Let $f(x) \in \mathbb{Z}[X]$ such that the sequence $(f(n))_{n\in\mathbb{N}}$ contains prime numbers infinitely often. Then
$f(x)$ is irreducible.
The leading coefficient of $f$ is positive.
The set $\{f(n):n \in \mathbb{Z}\}$ has no common divisor $>1. \ (\Rightarrow$ The coefficients of $f(x)$ should be relatively prime.)
My Thoughts:
- Let $f(x)=g(x)h(x)$ with $g(x)$ and $h(x)$ in $\mathbb{Z}[X]$ of positive degree. Now we note that
$f(x)$ takes prime values infinitely often $\implies$ Either $g(x)$ or $h(x)$ takes the value $\pm 1$ infinitely often. WLOG let
$g(x)$ be the polynomial taking the values $\pm 1$ infinitely often $\implies$ $g(x)$ takes at least one of $1$ or $-1$ infinitely often, let's say it takes $1$ infinitely often.
Then consider a polynomial $G(x)=g(x)-1$. From the above arguments $G(x)$ has infinitely many zeroes in $\mathbb{N}$. Since $\operatorname{deg}(G)=\operatorname{deg}(g)$ which is finite and $\mathbb{Z}$ is an integral domain, we can tell $G$ contains atmost $\operatorname{deg}G$ many zeroes. Contradiction!
Hence $f(x) \ne g(x)h(x)$ with $g(x)$ and $h(x)$ of positive degree $\implies f(x)$ is irreducible in $\mathbb{Z}[X]$.
How do I prove this part?
If $\{f(n):n\in\mathbb{Z}\}$ is an infinite set. What is the definition of $\operatorname{gcd}$ of infinitely many numbers? How do I show that the set $\{f(n):n \in \mathbb{N}\}$ has gcd $1$? Also, How to obtain that the coefficients of $f(x)$ are relatively prime?
