Poly. in integer coeff. takes infinitely many integers to composites.

[caffeinemachine]

Let $f(x)$ be a polynomial with integer coefficients. Show that $f(n)$ is composite for infinitely many integers $n$.

EDIT: As Bacterius has pointed out we need to assume that $f(x)$ is a non-constant polynomial.

 

[Bacterius]

Spoiler

[JUSTIFY]Refuted: $f(n) = 3$. But let's assume a non-constant polynomial

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Lemma 1

If $f$ is an integer polynomial (of degree $s$) then for all $m$, $p$, $k$:
$$f(m) \equiv f(m + kp) \pmod{p}$$

Proof
$$f(m + kp) \equiv a_0 + a_1 (m + kp) + a_2 (m + kp)^2 + \cdots + a_s (m + kp)^s \equiv a_0 + a_1 m + a_2 m^2 + \cdots + a_s m^s \equiv f(m) \pmod{p}$$

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Lemma 2

If an integer polynomial $f$ satisfies the property that $f(m) = f(m + kp)$ for all $k$, some $m$ and some $p > 0$, $f$ must be constant.

Proof

If such an $f$ was not constant, its graph would have to intersect the line $y = f(m)$ infinitely many times, which is clearly absurd as a polynomial must have finitely many roots.

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Lemma 3

Any non-constant integer polynomial $f(n)$ must be composite for at least one $n$. In other words, $f(n)$ cannot be prime for all $n$.

Proof

Assume $f(n)$ generates only primes, that is, $f(n)$ is prime for all $n$. Then $f(1) = p$ for some prime $p$, so $f(1) \equiv 0 \pmod{p}$. Thus $f(1 + kp) \equiv 0 \pmod{p}$ for all integers $k$ by Lemma 1. But clearly this implies $f(1 + kp) = p$, otherwise $f(1 + kp)$ would be a multiple of $p$ (hence, a composite, contradicting the initial assumption). We are left with the implication that $f(1) = f(1 + kp)$ for all integers $k$, hence $f$ must be constant by Lemma 2, a contradiction.

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Theorem

Let $f$ be a non-constant integer polynomial. Then $f(n)$ is composite for infinitely many $n$.

Proof

There exists at least one $m$ such that $f(m)$ is composite by Lemma 3, let $f(m) = c$. Then $f(m) \equiv 0 \pmod{c}$. This implies $f(m + kc) \equiv 0 \pmod{c}$ for all integers $k$ by Lemma 1. Hence $f$ generates infinitely many integers divisible by $c$. Those integers must be composite since $c$ is composite. $\blacksquare$

[HR][/HR]

A less rigorous version of the argument is that if $f(m)$ is composite, then so is $f(m + k \cdot f(m))$ for all integers $k$.[/JUSTIFY]