$x^2+y^2=2z^2$, positive integer solutions

Question

Determine all positive integer solutions of the equation $x^2+y^2=2z^2$.

First I assume $x \geq y$, and I have $x^2-z^2=z^2-y^2$. Then I have $(x-z)(x+z)=(z-y)(z+y)$, but from here, I don't know how it can help me to describe solutions (I know that there are infinitely many).

Answer 1

Just a start:

Note that $$\left(\frac{x-y}{2}\right)^2 + \left(\frac{x+y}{2}\right)^2=\frac{x^2+y^2}{2}=z^2.$$

And $\frac{x-y}{2}$ and $\frac{x+y}{2}$ are integers (why?)

So you need to find solutions to $u^2+v^2=z^2$.

Answer 2

In General formula generic for Pythagorean triples looks a little different.

$$x^2+y^2=az^2$$

If the number can be represented as a sum of squares. $a=t^2+k^2$

The solution has the form:

$$x=-tp^2+2kps+ts^2$$

$$y=kp^2+2tps-ks^2$$

$$z=p^2+s^2$$

Answer 3

From Brahmagupta-Fibonacci Identity: $${\left( {p}_{2}\,{s}_{2}+{p}_{1}\,{s}_{1}\right) }^{2}+{\left( {p}_{1}\,{s}_{2}-{s}_{1}\,{p}_{2}\right) }^{2}=\left( {p}_{2}^{2}+{p}_{1}^{2}\right) \,\left( {s}_{2}^{2}+{s}_{1}^{2}\right)$$ get: $${\left( {p}_{2}\,{s}_{2}\,{t}_{2}+{p}_{1}\,{s}_{1}\,{t}_{2}+{p}_{1}\,{t}_{1}\,{s}_{2}-{s}_{1}\,{t}_{1}\,{p}_{2}\right) }^{2}+{\left( {p}_{1}\,{s}_{2}\,{t}_{2}-{s}_{1}\,{p}_{2}\,{t}_{2}-{t}_{1}\,{p}_{2}\,{s}_{2}-{p}_{1}\,{s}_{1}\,{t}_{1}\right) }^{2}=\left( {p}_{2}^{2}+{p}_{1}^{2}\right) \,\left( {s}_{2}^{2}+{s}_{1}^{2}\right) \,\left( {t}_{2}^{2}+{t}_{1}^{2}\right)$$ and get solution: $${\left( {p}_{2}\,{t}_{2}^{2}+2\,{p}_{1}\,{t}_{1}\,{t}_{2}-{t}_{1}^{2}\,{p}_{2}\right) }^{2}+{\left( {p}_{1}\,{t}_{2}^{2}-2\,{t}_{1}\,{p}_{2}\,{t}_{2}-{p}_{1}\,{t}_{1}^{2}\right) }^{2}=\left( {p}_{2}^{2}+{p}_{1}^{2}\right) \,{\left( {t}_{2}^{2}+{t}_{1}^{2}\right) }^{2}$$