# A density zero set of primes dividing the values of a non-constant integer polynomial

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?

• – Greg Martin Feb 6 '20 at 21:44

## 1 Answer

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 $$P(x)$$.

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