Prove that $n^4+4$ is composite

Question

Prove that for all integers $n>1$, $n^4+4$ is composite.

I got $n^4+4=(n^2-2n+2)(n^2+2n+2)$ What should be my next step?

Answer 1

Your inputs are $n\in\Bbb Z, n>1$

Note for $n>1, n\in\Bbb Z, (n^2-2n+2)\ne(n^2+2n+2)$, so there's no way it's a square number,

and that for $n\in \Bbb Z, n^2-2n+2\in\Bbb Z$ and $n^2+2n+2\in \Bbb Z$. So you have two integer factors, both of which are $>1$ for $n>1$, and thus your $n^4+4$ is composite

Answer 2

If you don’t now the factorization of $X^4+4$ as a $\Bbb Z$-polynomial, you’re in good company: Leibniz didn’t, either (or the similar real factorization of $X^4+1$).

This should be taught in high schools: $$ X^4+4=(X^2-2X+2)(X^2+2X+2) $$

Answer 3

Clearly, $n^4+4=(n^2-2n+2)(n^2+2n+2)$. Moreover, $n^2-2n+2=(n-1)^2+1$ and $n^2+2n+2=(n+1)^2+1$; both these integers are greater than $1$ as $n>1$. Thus, $n^4+4$ is a composite number whenever $n>1$.