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Let $n$ be a square-free natural number, and let $f\in\mathbb{Z}[x]$ be monic and irreducible of degree $\geq2$. I am trying to determine whether there always exists a prime $p$, $p\nmid n$, satisfying both of the following conditions:

  1. $\left(\frac{-n}{p}\right)=1$;
  2. $f(x)=0$ is solvable modulo $p$.

I know that the Chebotarev Density Theorem implies that primes satisfying 1. have density 1/2, and I also know that there are infinitely primes satisfying 2. I'm not sure if this is known, or how hard this problem is, but any references would be very welcome.

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    $\begingroup$ With 2 you mean that there is a root of $f$ modulo $p$? You get a positive answer by applying Chebotarev to the compositum of the splitting field of $f$ and $\Bbb Q(\sqrt{-n})$. $\endgroup$
    – Wojowu
    Commented Apr 20, 2023 at 23:15
  • $\begingroup$ @Wojowu yes, that's what I mean for 2. I edited the question, thanks. Would you mind expanding a bit on your answer? If $K$ is the composite field, then what would the conjugation invariant subset $X$ of $Gal(K/\mathbb{Q})$ that I apply Chebotarev to be? $\endgroup$
    – Jack
    Commented Apr 21, 2023 at 3:28
  • $\begingroup$ Just the identity. A prime with trivial Frobenius in that compositum will have trivial Frobenius in both subfields, which implies the respective conditions. $\endgroup$
    – Wojowu
    Commented Apr 21, 2023 at 3:40
  • $\begingroup$ I should add that for the correct (positive) logarithmic density of splitting primes one does not need Chebotarev. One simply needs that the Dedekind zeta function has a simple pole at $s=1$, then take the logarithmic derivative, and observe that non-splitting primes only contribute $O(1)$ around $s=1$. $\endgroup$
    – GH from MO
    Commented Apr 21, 2023 at 6:02
  • $\begingroup$ @Wojowu thanks, that makes sense. $\endgroup$
    – Jack
    Commented Apr 21, 2023 at 15:21

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