Pairs $(0,1)$ and $(-1,0)$ are obvious solutions Let $ab \neq 0$ then one of the fractions $\frac{a^2}{|a+1|}$ or $\frac{b^2}{|b-1|}$ is less than $1$ Case 1: $a^2 \leq |a+1| \to a=1 \to 2(b-1)=b^2$ but there are no solutions Case 2: $b^2 \leq |b-1| \to b=-1 \to -2(a+1)=a^2$ but there are no solutions