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What is the proof that for any integer $n$ and any non-constant, integer coefficient polynomial $P(x)$, there are infinitely many primes congruent to $1 \pmod{n}$ that divide $P(x)$ for some $x$?

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Neglecting the $p \equiv 1 \pmod{n}$ condition, here is a nice explanation of why there are infinitely many primes dividing $P(x)$ for some $x$. –  Dan Brumleve Aug 21 '12 at 10:08
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1 Answer

This is a pretty standard application of Cebatarov density: Let $K$ be the splitting field of $P(x)$. Let $L$ be the composite field of $K$ and $\mathbb{Q}(\zeta_n)$, where $\zeta_n$ is a primitive $n$-th root of unity. A prime $p$ splits in $K$ if and only if1 $P(x)$ splits into distinct linear factors modulo $p$; it splits in $\mathbb{Q}(\zeta_n)$ if and $p \equiv 1 \bmod n$; it splits in $L$ if and only if both are true. By Cebatarov density (or the weaker Theorem of Frobenius) the primes that split in $L$ have density $1/\dim L$.

I can think of ways to make various parts of this argument more elementary, but I don't see how to get away from using algebraic number theory; I'd be curious to see an elementary proof.

1 With finitely many exceptions, related to the fact that the ring of integers in $K$ may not be $\mathbb{Z}[x]/P(x)$.

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