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Prove that this equation has no solutions in the set of positive integers. $x^2 − y^2 = 2xyz$.

I know $x^2$ and $y^2$ must be both odd or even. I know how to solve the question in the case that they are even, but I don't know how to prove the equation if they are odd.

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Dividing $x$ and $y$ by their greatest common divisor, we may assume $x$ and $y$ are coprime. But $y^2 = x (x - 2 y z)$ is divisible by $x$, and similarly $x^2$ is divisible by $y$, so ...

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  • $\begingroup$ I understand the second part, but how may we assume x and y are coprime? i.e. What is the proof for x and y having a gcd of 1 $\endgroup$ – Wizard Nov 6 '16 at 7:36
  • $\begingroup$ Let $a = \gcd(x,y)$. Then $a^2$ divides $x^2 - y^2$ and $2xyz$, i.e., replace $x$ with $x/a$ and $y$ with $y/a$. The new $x$ and $y$ have a gcd of $1$. $\endgroup$ – Simon S Nov 6 '16 at 7:41
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    $\begingroup$ I see now, thanks so much Simon S and Robert Israel $\endgroup$ – Wizard Nov 6 '16 at 7:49
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Bring $y^2$ to the the right hand side and add $x^2z^2$ to both sides. You end up with $$x^2(1+z^2)=(y+xz)^2$$ thus $1+z^2$ must be a square, which forces $z=0$.

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  • $\begingroup$ Nice alternative. $\endgroup$ – Simon S Nov 6 '16 at 7:52

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