $\begingroup$ In particular, taking the product of a prime $p$ between $X^{1/2-\varepsilon}$ and $X^{1/2}$ together with a quadratic residue modulo that prime of size at most $X/p$ already gives a positive density of integers with the required property. (One can also take some primes above $X^{1/2}$, but one starts needing information about the distribution of quadratic residues in short intervals, e.g. Polya-Vinogradov or Burgess bounds.) $\endgroup$
    – Terry Tao
    Commented Oct 4, 2014 at 16:11
  • $\begingroup$ Nice! Thanks much, all. Sorry for the far too optimistic guess. $\endgroup$
    – alpoge
    Commented Oct 4, 2014 at 16:49

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