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?

In order to get a prime, we need $n$ odd. So $4^n=4\cdot 4^{2k}$ for some $k$, and therefore $4^n=4\cdot (2^k)^4$ where $k=\frac{n-1}{2}$. 
Now use the factorization
$$x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2),$$
with $x=n$ and $y=2^{(n-1)/2}$. 
The case $n=1$ gives the lone prime. For all other $n$, we have $x^2-2xy+2y^2\gt 1$.
Remark: It is hard to judge whether the above factorization is "natural." Perhaps it will look more reasonable if we express $x^4+4y^4$ as a difference of squares:
$$x^4+4y^4=(x^2+2y^2)^2-4x^2y^2.$$
