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For a given $P\in \mathbb{Z}[x]$ call a positive prime $p$ good if there exists $n\in \mathbb{Z}$ such that $p$ divides $P(n)$. Does there exist a non-constant $P$ such that the set of good primes has well-defined Dirichlet density (with respect to the set of all positive primes) and that density is equal to zero?

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No. The number of roots of $P(x)$ modulo a prime $p$, when averaged over $p$, asymptotically equals the number of irreducible factors of $P(x)$ by the prime ideal theorem. Together with the fact that this number of roots is at most the degree of $P(x)$, this shows that a positive density of primes $p$ have the property that $P(x)$ has a root modulo $p$; in other words, a positive density of primes $p$ divide some value of $P(x)$.

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    $\begingroup$ I presume you mean this prime ideal theorem (you may want to include the link or other reference, just for completeness). $\endgroup$ – Wojowu Feb 6 '20 at 21:51

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