SPANISH MATHEMATICAL OLYMPIAD (FINAL ROUND)

First note that if $xy=0$, then $x^2+y^2=0$ and $x=y=0$ is a solution. Also note that if $xy<0$, then $x^2+y^2<0$, impossible; thus $x,y$ are both positive or both negative. Changing simultaneously the sign of $x$ and $y$ does not change the equation, hence we may assume without loss of generality that $x,y$ are both positive. Let $m,n$ ($m\ge n$) be non-negative integers, and let $a,b$ be coprimes integers, not divisible by $3$. Consider $x=3^{m}a$ and $y=3^{n}b$. Then the given equation is $$8((3^{m-n}a{)}^2+b^2)=3^{3m+n-4}{a}^3{b}^3.$$ Because any perfect square has remainder $0$ or $1$ when divided by $3$, the left hand side member is not divisible by $3$, thus looking at the right hand side we get $3m+n-4=0$, and because $m\ge n\ge 0$ it follows $m=n=1$. The equation takes the form $$8(a^2+b^2)=a^3{b}^3.$$ By symmetry we may assume $a\ge b$. Then $$16a^2\ge a^3{b}^3\iff 16\ge a{b}^3$$ Hence we have either (1) $b=2$ which implies $a=2$, or (2) $b=1$, but in this case the only possible values of $a$ are $1,2,4,8$ and none of them satisfies the equation $a^3-8a^2-8=0$. It is easy to see that $(a,b)=(2,2)$ and $(x,y)=(6,6)$. So, the only solutions are $$(x,y)=(-6,-6),(x,y)=(0,0),(x,y)=(6,6).$$