## Task 11/88

Determine all natural numbers $n$ for which $n^{4}+4$ is a prime number!
One quickly recognizes that for $n=1$ the prime number 5 is obtained: $1^{4}+4=5$. Now it holds that

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

If $p$ is to be a prime number, the smaller of the two factors must be equal to 1: $\left(n^{2}+2-2 n\right)=1$ and $\left(n^{2}-2 n+1\right)=(n-1)^{2}=0$, i.e., $n=1$. Therefore, $n=1$ is the only solution.