Determine all prime numbers $p, q$ and $r$ such that $p+q^{2}=r^{4}$.

(Karl Czakler)

Answer. The only solution is $p=7, q=3, r=2$.
The given equation can be written in the form $p=r^{4}-q^{2}=\left(r^{2}-q\right)\left(r^{2}+q\right)$. Since $p$ is a prime number, it follows that $r^{2}-q=1$ and $r^{2}+q=p$. From $r^{2}-q=1$ we get $q=r^{2}-1=(r-1)(r+1)$. Since $q$ is a prime number, it follows that $r-1=1$, hence $r=2$ and $q=3$. With $r^{2}+q=p$ we obtain $p=7$.

(Karl Czakler)