Determine for what numbers $n$ the number $n^4 + 4$ is a composite number.
Sorry about my English. I found $n^4 + 4 = (n^2 + 2n +2)(n^2 -2n + 2)$, but i don't know what to do from here.
You did the non-obvious part, and are now essentially finished! Determine the $n$ for which one of your terms could be $\pm 1$. Not many! (And neither can be $-1$.) For all other $n$, you will have a non-trivial factorization. It may be useful to note that $n^2-2n+2=(n-1)^2+1$, with something similar for the other one.
Added: It turns out that in effect the OP wondered whether $n^4+4$ can be a prime power. Except when $n=0$, it cannot.
If a prime $p \gt 2$ divides both $n^2-2n+2$ and $n^2+2n+2$, then $p$ divides $n$, but then $p$ divides $2$, contradiction. So the only possibilities are $n^2+2n+2$, $n^2-2n+2$ both a non-trivial power of $2$, Then $n$ has to be even. It follows that each of $n^2-2n+2$ and $n^2+2n+2$ is congruent to $2$ modulo $4$, so if each is a power of $2$, each must be $2$, giving $n=0$.