Problem 5. Solve the system of equations

$$
\left\{\begin{array}{rl}
x^{2}+x y+y^{2} & =23 \\
x^{4}+x^{2} y^{2}+y^{4} & =253
\end{array} .\right.
$$
Answer: $\left(-\frac{\sqrt{29}+\sqrt{5}}{2} ;-\frac{\sqrt{29}-\sqrt{5}}{2}\right),\left(-\frac{\sqrt{29}-\sqrt{5}}{2} ;-\frac{\sqrt{29}+\sqrt{5}}{2}\right),\left(\frac{\sqrt{29}-\sqrt{5}}{2} ; \frac{\sqrt{29}+\sqrt{5}}{2}\right),\left(\frac{\sqrt{29}+\sqrt{5}}{2} ; \frac{\sqrt{29}-\sqrt{5}}{2}\right)$

Solution. Note that $x^{4}+x^{2} y^{2}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-x^{2} y^{2}=\left(x^{2}+x y+y^{2}\right)\left(x^{2}-x y+y^{2}\right)$. Thus, the original system is equivalent to $\left\{\begin{array}{l}x^{2}-x y+y^{2}=11 \\ x^{2}+x y+y^{2}=23\end{array}\right.$.

Let's make a substitution: $u=x+y, v=x y$. Then from the first equation we get that $u^{2}-3 v=11$, and from the second $-u^{2}-v=23$. Therefore, $v=6$, and $u^{2}=29, u= \pm \sqrt{29}$.

Thus, according to Vieta's theorem, the pair $(x, y)$ are the roots of the equation $t^{2} \pm \sqrt{29} t+6=0$. Therefore, $(x, y)$ can only be $\left(-\frac{\sqrt{29}+\sqrt{5}}{2} ;-\frac{\sqrt{29}-\sqrt{5}}{2}\right),\left(-\frac{\sqrt{29}-\sqrt{5}}{2} ;-\frac{\sqrt{29}+\sqrt{5}}{2}\right),\left(\frac{\sqrt{29}-\sqrt{5}}{2} ; \frac{\sqrt{29}+\sqrt{5}}{2}\right)$, $\left(\frac{\sqrt{29}+\sqrt{5}}{2} ; \frac{\sqrt{29}-\sqrt{5}}{2}\right)$. Each of these options satisfies the original system.