
1. Solve the system of equations

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

in the domain of the reals.


Solution. 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 $x$ 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 three squares cannot simultaneously vanish. Any solution $(x, y)$ of the given system must therefore satisfy $x-y=0$, or $y=x$, which reduces the system to the single equation $x+x^{2}=x^{3}$ with roots $x_{1}=0$ and $x_{2,3}=\frac{1}{2}(1 \pm \sqrt{5})$.

Thus there are exactly three solutions $(x, y)$, namely $y=x \in\left\{0, \frac{1}{2}(1+\sqrt{5})\right.$, $\left.\frac{1}{2}(1-\sqrt{5})\right\}$.
