# How to prove that $x^2+y^2+z^2 = 2xyz$ has no solutions over the positive integers? [duplicate]

My strategy is proof by contradiction. So assume the opposite of the proof statement is true. I could only think of $x^2 - y^2 + y^2 + z^2 = 2xyz$ $=(x+y)(x-y) + y^2 + z^2$ but I don't think that is a helpful simplification. Please help!

Thanks!

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• Note: you can treat this as a quadratic in $x$ – Mark Bennet Oct 14 '15 at 12:50
Here's a hint. Rearrange to get $z^2 + z(-2xy) + (x^2+y^2) = 0$. By the quadratic formula, $z = \frac{2xy \pm \sqrt{4 x^2 y^2 - 4(x^2+y^2)} }{2}$. Your proposition is true iff there are no $x$ and $y$ for which the right hand side is a positive integer.
• but the problems is that I have seemed to prove the assumption true...because $\sqrt{x^2y^2-x^2-y^2} \ge 0$ is true and there are positive solutions for $x,y,z$ – shoestringfries Oct 14 '15 at 13:53