808. Solve the equation in integers

$$
(x+y)^{2}=x-y
$$
$\triangle$ It is hardly expedient to expand the expression $(x+y)^{2}$; instead, this expression should be utilized.

Let $x+y=t$, where $t$ is some integer. Then $x-y=t^{2}$.

We obtain the system of equations:

$$
\left\{\begin{array}{l}
x+y=t \\
x-y=t^{2}
\end{array}\right.
$$

By adding and subtracting the equations, we express $x$ and $y$ in terms of $t$:

$$
x=\frac{t(1+t)}{2}, \quad y=\frac{t(1-t)}{2}
$$

Can $t$ take any integer values here? First of all, the obtained fractions are equal to integers since their numerators are even. Moreover, if we substitute these values of $x$ and $y$ into the original equations, we get an identity. Therefore, they define the set of all solutions to the equation. Since $t$ is any integer, the equation has an infinite number of solutions in integers.

Answer: $\left[\frac{t(1+t)}{2} ; \frac{t(1-t)}{2}\right), t \in \mathrm{Z}$.