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This is a self-answered question. I post it here since it wasn't trivial for me.

Consider the equality $$ 2\frac{x^4-4 x^3+2 x^2 (y^2+2)-4 x y^2+y^2 (y^2+8)}{(x^2-2 x+y^2)^2}=2+\frac{4y^2}{(1-x)^2}, \tag{1} $$ where $x,y$ are real numbers.

Mathematica claims that if this equality holds, then if $y \neq 0$, then $$ x\in \{\pm\sqrt{2-y^2}, 2 \pm \sqrt{2-y^2}\}. $$

Is there any chance we could verify this "by hand"?

(If $y=0$, then the equality holds for any $x \neq 2$.)

The motivation for considering this equality comes from this problem.

A-Level Student
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Asaf Shachar
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    When you post a question, to answer it yourself, Asaf, you are required to write the question as though you do not know an answer, including far more context, than you did here. Linking is not the same as you telling us yourself what your motivation for asking this question, nor telling us at what level of study you are at. A question should be asked inviting others to know what you know, to answer as well. – amWhy May 20 '21 at 21:13
  • Thanks. I generally agree with you. Specifically, I think that the motivation for the question should be somewhat independent of the question whether I intend to answer myself or not. If the question seems sufficiently "natural", then that's good. If not, then further motivation needs to be added. In this case, I agree that I did not add sufficient motivation. (I thought it was a too long and complicated detour from the question itself, but perhaps I was wrong; I definitely could add some more). – Asaf Shachar May 27 '21 at 09:59
  • However, the external motivation did not actually play any role in the solution, which was after all just routine manipulations. – Asaf Shachar May 27 '21 at 09:59

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Multiplying equation $(1)$ by $(x^2-2 x+y^2)^2$, we obtain

$$ 2x^4-8 x^3+4x^2y^2+8x^2-8xy^2+2y^2(y^2+8)= $$ $$2x^4+8x^2+2y^4-8x^3-8xy^2+4x^2y^2+\frac{4y^2(x^2-2 x+y^2)^2}{(1-x)^2}, $$ which after cancellations becomes $$ 16y^2=\frac{4y^2(x^2-2 x+y^2)^2}{(1-x)^2}. $$ Thus, if $y \neq 0$, we can divide by $4y^2$, and get $$ 4=\big(\frac{x^2-2 x+y^2}{1-x}\big)^2, $$ so $$ x^2-2 x+y^2=\pm 2(1-x). $$ If $x^2-2 x+y^2=2(1-x)$, then $x^2+y^2=2 \Rightarrow x=\pm \sqrt{2-y^2}$.

If $x^2-2 x+y^2=2(x-1)$, then solving the quadratic for $x$ one get $x=2 \pm \sqrt{2-y^2}$.

Asaf Shachar
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