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Iwaniec proved (and many people extended) that the number of $n \le x$ for which $n^2+1=P_2$ (product of at most two primes) is $\gg x/\log x$. I am wondering what is known/can be proved for the counting function of a localised version of this problem $$\{n\le x \mathrel\vert\exists p,q\geq x^\alpha\text{ primes}:pq \mid n^2+1\}$$ for $\alpha> 1/2$ (or say $\alpha=1-\varepsilon$).

In particular, is it possible to prove a cheap upper bound of the (nearly) correct order of magnitude?

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    $\begingroup$ I found it hard to understand some of the quantification, so I edited, I hope in a way that clarified rather than obscuring or changing meaning. $\endgroup$
    – LSpice
    Commented Aug 5, 2021 at 16:50
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    $\begingroup$ Do you mean to count $n\leq x$ such that there exist two primes $p,q\geq x^\alpha$ with $pq\mid n^2+1$? Note that this condition does not imply that $n^2+1$ is $P_2$. $\endgroup$
    – GH from MO
    Commented Aug 5, 2021 at 17:45
  • $\begingroup$ @GH: indeed, that is what I mean. You are of course right that this condition does not imply that $n^2+1\in P_2.$ The example is merely to illustrate that, some lower bound could possibly be given provided one can better localise primes in Iwaniec's proof, say. $\endgroup$
    – user334097
    Commented Aug 5, 2021 at 18:39
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    $\begingroup$ This is related to the equidistribution of roots of $n^2+1$ modulo $pq$ as $p,q$ vary. If one replaced $pq$ by a single prime $p$ then there is a famous paper of Duke, Friedlander and Iwaniec in this area mathscinet.ams.org/mathscinet-getitem?mr=1324141 . Perhaps some of the methods in this paper or followup work can be adapted to your question, though I would imagine the specific question itself has not been explicitly addressed in the literature. $\endgroup$
    – Terry Tao
    Commented Aug 5, 2021 at 21:48
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    $\begingroup$ @Terry Tao: thanks! I was indeed aware of this result and the equidistribution mod pq should in principle be doable. Is there perhaps a cheap way to prove any decent upper bound for my question? $\endgroup$
    – user334097
    Commented Aug 5, 2021 at 22:08

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