# Counting solutions modulo primes

Let $P(x)$ be an irreducible polynomial in $\mathbb{Z}[x]$ of degree $n.$ By $N(k,x)$ we denote the number of primes up to $x,$ such that $P(x)$ has exactly $k$ solutions in $\mathbb{Z}_p.$ Is it possible to get the asymptotic for $N(k,x)?$ I am mostly interested in the case when $k=1$ and Galois group of $P(x)$ is not $S_n.$

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Sorry, if I sound dumb. What is the meaning of "up to $x$"? As $x$ is a symbolic variable, and this upto $x$ sounds like bounded above by $x$, I am confused. –  P Vanchinathan Apr 4 '14 at 2:49
For that matter, what does it mean for a polynomial to have a solution? –  Gerry Myerson Apr 4 '14 at 4:08

Yes, this is possible. The more natural method is by using Chebotarev's density theorem. If $G$ is the Galois group of $P(x)$, and $p$ is a prime not dividing the discriminant of $P$, the number of roots $k$ of $P(x)$ in $\mathbb Z/p \mathbb Z$ is the number of fixed points of the Frobenius element $\sigma_p$ of $G$. So let $D_k$ be the set of elements of $G$ which have exactly $k$-fixed prime. Cheobotarev density theorem tells you that the number $N(k,x)$ of primes up to $x$ with $\sigma_p$ in $G$ is $\sim \frac{|D_k|}{|G|} Li(x) \sim \frac{|D_k|x}{|G|\log x}$.If you want more precise result with an error term, you need to apply an effective version of Chebotarev's density theorem. Serre's paper "Quelques applications du théorème de Chebotarev" is a good place to start, if you read French.
Thanks for the reference. So in fact, |D_1| can be $0.$ Would it be possible to estimate the number of the primes from below then? –  Bob Apr 4 '14 at 1:01
"$k$-fixed prime" supposed to be "$k$ fixed points"? –  Gerry Myerson Apr 4 '14 at 4:11