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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!



marked as duplicate by lab bhattacharjee algebra-precalculus Oct 14 '15 at 12:54

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  • $\begingroup$ Note: you can treat this as a quadratic in $x$ $\endgroup$ – Mark Bennet Oct 14 '15 at 12:50
  • $\begingroup$ Sorry there's a typo in my question. $\endgroup$ – shoestringfries Oct 14 '15 at 12:52
  • $\begingroup$ @lab bhattacharjee I don't think this is a duplicate-I'm proving there is no solutions and the answers there don't really help me. $\endgroup$ – shoestringfries Oct 14 '15 at 14:03
  • $\begingroup$ The accepted answer in the link shows that all solutions are (0,0,0). Therefore, it shows that there is no solution over the positive integers. There are also several other good proofs on that page, plus the one I hinted at below before I realized the question was a duplicate. $\endgroup$ – Eric Brooks Oct 14 '15 at 14:19

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.

(Edited to reflect the edit in the question)

  • $\begingroup$ 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 $x,y,z$ $\endgroup$ – shoestringfries Oct 14 '15 at 13:53

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