Compositeness of $n^4+4^n$ My coach said that for all positive integers $n$, $n^4+4^n$ is never a prime number.
So we memorized this for future use in math competition.  But I don't understand why is it?

You can work $\bmod 5$:
As Jossie said, if $n$ is even, then both numbers are even. If $n$ is odd, set $n = 5k + r$;
If not, you can repeatedly use the fact that for $p$ a prime and $(a, p) = 1, a^{p - 1} = 1 \pmod p$ and so $a^p = a \pmod p$;
in this case, $(a, 5) = 1$ , then $a^{4n} = 1 \pmod 5$
$0 \leq r <5$ . Then 
$4^n + n^4 = 4^{5k + r} + r^4 \pmod 5 = 4^{5k} 4^r + r^4 = 4^{r + 1} + 1 = 4$. $4^r + 1 = 4 + 1 = 5 \pmod 5$.
