2. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
\left(x^{2}+1\right)\left(y^{2}+1\right)+24 x y & =0 \\
\frac{12 x}{x^{2}+1}+\frac{12 y}{y^{2}+1}+1 & =0
\end{aligned}
$$
2. Since for any real numbers $x, y$ both numbers $\left(x^{2}+1\right)$ and $\left(y^{2}+1\right)$ are non-zero (indeed positive), we can transition to new variables

$$
u=\frac{x}{x^{2}+1} \quad \text { and } \quad v=\frac{y}{y^{2}+1}
$$

in which the original system of equations apparently takes the form

$$
1+24 u v=0 \quad \text { and } \quad 12 u+12 v+1=0 \text {. }
$$

From this, for the unknown $u$ we easily obtain the quadratic equation

$$
24 u^{2}+2 u-1=0
$$

with roots $u_{1}=\frac{1}{6}$ and $u_{2}=-\frac{1}{4}$, to which "symmetrically" correspond the values $v_{1}=-\frac{1}{4}$ and $v_{2}=\frac{1}{6}$. Since (quadratic) equations

$$
\frac{s}{s^{2}+1}=\frac{1}{6} \quad \text { and } \quad \frac{t}{t^{2}+1}=-\frac{1}{4}
$$

have solutions

$$
s_{1,2}=3 \pm \sqrt{8} \quad \text { and } \quad t_{1,2}=-2 \pm \sqrt{3}
$$

the original system has exactly eight solutions, namely pairs of the form $(x, y)=(3 \pm \sqrt{8},-2 \pm \sqrt{3})$ and $(x, y)=(-2 \pm \sqrt{3}, 3 \pm \sqrt{8})$, where the signs are combined arbitrarily.

Award 6 points for a complete solution.