4. [4 points] Solve the system of equations

$$
\left\{\begin{array}{l}
3 x^{2}+3 y^{2}-x^{2} y^{2}=3 \\
x^{4}+y^{4}-x^{2} y^{2}=31
\end{array}\right.
$$
Answer: $(\sqrt{5} ; \pm \sqrt{6}),(-\sqrt{5} ; \pm \sqrt{6}),(\sqrt{6} ; \pm \sqrt{5}),(-\sqrt{6} ; \pm \sqrt{5})$.

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

$$
\left\{\begin{array}{l}
3 u-v=3 \\
u^{2}-2 v-v=31
\end{array}\right.
$$

From the first equation $v=3 u-3$; substitute this expression into the second equation and solve it:

$$
u^{2}-3(3 u-3)=31 \Leftrightarrow u^{2}-9 u-22=0 \Leftrightarrow\left[\begin{array}{l}
u=11 \\
u=-2
\end{array}\right.
$$

Since $u>0$, only the value $u=11$ is valid. Thus, $v=30$, and in the original variables we get

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 1 1 , } \\
{ x ^ { 2 } y ^ { 2 } = 3 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x ^ { 2 } = 1 1 - y ^ { 2 } } \\
{ ( 1 1 - y ^ { 2 } ) y ^ { 2 } = 3 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
y^{2}=5 \text { and } x^{2}=6 \\
y^{2}=6 \text { and } x^{2}=5
\end{array}\right.\right.\right.
$$

From this, it follows that the system has 8 solutions: $(\sqrt{5} ; \pm \sqrt{6}),(-\sqrt{5} ; \pm \sqrt{6}),(\sqrt{6} ; \pm \sqrt{5}),(-\sqrt{6} ; \pm \sqrt{5})$.