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

$$
\begin{aligned}
& x+y^{2}=y^{3} \\
& y+x^{2}=x^{3}
\end{aligned}
$$

(Jaroslav Švrček)
Solution. By subtracting the first equation from the second, we get

$$
\begin{aligned}
\left(x^{3}-y^{3}\right)-\left(x^{2}-y^{2}\right)+(x-y) & =0 \\
(x-y)\left(x^{2}+x y+y^{2}-x-y+1\right) & =0 .
\end{aligned}
$$

The second factor is positive for any real numbers $x$ and $y$, since

$$
x^{2}+x y+y^{2}-x-y+1=\frac{1}{2}(x+y)^{2}+\frac{1}{2}(x-1)^{2}+\frac{1}{2}(y-1)^{2}
$$

and the bases of all three squares cannot be zero simultaneously. Therefore, for any solution $(x, y)$ of the given system, it must be that $x-y=0$ or $y=x$, which reduces the system to a single equation $x+x^{2}=x^{3}$ with roots $x_{1}=0$ and $x_{2,3}=\frac{1}{2}(1 \pm \sqrt{5})$.

The solutions are precisely three pairs $(x, y)$, where $y=x \in\left\{0, \frac{1}{2}(1+\sqrt{5}), \frac{1}{2}(1-\sqrt{5})\right\}$.