2016/002) Let n be any positive integer $>1$ then show that $n^4+4^n$ is a composite number

there are 2 cases
1) n is even.
$n^4+4^n > 2$ and even so composite
2) n is odd
$n^4 + 4^n= n^4 + 2n^2*2^n + 4^n - 2n^2*2^n$
$=(n^2 + 2^n)^2 - 2n^2*2^n$
$=(n^2 + 2^n)^2 - n^2*2^{n+1}$
$= (n^2 + 2^n - n * 2^{\frac{n+1}{2}})(n^2 + 2^n + n * 2^{\frac{n+1}{2}})$
Therefore, if n is odd it has 2 factors so composite

So it is composite for any n