$x^2+y^2=2z^2$, positive integer solutions 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). 
 A: 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 $u^2+v^2=z^2$.
A: 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:
$$x=-tp^2+2kps+ts^2$$
$$y=kp^2+2tps-ks^2$$
$$z=p^2+s^2$$
A: Though others have posted simple solutions to the problem but if you want to proceed the way you did then here it is:
First let $\dfrac xz = a$ and $\dfrac yz = b$. This reduces problem to two variables (which are rational numbers).
Now we have $(b+1)(b-1)=(1+a)(1-a)$
which means $${b+1\over1+a} ={ 1-a\over b-1}$$
Let this fraction be $\dfrac mn$ in reduced form. From this we get 2 linear equations in 2 variables (in terms of a and b). 
Solving those will give
$$a = \frac{(n² + 2mn - m²)}{(m² + n²)}$$
$$b = {(m² + 2mn - n²)\over(m² + n²)}$$
 Substituting values of a and b back gives the solution to original problem:
$$x = n² + 2mn - m²$$ $$y = m² + 2mn - n²$$ $$z = m² + n²$$
A: 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:
$${\left( {p}_{2}\,{t}_{2}^{2}+2\,{p}_{1}\,{t}_{1}\,{t}_{2}-{t}_{1}^{2}\,{p}_{2}\right) }^{2}+{\left( {p}_{1}\,{t}_{2}^{2}-2\,{t}_{1}\,{p}_{2}\,{t}_{2}-{p}_{1}\,{t}_{1}^{2}\right) }^{2}=\left( {p}_{2}^{2}+{p}_{1}^{2}\right) \,{\left( {t}_{2}^{2}+{t}_{1}^{2}\right) }^{2}$$
