Finding all real solutions to $y^2=x^3-3x^2+2x$ and $x^2=y^3-3y^2+2y$

Question

Find the all real solutions following system equations: $$y^2=x^3-3x^2+2x$$ $$x^2=y^3-3y^2+2y$$

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I was only able to edit and get $x^3-2x^2+2x=y^3-2y^2+2y$.

Answer 1

Subtracting the two equations yields $$(x-y)\left(x^2+xy+y^2-2x-2y+2\right)=0\,.$$ We have $$x^2+xy+y^2-2x-2y+2=\frac{1}{2}x^2+\frac{1}{2}y^2+\frac{1}{2}(x+y-2)^2>0\,,$$ as $x=0$, $y=0$, and $x+y-2=0$ cannot be satisfied simultaneously. (In fact, the minimum value of $x^2+xy+y^2-2x-2y+2$ is $\dfrac23$, which is met if and only if $(x,y)=\left(\dfrac23,\dfrac23\right)$.) Thus, $x=y$ must hold.

Answer 2

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

If $x\ne y$

$$x^2+x(y-2)+y^2-2y+2=0$$

The discriminant is $$-[4(y^2-2y+2)-(y-2)^2]=-[3y^2-4y+4]=-[2y^2+(y-2)^2]<0$$

Answer 3

We have

$$ f(x,y) = y^2-x^3+3x^2-2x\\ f(y,x) = x^2-y^3+3y^2-2y $$

so

$$ f(x,x) = x^2-x^3+3x^2-2x = 0\Rightarrow {x = 0}\cup{x = 2\pm \sqrt2} $$

Attached a plot showing the solutions

Answer 4

@Cesareo, You give me hint.

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We get the following equations if we take sides. $$x^3-2x^2+2x=y^3-2y^2+2y$$

Let $f(t)=t^3-2t^2+2t$. So $f(x)=f(y)$.

$f'(x)=3x^2-4x+2>2x^2-4x+2=2(x^2-2x+1)=2(x-1)^2\geq 0\Rightarrow f'(x)>0$

Hence, $f$ is a strictly increasing function. So $f$ is $1-1$.

we obtaion $f(x)=f(y)\Rightarrow x=y$

Now, if we take $y=x$ in given system equations: $$x^2=x^3-3x^2+2x\iff x^3-4x^2+2x=0\iff x(x^2-4x+2)=0$$ Therefore, all solutions are $(0,0),(2+\sqrt{2},2+\sqrt{2}),(2-\sqrt{2},2-\sqrt{2})$.