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Let $p$ be a prime such that $p\equiv1\mod4$. Is it true that there will always exist a prime $q$ that satisfies $q\vert(p+a^2)$ and $q\equiv3\mod4$, for some integer $a$?

I have tried proceeding by contradiction, assuming that for every prime $q$ such that $q\vert(p+a^2)$ it must satisfy $q\equiv1\mod4$, however I have not been able to find a contradiction. Is this even true, and is it a much more complicated proof than I think? It seems to me like it must be true as there are infinitely many primes congruent to 3 modulo 4.

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    $\begingroup$ I think the question is not clear. I believe you are asking: "Given a prime $p\equiv 1 \pmod 4$ does there exist a prime $q\equiv 3\pmod 4$ such that $q$ divides $p+a^2$ for some integer $a$". In any case, that's the question I attempted to answer below. $\endgroup$
    – lulu
    Commented May 11, 2020 at 18:51
  • $\begingroup$ Yes that was the question I was asking, sorry for any confusion. $\endgroup$
    – user520830
    Commented May 11, 2020 at 19:28

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If $q\equiv -1 \pmod 4$, By quadratic reciprocity we have $$\left(\frac {-p}q\right)=\left(\frac {-1}q\right)\times\left(\frac qp\right)=-\left(\frac qp\right)$$

So choose any quadratic non-residue $n$ $\pmod p$ and let $q$ be a prime satisfying $$q\equiv n\pmod p\quad \&\quad q\equiv -1\pmod 4$$ (infinitely many such $q$ exist by Dirichlet's Theorem). It follows that $q$ is of the form you want.

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  • $\begingroup$ I figured this question related to quadratic reciprocity, but, wait, doesn't $\left(\frac pq\right)=\left(\frac qp\right)$ if $p\equiv1\pmod4$? $\endgroup$
    – J. W. Tanner
    Commented May 11, 2020 at 18:53
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    $\begingroup$ @J.W.Tanner Yes, but I omitted the $-1$ multiplier for $p$. I will edit. The congruence $p+a^2\equiv 0 \pmod q$ is the same as requiring that $-p$ is a square $\pmod q$. $\endgroup$
    – lulu
    Commented May 11, 2020 at 18:54
  • $\begingroup$ because it's $a^2+p$ not $a^2-p\equiv0\pmod q$ $\endgroup$
    – J. W. Tanner
    Commented May 11, 2020 at 18:56
  • $\begingroup$ @J.W.Tanner Right. We have $a^2\equiv -p \pmod q$ so we need $\left(\frac {-p}q\right)=1$, which ends up being the same as requiring that $\left( \frac qp\right)=-1$. $\endgroup$
    – lulu
    Commented May 11, 2020 at 18:57
  • $\begingroup$ I got it, @lulu, thanks; +$1$ $\endgroup$
    – J. W. Tanner
    Commented May 11, 2020 at 18:58

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