Estonian Math Competitions

Answer: $(\frac{z^3(z-1)}{2},\frac{z^6(z-1)}{2})$ where $z$ is arbitrary integer.

Solution: If $x=0$ then $y=0$. We now assume that $x\ne 0$. Let $d=\gcd (x,y)>0$ and $x=da,y=db$. Dividing the sides of the equation by $d^4$, we get $b^4=a(2a^{2}d+b{)}^3$. Thus $a\mid b^4$. Since $a$ and $b$ are coprime, the only possibility is $|a|=1$ and the equation reduces to $$b^4=\pm (2d+b{)}^3.$$ As the same integer is simultaneously a cube and a fourth power, it is a twelfth power of some integer, yielding $b=z^3$ for some integer $z$. Substituting $b=z^3$ into the equation and taking cube root gives $z^4=\pm (2d+z^3)$ or, equivalently, $\pm z^4=2d+z^3$. Since $-z^4\le -|z^3|\le z^3<2d+z^3$, the minus sign is not possible, therefore $d=\frac{z^4-z^3}{2}=\frac{z^3(z-1)}{2}$. We conclude that $x=\frac{z^3(z-1)}{2}$ and $y=\frac{z^6(z-1)}{2}$, where $z$ is an arbitrary integer different from $0$ and $1$. In this case $d>0$ and the solution satisfies the original equation. If $z=0$ or $z=1$, we get the initial solution $(0,0)$.