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Let $k$ be a squarefree positive integer. We know that a prime $p$ splits in $K=\mathbb{Q}(\sqrt{-k})$ if and only if $-k$ is a quadratic residue mod $p$.

My question is: can we explicitly determine the set of all primes that split in $K$ for a certain parametrized $k$?

Thanks!

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The following answer concerns the original version of the question, while my comment below addresses the updated version.

For $p>3$ the Chinese Remainder Theorem shows that there are integers $k\equiv 11\pmod{24}$ such that $-k$ is not a quadratic residue modulo $p$.

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  • $\begingroup$ Oh, so $3$ is the only prime that splits in $K$ for $k\equiv11\pmod{24}$? $\endgroup$
    – user491084
    Commented Sep 12, 2022 at 10:04
  • $\begingroup$ @user491084 Yes, $p=3$ is the only prime that splits simultaneously in every $\mathbb{Q}(\sqrt{-k})$ for $k\equiv 11\pmod{24}$. If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Commented Sep 12, 2022 at 10:06
  • $\begingroup$ Sure, I have now edited the question and asked in a more general context :) $\endgroup$
    – user491084
    Commented Sep 12, 2022 at 10:07
  • $\begingroup$ @user491084 My answer should make it clear that the answer to your general question is also yes. The prime $p$ must divide the modulus of the arithmetic progression of the $k$'s, and for such a prime either every $-k$ is a quadratic residue modulo $p$ or every $-k$ is a quadratic non-residue modulo $p$ (which can be checked by plugging in the relevant residue into the quadratic residue symbol modulo $p$). $\endgroup$
    – GH from MO
    Commented Sep 12, 2022 at 10:11
  • $\begingroup$ Okay, that makes sense. Thanks!!! $\endgroup$
    – user491084
    Commented Sep 12, 2022 at 10:13

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