Let $P$.

Find all such prime numbers $p$ and $q$ that $p+q=(p-q)^{3}$.
Let $p-q=n$, then $p+q=n^{3}$. From this, $q=1 / 2\left(n^{3}-n\right)=1 / 2(n-1) n(n+1)$.

Among three consecutive integers, one is divisible by 3, so $q$ is divisible by 3. Therefore, $q=3$. This

## Answer

$p=5, q=3$.