8.5. Find all integer solutions \(x\) and \(y\) for which \(x^{2} + xy - y = 2\). Justify your answer.
8.5. Answer: $(2,-2),(0,-2)$

Let's transform the original equation as follows:

$$
\begin{gathered}
x^{2}+x y-y=2 \\
x^{2}-1+y(x-1)=1 \\
(x-1)(x+1)+y(x-1)=1 \\
(x-1)(x+1+y)=1
\end{gathered}
$$

The product of two integers equals 1 if both numbers are 1 or if both numbers are -1.

In the first case, we have the solution $(2,-2)$, and in the second case, $(0,-2)$.

## Comments on Evaluation.

Full answer only: 1 point. The expression $(x-1)(x+1+y)$ is obtained, but there is no further progress: 3 points. The expression $(x-1)(x+1+y)$ is obtained, but only one answer is found: 4 points.