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Find all the solutions to the Diophantine equation $x^2+y^2=2z^2$. I do not have a lot of experience on Diophantine equations and I do not know how to approximate them. I can see that the triples of the form $(x,x,x)$, example $(0,0,0), (1,1,1),\ldots,$ etc, gives solutions to my equation but that is all, how do I define that is all of them or if they are not how I find the rest?

Angelo
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Jam
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  • http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/740425#740425 – individ May 05 '14 at 03:56
  • Go to this thread and see the formula. This is a general Legendre equation and solve standard. – individ May 05 '14 at 03:57
  • See: https://math.stackexchange.com/questions/1250912/diophantine-equations-solving-a2-b2-2c2, https://math.stackexchange.com/questions/1767109/x2y2-2z2-positive-integer-solutions https://math.stackexchange.com/questions/1282600/parametric-characterization-for-x2-y2-2z2, – Martin Sleziak Jun 07 '19 at 09:16

1 Answers1

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Suppose that $x^2+y^2=2z^2$. Then $x$ and $y$ are both even or both odd.

Let $s=\frac{x-y}{2}$ and $t=\frac{x+y}{2}$. Then $s$ and $t$ are integers, and $$s^2+t^2=\frac{1}{4}((x-y)^2+(x+y)^2)=\frac{1}{2}(x^2+y^2)=z^2.$$

Conversely, from any solution of $s^2+t^2=z^2$, we can set $x=s+t$ and $y=s-t$, and find that $x^2+y^2=2z^2$.

So the solutions of our equation are closely related to solutions of $s^2+t^2=z^2$. There is a pretty complete standard theory of these, which it is likely that you have been exposed to.

Apart from the order of $x$ and $y$, we therefore obtain all solutions as follows. Let $u$, $v$, and $k$ be integers. Then if we let $x=k(u^2-v^2) +2kuv$ and $y=k(u^2-v^2)-2kuv$ we get a solution of our equation, and all solutions can be obtained in this way.

André Nicolas
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  • I know the standar formulas for the pythagorean tripples.So you set s and t a standard formula .Is there something that gave away the formula or just experience.And so the solutions are all integers x = s+t, y=s-t where s,t belong to Z – Jam May 04 '14 at 18:34
  • I am sure I have seen it before, indeed assigned it before. The solutions are (sort of) given by $x=s+t$, $y=s-t$, where $s^2+t^2$ is a perfect square, and we know how to generate these. If we are working over the integers the signs of $x$, $y$, $z$ can then be modified arbitrarily. – André Nicolas May 04 '14 at 18:40
  • So i must continue so that s^2+t^2=z^2 and that i can solve it by the already formulas.hm.. i do not know if the "set" part would come to my mind in a test but the thought is that you are trying to find s,t numbers depending on x,y that will give you a perfect square with a certain coeficient is that right? – Jam May 04 '14 at 18:45
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    The last paragraph has the generating formula. – André Nicolas May 04 '14 at 19:16
  • And you found that formula from the hypothesis that s^2+t^2=z^2? – Jam May 04 '14 at 19:19
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    Yes, I used the standard generating formula for Pythagorean triples. – André Nicolas May 04 '14 at 19:27