Polynomial is always composite for all integers x.

Question

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!

Also: https://math.stackexchange.com/q/304330/42969, https://math.stackexchange.com/q/817062/42969, https://math.stackexchange.com/q/2416348/42969 — Martin R, Oct 14 '19 at 07:06

score 0 · Answer 1 · answered Oct 14 '19 at 07:24

0

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.

answered Oct 14 '19 at 07:24

Kaligule

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Is that different from https://math.stackexchange.com/a/193804/42969 (the answer to the first duplicate target)? – Martin R Oct 14 '19 at 07:41
@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 '19 at 10:08

Polynomial is always composite for all integers x.

1 Answers1