12.9. Solve the equation in integers

$$
x+y=x^{2}-x y+y^{2}
$$
12.9. Answer: $(0,0),(0,1),(1,0),(1,2),(2,1),(2,2)$. Let's consider the given equation as a quadratic equation in terms of $x$:

$$
x^{2}-(y+1) x+y^{2}-y=0 \text {. }
$$

The discriminant of this equation is $-3 y^{2}+6 y+1$. It is negative for $y \geqslant 3$ and for $y \leqslant-1$. Therefore, for $y$ we get three possible values: $0,1,2$. For each of these values, we get an equation that is easily solved.