4. [4 points] Solve the system of equations

$$
\left\{\begin{array}{l}
3 x^{2}+3 y^{2}-2 x^{2} y^{2}=3 \\
x^{4}+y^{4}+\frac{2}{3} x^{2} y^{2}=17
\end{array}\right.
$$
Answer: $(\sqrt{2} ; \pm \sqrt{3}),(-\sqrt{2} ; \pm \sqrt{3}),(\sqrt{3} ; \pm \sqrt{2}),(-\sqrt{3} ; \pm \sqrt{2})$.

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 becomes

$$
\left\{\begin{array}{l}
3 u-2 v=3 \\
u^{2}-2 v+\frac{2}{3} v=17
\end{array}\right.
$$

From the first equation, $v=\frac{3 u}{2}-\frac{3}{2}$; substitute this expression into the second equation and solve it:

$$
u^{2}-\frac{4}{3}\left(\frac{3 u}{2}-\frac{3}{2}\right)=17 \Leftrightarrow u^{2}-2 u-15=0 \Leftrightarrow\left[\begin{array}{l}
u=5 \\
u=-3
\end{array}\right.
$$

Since $u>0$, only the value $u=5$ is valid. Thus, $v=6$, and in the original variables we get

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

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