Problem 5. Solve the system of equations

$$
\left\{\begin{aligned}
x^{2}-x y+y^{2} & =7 \\
x^{4}+x^{2} y^{2}+y^{4} & =91
\end{aligned}\right.
$$
Answer: $(-3 ;-1),(-1 ;-3),(1 ; 3),(3 ; 1)$.

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 $\left\{\begin{array}{l}x^{2}-x y+y^{2}=7 \\ x^{2}+x y+y^{2}=13\end{array}\right.$.

Let's make a substitution: $u=x+y, v=x y$. Then from the first equation we get that $u^{2}-3 v=7$, and from the second $-u^{2}-v=13$. Therefore, $v=3$, and $u^{2}=16, u= \pm 4$.

Thus, according to Vieta's theorem, the pair $(x, y)$ are the roots of the equation $t^{2} \pm 4 t+3=0$. Therefore, $(x, y)$ can only be $(-3 ;-1),(-1 ;-3),(1 ; 3),(3 ; 1)$. It is easy to see that each of these options satisfies the original system.