4. [5 points] Solve the system of equations

$$
\left\{\begin{array}{l}
\frac{1}{x^{2}+y^{2}}+x^{2} y^{2}=\frac{5}{4} \\
2 x^{4}+2 y^{4}+5 x^{2} y^{2}=\frac{9}{4}
\end{array}\right.
$$
Answer: $\left(\frac{1}{\sqrt{2}} ; \pm \frac{1}{\sqrt{2}}\right),\left(-\frac{1}{\sqrt{2}} ; \pm \frac{1}{\sqrt{2}}\right)$.

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

$$
\left\{\begin{array}{l}
\frac{1}{u}+v=\frac{5}{4} \\
2 u^{2}+v=\frac{9}{4}
\end{array}\right.
$$

From the first equation $v=\frac{5}{4}-\frac{1}{u}$; substitute this expression into the second equation and solve it:

$$
\begin{aligned}
2 u^{2}+\left(\frac{5}{4}-\frac{1}{u}\right)=\frac{9}{4} \Leftrightarrow 2 u^{2}-1-\frac{1}{u}=0 & \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 2 u ^ { 3 } - u - 1 = 0 , } \\
{ u \neq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(u-1)\left(2 u^{2}+2 u+1\right)=0, \\
u \neq 0
\end{array} \quad \Leftrightarrow u=1\right.\right.
\end{aligned}
$$

Then we find that $v=\frac{1}{4}$, and in the original variables we get

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 1 , } \\
{ x ^ { 2 } y ^ { 2 } = \frac { 1 } { 4 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}=1-y^{2} \\
\left(1-y^{2}\right) y^{2}=\frac{1}{4}
\end{array} \Leftrightarrow x^{2}=y^{2}=\frac{1}{2}\right.\right.
$$

From this, it follows that the system has 4 solutions: $\left(\frac{1}{\sqrt{2}} ; \pm \frac{1}{\sqrt{2}}\right),\left(-\frac{1}{\sqrt{2}} ; \pm \frac{1}{\sqrt{2}}\right)$.