How to solve a system of equations?

Question

How can I solve this system of equations?

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

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

Answer 1

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

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

Subtracring $(1.)$ from $(2.)$ ,

$$\begin{align} x^2 - y^2 & = y^3 - x^3 + 3x^2 - 3y^2 + 2y-2x \\ \color{#f04}{(x+y)(x-y)}& \color{#f04}{= (y-x)(y^2+x^2+xy) +3(x+y)(x-y) + 2(y-x)} \\ \end{align}$$

If $(x-y)=0 \implies x=y$ , Which gives $\color{#39f}{x^2 = x^3 - 3x^2 + 2x}$ , Which yields $\color{#4b9}{x =0 , 2\pm \sqrt 2}$

If $x \ne y$ , Then :

$$x+y = -(x^2 + y^2 +xy) + 3(x+y) - 2$$ $$(x^2+y^2+xy) = 2(x+y-1)$$ $$x^2 + (y-2)x + (y^2-2y+2) = 0$$

Using Quadratic formula , we have obtain the discriminant : $$D = (y-2)^2 - 4(y^2-2y+2)$$

$$D = -3y^2 + 4y - 4 \implies \boxed{D = -3 \left(y - \frac23 \right)^2 - \frac83}$$

We can observe that $\color{#da0}{D\lt 0}$ for all values of $y$. Hence the equation has no solutions for real $x,y$.

Hence the only solutions are $(x , y) = (0,0) \,\, , (2\pm\sqrt2,2\pm\sqrt2)$