Prove that $x\mathcal{R}y\iff x^2-y^2=2(y-x)$ is an equivalence relation

Question

Prove that $$x\mathcal{R}y\iff x^2-y^2=2(y-x)$$ is an equivalence relation.

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Reflexive. For all $x$ we have $x^2-x^2=2(x-x)$, so $x\mathcal{R}x$.

Symmetric. For all $x,y$ we have \begin{align}x\mathcal{R}y&\implies x^2-y^2=2(y-x)\\&\implies(-1)(x^2-y^2)=(-1)2(y-x)\\&\implies y^2-x^2=2(x-y)\\&\implies y\mathcal{R}x.\end{align}

Transitive. For all $x,y,z$ we have $$\begin{cases}x\mathcal{R}y\\y\mathcal{R}z\end{cases}\implies\begin{cases}x^2-y^2=2(y-x)\\y^2-z^2=2(z-y)\end{cases}\implies x^2-y^2+y^2-z^2=2(y-x)+2(z-y),$$ so $$x^2-z^2=2(y-x+z-y)=2(z-x)\implies x\mathcal{R}z.$$

Is it correct?

Answer 1

Your proof is correct and well-presented. Good job!

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Mostly posting this so this question can be finally considered to have an answer. Made it Community Wiki since I have nothing of substance to add.