2. Find all pairs of real numbers $x, y$, satisfying the equation

$$
\frac{2 x}{1+x^{2}}=\frac{1+y^{2}}{2 y}
$$
Solution. Transform the given equation to the form: $4 x y=\left(1+y^{2}\right)\left(1+x^{2}\right)$, which is equivalent to $1-2 x y+x^{2} y^{2}+x^{2}-2 x y+y^{2}=0$ or $(1-x y)^{2}+(x-y)^2=0$. From this, it follows that $x=y$ and $x y=1$, that is, the solutions to the given equation are two pairs: $x=y=1$ and $x=y=-1$.

Evaluation Criteria. Found one of the two pairs without further progress - 0 points. Found both pairs - at least 1 point. Obtained the equation $(1-x y)^{2}+(x-y)^{2}=0$, but solved it incorrectly -4 points.