4. [4 points] Solve the system of equations

$$
\left\{\begin{array}{l}
2 x^{2}+2 y^{2}-x^{2} y^{2}=2 \\
x^{4}+y^{4}-\frac{1}{2} x^{2} y^{2}=19
\end{array}\right.
$$
Answer: $(2 ; \pm \sqrt{3}),(-2 ; \pm \sqrt{3}),(\sqrt{3} ; \pm 2),(-\sqrt{3} ; \pm 2)$.

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 becomes

$$
\left\{\begin{array}{l}
2 u-v=2 \\
u^{2}-2 v-\frac{1}{2} v=19
\end{array}\right.
$$

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

$$
u^{2}-\frac{5}{2}(2 u-2)=19 \Leftrightarrow u^{2}-5 u-14=0 \Leftrightarrow\left[\begin{array}{l}
u=-2 \\
u=7
\end{array}\right.
$$

Since $u>0$, only the value $u=7$ is valid. Thus, $v=12$, and in the original variables we get

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 7 , } \\
{ x ^ { 2 } y ^ { 2 } = 1 2 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x ^ { 2 } = 7 - y ^ { 2 } } \\
{ ( 7 - y ^ { 2 } ) y ^ { 2 } = 1 2 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
y^{2}=3 \text { and } x^{2}=4 \\
y^{2}=4 \text { and } x^{2}=3
\end{array}\right.\right.\right.
$$

From this, it follows that the system has 8 solutions: $(2 ; \pm \sqrt{3}),(-2 ; \pm \sqrt{3}),(\sqrt{3} ; \pm 2),(-\sqrt{3} ; \pm 2)$.