Solving a diophantine equation x²+y²=2z²

Asked Oct 25 '22 at 19:54

Active Oct 25 '22 at 19:54

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solve the equation x²+y²=2z² in positive integers, Found this question in a book named Challenge and thrill of pre college mathematics.

Its kind of obvious to prove when x=y=z but when z=√(x²+y²)/√2, i am stuck on this step and would appreciate any hint or solutions `

asked Oct 25 '22 at 19:54

Unique_1

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I suggest: look at some of the ways to enumerate Pythagorean triples and try to adapt those methods. – lulu Oct 25 '22 at 20:03
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do you know the parametrization for (primitive) Pythagorean triples ? If $a^2 + b^2 = c^2$ then $(a-b)^2 + (a+b)^2 = 2 c^2$ Oh, and primitive solutions to your problem have $x,y,z$ all odd – Will Jagy Oct 25 '22 at 20:03
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Does this answer your question? Diophantine Equations problem 2 (with several other duplicates linked to it, as indicated in the "Linked" section on the right side) - found using an Approach0 search. There's also the closely related (which uses what @WillJagy indicated in the comment above) Parametric characterization for $x^2 + y^2 = 2z^2$. – John Omielan Oct 25 '22 at 20:06
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Yep i got that now , thanks everyone for suggesting things and providing a direct solution :) – Unique_1 Oct 25 '22 at 20:19
$(\frac{m^2-2m-1}{m^2+1},\frac{-m^2-2m+1}{m^2+1})$... – Bob Dobbs Oct 25 '22 at 21:36

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