# Polynomial is always composite for all integers x. [duplicate]

Let f(x) be a nonconstant polynomial with integer coefficients. Prove that there is some integer n such that |f(n)| is composite.

I'm just a little confused on how to begin. I am fairly certain we start by assuming that f(x) outputs only primes for all integers x, but I don't know where to go from there. Any help would be greatly appreciated.

Thanks!

## marked as duplicate by Martin R, Larry, Vinyl_cape_jawa, José Carlos Santos, Arnaud D.Oct 14 at 15:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• – Martin R Oct 14 at 7:06

## 1 Answer

Since you are only looking for a strategy to start with, not a solution:

$$p:=f(0)$$ and by assumption $$p$$ is prime. Now look at the values of $$f(k\cdot p)$$ and see if you can proove they are not prime.

• Is that different from math.stackexchange.com/a/193804/42969 (the answer to the first duplicate target)? – Martin R Oct 14 at 7:41
• @MartinR It is the same solution, but since OP only asked for an idea, not a complete proof, I figured this would be more what he needs. – Kaligule Oct 14 at 10:08