Find all pairs of primes $(p, q)$ such that

$$
p^{3}-q^{5}=(p+q)^{2} .
$$
Assume first that neither of the numbers equals 3. Then, if $p \equiv q \bmod 3$, the left hand side is divisible by 3 , but the right hand side is not. But if $p \equiv-q \bmod 3$, the left hand side is not divisible by 3 , while the right hand side is. So this is not possible.

If $p=3$, then $q^{5}3$ then $p^{2}-p-6$ is positive and increases with $p$. So the equation has at most one solution. It is easy to see that $p=7$ is the one and $(7,3)$ is a solution to the given equation.