For which integers $n$ is the number $n^{4}+4$ composite?

#
$n^{4}+4=n^{4}+4 n^{2}+4-4 n^{2}$.

## Solution

$n^{4}+4=n^{4}+4 n^{2}+4-4 n^{2}=\left(n^{2}+2\right)^{2}-(2 n)^{2}=\left(n^{2}-2 n+2\right)\left(n^{2}+2 n+2\right)$.

The numbers $n^{2}+2 n+2=(n+1)^{2}+1$ and $n^{2}-2 n+2=(n-1)^{2}+1$ are greater than 1 for $n \neq \pm 1$. For $n= \pm 1 \quad n^{4}+4=5$.

## Answer

For $n \neq \pm 1$.