Primes dividing a polynomial Let $g(x)\in \mathbb{Z}[x]$, a nonconstant polynomial. Show that the set of primes $p$ such that $p\mid g(n)$ for some $n\in \mathbb{Z}$ is infinite.
I don't know how to start. I have tried asuming that the set is finite but i haven't reached any contradiction.
 A: Assume that there are only finitely many such primes $p_1$, $p_2$, ..., $p_r$, and let $q = p_1 p_2 \dots p_r$. Let $g(0) = c$. If $c = 0$, the result is obvious, so assume $c \ne 0$. For integer values of $x$, we have 
$$g(cqx) \equiv c \pmod{cq}.$$
For large enough values of $x$, we have $g(cqx) \ne \pm c$, and then $g(cqx)$ must have a prime divisor that is prime to $q$. 
A: Outline: If the constant term of the polynomial is $0$, the result is obvious. The rest of the proof imitates the standard Euclid-style proof that there are infinitely many primes.
So let the constant term be $a\ne 0$. It follows that the polynomial  $g(n)$ has the shape 
$$g(n)=nq(n)+a,$$
where $q(n)$ is a polynomial with integer coefficients.
As $n$ gets large, $g(n)$ becomes very large positive or very large negative. Without loss of generality we can assume it becomes very large positive. In particular, for $n$ large enough we have $g(n)\gt |a|$.
Now let $k$ be large, and look at 
$g(k!a^2)=a^2k!q(a^2k!)+a=a(ak!q(a^2k!)+1)$. Then $ak!q(a^2k!)+1$ is divisible by some prime, and that prime must be greater than $k$. 
A: HINT:
Like Euler: for  $P$ of degree $d\ge 1$ we have
$$\lim_{n\to \infty} \frac{|P(n)|^{\frac{1}{d}} }{n} = c \ne 0$$
so
$$\sum_{n\ge n_0} \frac{1}{|P(n)|^{\frac{1}{d}}} = \infty$$
If only the primes $p_1$, $\ldots$, $p_l$ divided any of the numbers $|P(n)|$, $n \ge n_0$ then the above sum would be
$$\le  \prod_{k=1}^l \left( \sum_{n\ge 0} \frac{1}{p_k^{\frac{n}{d}} }\right) = \prod_{k=1}^l \frac{1}{ 1 - \frac{1}{p_k^{1/d}} }< \infty $$
which is not possible.  In fact we showed that if $\mathcal{P}$ is the set of primes that divide at least one of the terms of a sequence $(a_n)_n$ with polynomial growth ( $a_n = \mathcal{O}(n^d)$ ),  then
$$\prod_{p \in P} \frac{1}{ 1 - \frac{1}{p^{1/d}} } = \infty$$
or, equivalently:
$$\sum_{p \in P} \frac{1}{p^{1/d}}=\infty$$
