Exercise A0.V.2.25

Proposition

Fix a non-constant polynomial $f \in \mathbb{Z}[x]$ and let $S$ be the image set of $f.$ Then the set of primes dividing at least one element of $S$ is infinite.

Proof

Suppose that $P$ is a finite set of primes, each dividing at least one element of $S.$ Let $q = \prod P.$

Suppose $f(0) = 0.$ If $f(x) = ax^n$ for some $a \in \mathbb{Z} \setminus \{0\}$ and $n \in \mathbb{Z}_{\geq 1}$ then each prime $p$ divides $f(p),$ contradicting finiteness of $P$. Suppose otherwise, that there exists a polynomial $g \in \mathbb{Z}[x]$ such that $f(x) = ax^ng(x)$ and $g(0) \neq 0.$ In that case, it would be sufficient to show that our proposition holds when $f(0) \neq 0.$

Suppose $f(0) = c$ and $c \neq 0.$ In this case, there exists a polynomial $g(x)$ such that $f(x) = g(x)x + c$ and

$$f(cx) = g(cx) \cdot cx + c = c \cdot (g(cx)x + 1).$$

In this case, it would be sufficient to show that our proposition holds when $f(0) = 1.$

Suppose $f(x) = g(x)\cdot x + 1$ for some $g \in \mathbb{Z}[x].$

Let $k \in \mathbb{Z}$ be such that $|g(kq)\cdot kq| > 10.$ Since $f$ is a non-constant polynomial, such $k$ always exists. Then $f(kq) = g(kq)\cdot kq + 1.$ In this case, $z = g(kq) \cdot kq + 1$ is not divisible by any $p \in P,$ as $z \equiv 1 \pmod p.$ So $z$ is either a prime or has a proper prime factor not present in $P.$ ⚠️

$\blacksquare$