59. Solve the equation $x-y=x^{2}+x y+y^{2}$ in integers.
Solution. We rewrite this as a quadratic equation in terms of $x: x^{2}+(y-1) x+y^{2}+y=0$. The discriminant is $1-3 y(y+2)$. It is clear that for $y \geqslant 1$ and $y \leqslant-3$ this number is negative, meaning there are no solutions. For $y=0$ we get $x=0$ or $x=1$. For $y=-1$ we get $x=0$ or $x=2$. Finally, for $y=-2$ we get $x=1$ or $x=2$. Thus, the answer is six pairs: $(0,0),(1,0),(0,-1),(2,-1),(1,-2)$ and $(2,-2)$.