prove that if $f(x) \in \mathbb{Z}[x]$ and $f$ is non-constant, that for infinitely many $n, f(n)$ is not prime prove that if $f(x) \in \mathbb{Z}[x]$, that for infinitely many $n, f(n)$ is not prime.
I tried looking at the polynomial of that form $f(x) = \sum_{i\in \mathbb{N}}  a_ix^i$. I am kinda stuck her though, is it helpfull to look at the roots?
Kes

"No non-constant polynomial function $P(n)$ exists that evaluates to a prime number for almost all integers $n$", says Formula for primes in Wikipedia, with a sketch of the proof.
For any integer $k$, $f(kf(0))  = 0\mod f(0)$. If $f(0)$ is not a prime, we have our first composite. If $f(0)$ is prime, then  $f(kf(0))$ is prime only if  $f(kf(0)) =  f(0)$. A polynomial that takes the same value an infinite number of times is constant, and ours is not by hypothesis.  So at least one $f(kf(0))$ is composite. You can repeat the process with $f(kf(n))  = 0\mod f(n)$.
