4. [5 points] Solve the system of equations

$$
\left\{\begin{array}{l}
\frac{4}{x^{2}+y^{2}}+x^{2} y^{2}=5 \\
x^{4}+y^{4}+3 x^{2} y^{2}=20
\end{array}\right.
$$
Answer: $(\sqrt{2} ; \pm \sqrt{2}),(-\sqrt{2} ; \pm \sqrt{2})$.

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

$$
\left\{\begin{array}{l}
\frac{4}{u}+v=5 \\
u^{2}+v=20
\end{array}\right.
$$

From the first equation $v=5-\frac{4}{u}$; substitute this expression into the second equation and solve it:

$$
\begin{aligned}
& u^{2}+\left(5-\frac{4}{u}\right)=20 \Leftrightarrow u^{2}-15-\frac{4}{u}=0 \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ u ^ { 3 } - 1 5 u - 4 = 0 , } \\
{ u \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ ( u - 4 ) ( u ^ { 2 } + 4 u + 1 ) = 0 , } \\
{ u \neq 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
u=4, \\
u=-2 \pm \sqrt{3}
\end{array}\right.\right.\right.
\end{aligned}
$$

Since $u>0$, only the value $u=4$ is suitable. Therefore, $v=4$, and in the original variables we get

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 4 , } \\
{ x ^ { 2 } y ^ { 2 } = 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}=4-y^{2}, \\
\left(4-y^{2}\right) y^{2}=4
\end{array} \Leftrightarrow x^{2}=y^{2}=2\right.\right.
$$

From this, it follows that the system has 4 solutions: $(\sqrt{2} ; \pm \sqrt{2}),(-\sqrt{2} ; \pm \sqrt{2})$.