Solve the following system of equations:

$$
\begin{aligned}
& x+y=x^{2}+2 x y+y^{2} \\
& x-y=x^{2}-2 x y+y^{2}
\end{aligned}
$$
# Solution.

$x+y=x^{2}+2xy+y^{2}$, which can be written as $x+y=(x+y)^{2}$.

$x-y=x^{2}-2xy+y^{2}$, which can be written as $x-y=(x-y)^{2}$.

Since both $x+y$ and $x-y$ are equal to their own squares, their values can independently only be 1 or 0.

If $x+y=0$ and $x-y=1$, then: $x=0.5, y=-0.5$.

If $x+y=0$ and $x-y=0$, then: $x=0, y=0$.

If $x+y=1$ and $x-y=1$, then: $x=1, y=0$.

If $x+y=1$ and $x-y=0$, then: $x=0.5, y=0.5$.

Therefore, the solutions are: $x_{1}=0.5, y_{1}=-0.5 ; x_{2}=0, y_{2}=0 ; x_{3}=1, y_{3}=0 ; x_{4}=0.5, y_{4}=0.5$.