Primes dividing a polynomial

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 numbers $P(n)$ then

$$\prod_{p \in P} \frac{1}{ 1 - \frac{1}{p^{1/d}} } = \infty$$ or, equivalently: $$\sum_{p \in P} \frac{1}{p^{1/d}}=\infty$$