Let $g(x)\in \mathbb{Z}[x]$ be 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 assuming that the set is finite but i haven't reached any contradiction.
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 $k$.
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: $$\sum_{p \in P} \frac{1}{p^{1/d}}=\infty$$
Assume that there are only finitely many such primes $p_1, p_2, ..., p_r$, and let $q = p_1 p_2 \cdots p_r$. 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 $q$.
