$p \ne q$, so let $p<q$ $(p+q)^2-pq=r^2$ $pq=(p+q-r)(p+q+r)$ Since $p,q$ are primes, $pq$ has exactly 4 divisors: $1,p,q,pq$ and $p+q+r>q$. So, $p+q-r=1,\;p+q+r=pq$ $2p+2q-1=pq$ $(p-2)(q-2)=3$ $p=3,q=5,r=7$