Konigsberg wrote: Find all integer solutions to $y^2(x^2+y^2-2xy-x-y)=(x+y)^2(x-y)$. $y=0$ implies $x^3=0$ and so the solution $(x,y)=(0,0)$ $x=0$ implies $y^2(y^2+y-1)=0$ and so $y=0$ and no new solution Let us from now consider $x,y\ne 0$ Let $x=au$ and $y=av$ with $\gcd(|u|,|v|)=1$ and $v>0$ and $a\ne 0$ Equation may be written $av^2(u-v)^2-v^2(u+v)=(u+v)^2(u-v)$ If $u=v$ or $u=-v$, this would imply $u=v=0$, impossible. So $|u|\ne |v|$ $u-v|v^2(u+v)$ and so $u-v|u+v$ and so $u-v|2u$ and so $u-v\in\{-2,-1,1,2\}$ If $u-v=2$, equation becomes $2av^2-v^2(v+1)=4(v+1)^2$ and so $v^2|4(v+1)^2$ and so $v\in\{1,2\}$ If $v=1$, we get then $a=9$ and the solution $(x,y)=(27,9)$ If $v=2$, we get $u=4$, impossible since $\gcd(u,v)\ne 1$ If $u-v=1$, equation becomes $av^2-v^2(2v+1)=(2v+1)^2$ and so $v^2|(2v+1)^2$ and so $v=1$ and $a=12$ and the solution $(x,y)=(24,12)$ If $u-v=-1$, equation becomes $av^2+v^2=-(2v-1)^2$ and so $v^2|(2v+1)^2$ and so $v=1$, impossible since then $u=0$ If $u-v=-2$, equation becomes $2av^2-v^2(v-1)=-4(v-1)^2$ and so $v^2|4(v-1)^2$ and so $v\in\{1,2\}$ If $v=1$, we get then $a=0$, impossible. If $v=2$, we get then $u=0$, impossible. Hence the solutions : $\boxed{(x,y)\in\{(0,0),(27,9),(24,12)\}}$