Romanian Mathematical Olympiad

The given relation is equivalent to $(p+q)^2 = r^2 + pq$, which can be written $(p+q+r)(p+q-r) = pq$.
The divisors of $pq$ are $1$, $p$, $q$ and $pq$. Since $p+q > \max\{p,q\}$, it follows that $p+q-r = 1$ and $p+q+r = pq$.

Adding the last two equalities yields $2p+2q = pq+1$, that is $(p-2)(q-2) = 3$.
This leads to $(p,q) \in \{(3,5), (5,3)\}$; in both cases $r=7$.