* Find all prime numbers $p, q$ and $r$ that satisfy the following equation: $p^{q}+q^{p}=r$.


Assume $p<q$. If $r$ is to be a prime number, then $p$ and $q$ cannot both be odd, so we can set $p=2$. Furthermore, when $q$ is a prime number not less than 5, we have
$$
2^{q}+q^{2}=\left(2^{q}+1\right)+\left(q^{2}-1\right)=3\left(2^{q-1}-2^{q-2}+2^{q-3}-\cdots+1\right)+(q-1)(q+1)
$$

which is a multiple of 3, and thus cannot be a prime number. When $q=3$, we get $r=17$. Therefore, the prime number solutions $p$, $q$, and $r$ of the original equation are:
$$
p=2, q=3, r=17 ; p=3, q=2, r=17 \text {. }
$$