NMO Selection Tests for the Junior Balkan Mathematical Olympiad

The given equality rewrites as $p^2+q^2-r^2=7(p+q-r)$. Since $(p+q-r)(p+q-r)=p^2+q^2-r^2+2pq$, it follows that $p+q-r$ divides $2pq$. If $p$, $q$, $r>2$, then $p+q-r$ is odd, so $p+q-r=p$, $q$ or $pq$. The first case gives $r=q$, then $p=7$, so $(p,q,r)=(7,q,q)$, $q$ being an arbitrary prime. Likewise, the second case gives $(p,q,r)=(p,7,p)$, $p$ being an arbitrary prime. If $p+q-r=pq$, then $1-r=(p-1)(q-1)$, impossible. If $p=2$ (or $q=2$) the equality becomes $(q-r)(q+r-7)=10$, implying $q=7$ and $r=5$, with two new solutions obtained, $(p,q,r)=(2,7,5)$ and $(p,q,r)=(7,2,5)$, or $r=2$, with no further solutions obtained. The solution triplets therefore are $(2,7,5)$, $(7,2,5)$, $(7,t,t)$, $(t,7,t)$, where $t$ is an arbitrary prime. (Notice the fact $r$ was known to be a prime turned out to be inconsequential).