## Task A-2.1.

Determine all pairs of real numbers $(x, y)$ that satisfy the system

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

By subtracting the given equations, we get $y-x+x^{2}-y^{2}=x^{3}-y^{3}$, which is equivalent to

$$
y-x+(x-y)(x+y)=(x-y)\left(x^{2}+x y+y^{2}\right)
$$

or

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

it follows that $x+y-1=x^{2}+x y+y^{2}$ or $x-y=0$.

In the first case, $x+y-1=x^{2}+x y+y^{2}$, which can be written as

$$
y^{2}+(x-1) y+x^{2}-x+1=0
$$

We consider this equation as a quadratic equation in $y$. Its discriminant is

$$
D=(x-1)^{2}-4\left(x^{2}-x+1\right)=-3 x^{2}+2 x-3
$$

The discriminant of the quadratic trinomial $-3 x^{2}+2 x-3$ is $-32<0$, so it holds that $D<0$ for every real number $x$, which implies that equation (ద) has no real solutions for $y$. Therefore, there are no solutions in this case.

In the second case, $y=x$ and both equations become $x+x^{2}=x^{3}, \operatorname{i.e} . x\left(x^{2}-x-1\right)=0$. It follows that $x_{1}=0$ or $x^{2}-x-1=0$. Solving the quadratic equation, we get the solutions $x_{2,3}=\frac{1 \pm \sqrt{5}}{2}$. The only solutions are $(x, y) \in\left\{(0,0),\left(\frac{1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right),\left(\frac{1-\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}\right)\right\}$.

Note: The equation $x+y-1=x^{2}+x y+y^{2}$ is equivalent to

$$
(x-1)^{2}+(y-1)^{2}+(x+y)^{2}=0
$$

which has no solutions since there do not exist $x$ and $y$ such that $x=1, y=1$ and $x+y=0$.