4. [5 points] Solve the system of equations

$$
\left\{\begin{array}{l}
\frac{2}{x^{2}+y^{2}}+x^{2} y^{2}=2 \\
x^{4}+y^{4}+3 x^{2} y^{2}=5
\end{array}\right.
$$
Answer: $(1 ; \pm 1),(-1 ; \pm 1)$.

Solution. Introduce new variables $u=x^{2}+y^{2}, v=x^{2} y^{2}$. Then $x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=u^{2}-2 v$ and the system takes the form

$$
\left\{\begin{array}{l}
\frac{2}{u}+v=2 \\
u^{2}+v=5
\end{array}\right.
$$

From the first equation $v=2-\frac{2}{u}$; substitute this expression into the second equation and solve it:

$$
\begin{aligned}
u^{2}+\left(2-\frac{2}{u}\right)=5 \Leftrightarrow u^{2}-3-\frac{2}{u}=0 & \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ u ^ { 3 } - 3 u - 2 = 0 , } \\
{ u \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ ( u + 1 ) ^ { 2 } ( u - 2 ) = 0 , } \\
{ u \neq 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
u=-1 \\
u=2
\end{array}\right.\right.\right.
\end{aligned}
$$

Since $u>0$, only the value $u=2$ is suitable. Therefore, $v=1$, and in the original variables we get

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 2 , } \\
{ x ^ { 2 } y ^ { 2 } = 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}=2-y^{2}, \\
\left(2-y^{2}\right) y^{2}=1
\end{array} \Leftrightarrow x^{2}=y^{2}=1\right.\right.
$$

From this, it follows that the system has 4 solutions: $(1 ; \pm 1),(-1 ; \pm 1)$.