1. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
x^{2}-x y+y^{2} & =7, \\
x^{2} y+x y^{2} & =-2 .
\end{aligned}
$$

(J. Földes)
Solution. Since the second equation can be rewritten as $x y(x+y)=-2$, let's similarly transform the first equation: $(x+y)^{2}-3 x y=7$. For the numbers $s=x+y, p=x y$, we then obtain the equivalent system

$$
\begin{aligned}
s^{2}-3 p & =7 \\
s p & =-2,
\end{aligned}
$$

which, after expressing $p=-2 / s$ (clearly $s$ cannot be $0$) from the second equation, leads to the cubic equation $s^{3}-7 s+6=0$. This equation has integer roots $s_{1}=1, s_{2}=2$, and $s_{3}=-3$. The corresponding values of the product $p=x y$ for these values of $s$ are: $p_{1}=-2, p_{2}=-1$, $p_{3}=\frac{2}{3}$. The numbers $x, y$ form a pair of roots of the quadratic equation $t^{2}-s t+p=0$, so they are one of the equations

$$
t^{2}-t-2=0, \quad t^{2}-2 t-1=0, \quad t^{2}+3 t+\frac{2}{3}=0
$$

Solving these equations, we obtain (all) six solutions of the given system:

$$
\{x, y\}=\{-1,2\},\{x, y\}=\{1+\sqrt{2}, 1-\sqrt{2}\},\{x, y\}=\left\{\frac{-9+\sqrt{57}}{6}, \frac{-9-\sqrt{57}}{6}\right\} .
$$