0
$\begingroup$

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?

$\endgroup$
1

1 Answer 1

2
$\begingroup$

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 $P(x)$.

$\endgroup$
1
  • 1
    $\begingroup$ 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, 2020 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.