Saudi Arabian Mathematical Competitions

From $3 \mid x^{2}+y^{2}$ it follows $3 \mid x$ and $3 \mid y$, hence $x=3u$ and $y=3v$, for some positive integers $u$ and $v$. Replace in the equation and get $3\left(u^{2}+v^{2}+11^{2}\right)=670 \sqrt{3(u-v)}$, hence $u-v=3k^{2}$, for some positive integer $k$, i.e. $u^{2}+v^{2}+11^{2}=670k$. We have $u^{2}+v^{2}>(u-v)^{2}=9k^{4}$, and obtain $9k^{4}+121<670k$, hence $k<5$.

Case 1. If $k=2$ or $k=4$, then $u^{2}+v^{2}+11^{2} \equiv 0\pmod{4}$, not possible since we have $11^{2} \equiv 1\pmod{4}$ and $u^{2}, v^{2} \equiv 0$ or $1\pmod{4}$.

Case 2. If $k=1$, then we get the system
$$\left\{\begin{array}{l} u^{2}+v^{2}=549 \\ u-v=3 \end{array}\right.$$
having integer solutions $u=18$ and $v=15$. In this case it follows $x=54$ and $y=45$.

Case 3. If $k=3$, then we get the system
$$\left\{\begin{array}{l} u^{2}+v^{2}=1882 \\ u-v=27 \end{array}\right.$$
having no solutions in integers.