Find integer solution $|x^2-y|=8|y^2-1| (1) $ I try to: If $y^2-1=0$ we have $y\in\{1,-1\}$ With $y=1\Longrightarrow x=\pm 1$ $y=-1\Longrightarrow x^2+1=0$ not satisfied. In other case: $\left| \dfrac{x^2-y}{y^2-1}\right|=8$.
But I can't to solve it.
As pointed out by several commenters, because of the absolute values, there are only two possible cases.
Case 1: $x^2-y=8(1-y^2)$. Then $8y^2-y=8-x^2$. The left-hand side is greater than or equal to $7$ for all $y ≠ 0$, while the right-hand side is less than or equal to $8$ for all $x$. The only possible equalities are $7$ or $8$, and it is easily seen that $(x,y)=(\pm 1,1)$ is the only solution.
Case 2: $x^2-y=8(y^2-1)$. This has an infinite number of solutions, which can be found using standard quadratic methods.