Problem 5. Solve the system of equations

$$
\left\{\begin{aligned}
x^{2}+x y+y^{2} & =37 \\
x^{4}+x^{2} y^{2}+y^{4} & =481
\end{aligned}\right.
$$
Answer: $(-4, -3), (-3, -4), (3, 4), (4, 3)$.

Solution. Note that $x^{4} + x^{2} y^{2} + y^{4} = (x^{2} + y^{2})^{2} - x^{2} y^{2} = (x^{2} + xy + y^{2})(x^{2} - xy + y^{2})$. Thus, the original system is equivalent to $\left\{\begin{array}{l}x^{2} - xy + y^{2} = 13 \\ x^{2} + xy + y^{2} = 37\end{array}\right.$.

Let's make a substitution: $u = x + y, v = xy$. Then from the first equation, we get $u^{2} - 3v = 13$, and from the second $-u^{2} - v = 37$. Therefore, $v = 12$, and $u^{2} = 49, u = \pm 7$.

Thus, according to Vieta's theorem, the pair $(x, y)$ are the roots of the equation $t^{2} \pm 7t + 12 = 0$. Therefore, $(x, y)$ can only be $(-4, -3), (-3, -4), (3, 4), (4, 3)$. It is easy to see that each of these options satisfies the original system.