Solve the following system of equations:

$$
\begin{aligned}
& y^{2}=x^{3}-3 x^{2}+2 x \\
& x^{2}=y^{3}-3 y^{2}+2 y
\end{aligned}
$$
Solution. Transforming the first equation: $y^{2}=x(x-1)(x-2)$. The left side is non-negative, so the right side must be as well: $x \in[0 ; 1] \cup[2 ;+\infty[$. Similarly from the second equation, $y \in[0 ; 1] \cup[2 ;+\infty[$.

Subtract one equation from the other:

$$
y^{2}-x^{2}=x^{3}-y^{3}-3 x^{2}+3 y^{2}+2 x-2 y
$$

Rearrange and factor the right side:

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

A product is zero if and only if one of its factors is zero.

Case I: $x-y=0$. Then $x=y$, and from the original equations:

$$
x^{2}=x^{3}-3 x^{2}+2 x, \quad \text { from which } \quad 0=x^{3}-4 x^{2}+2 x=x\left(x^{2}-4 x+2\right)
$$

From this, $x_{1}=y_{1}=0$, or

$$
x_{2,3}=\frac{4 \pm \sqrt{16-8}}{2}=2 \pm \sqrt{2}
$$

Thus, $x_{2}=y_{2}=2-\sqrt{2}$ and $x_{3}=y_{3}=2+\sqrt{2}$.

Case II: $(x-1)^{2}+(y-1)^{2}+x y=0$. We know that $(x-1)^{2} \geq 0,(y-1)^{2} \geq 0$ and since $x$ and $y$ are non-negative, $x y \geq 0$. The sum of the three terms can only be zero if each term is zero. From the first two terms, $x=y=0$, but then $x y=1 \neq 0$.

In this case, we did not find a solution.

Remarks. 1. Many wrote that since the two equations are symmetric, $x=y$. This is not necessarily true. See, for example, Ábrahám Gábor: "On equations of the type $f^{-1}(x)=f(x)$, or the responsibility of examiners and other interesting facts" (KöMaL, December 2010, and www.komal.hu/cikkek/abraham/abraham.h.shtml).

2. After factoring, some solved the second factor for $x$, then wrote the quadratic formula: $x^{2}+(y-2) x+\left(y^{2}-2 y+2\right)=0$, from which

$$
x=\frac{-y+2 \pm \sqrt{(y-2)^{2}-4\left(y^{2}-2 y+2\right)}}{2}=\frac{-y+2 \pm \sqrt{-3 y^{2}+4 y-4}}{2}
$$

The discriminant under the square root is $D=16-48<0$, so there is no solution in this case.

3. To show that $x$ and $y$ cannot be negative, many reasoned that if, for example, $x<0$, then $x^{3}<0,-3 x^{2}<0$ and $2 x<0$, so their sum would also be negative, but $y^{2} \geq 0$, which is not possible.