Estonian Mathematical Olympiad

Suppose that both $p$ and $q$ are odd. Then $p^2+q^2$ is even and the l.h.s. of the equation is divisible by 4. Squares of integers are congruent to 0 or 1 modulo 4 whence the r.h.s. is congruent to 1 or 2 modulo 4. The contradiction shows that one of $p$ and $q$ equals 2; let w.l.o.g. $p=2$. Squares of integers are congruent to 0 or 1 modulo 3. Obviously $r>3$ as the l.h.s. is greater than 10. As $r$ is prime, $r$ is not divisible by 3. Thus $r^2\equiv 1(\mathrm{mod}3)$, whence $r^2+1\equiv 2(\mathrm{mod}3)$. Now $2018\equiv 2(\mathrm{mod}3)$ implies $p^2+q^2\equiv 1(\mathrm{mod}3)$ and $p^2=4\equiv 1(\mathrm{mod}3)$ in turn implies $q^2\equiv 0(\mathrm{mod}3)$. Hence $q$ is divisible by 3, i.e., $q=3$. Therefore the l.h.s. is $2018\cdot 13$. As $2018\cdot 13\equiv 3\cdot 3=9\equiv 4(\mathrm{mod}5)$, we must have $r^2\equiv 3(\mathrm{mod}5)$. But 3 is not a quadratic residue modulo 5.