(factorisation) Find all natural numbers $n$ such that $n^{4}+4^{n}$ is prime.
For $n \geqslant 1, n^{4}+4^{n}=n^{4}+4 \cdot\left(2^{n-1}\right)^{2}=\left(n^{2}-n 2^{n}+2 \cdot 4^{n-1}\right)\left(n^{2}+n 2^{n}+2 \cdot 4^{n-1}\right)$ Thus, since $n^{2}+n 2^{n}+2 \cdot 4^{n-1} \geqslant 2$, if $n^{4}+4^{n}$ is prime, then $n^{2}+n 2^{n}+2=n^{4}+4^{n}$, and $n^{2}-n 2^{n}+2 \cdot 4^{n-1}=1$. Hence $\left(n-2^{n-1}\right)^{2}+4^{n-1}=1$, but $\left(n-2^{n-1}\right)^{2} \geqslant 0$, so since $4^{n-1} \geqslant 1$, it is necessarily true that $n=1$. And for $n=1$, we indeed have $n^{4}+4^{n}=5$ which is prime. And for $n=0, n^{4}+4^{n}=1$ and is therefore not prime.