<s>[INST] Integer solutions for $\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$? 
Find all triples $(x, y, z)$ where $x, y, z$ are coprime integers such that $$\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{z^2}$$

I did the following:
$$\begin{split}\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{z^2} &\Rightarrow \dfrac{x^2y^2}{x^2+y^2}=z^2 \\ & \Rightarrow x^2y^2-x^2z^2-y^2z^2=0 \\ & \Rightarrow x^2y^2-x^2z^2-y^2z^2+z^4=z^4 \\ & \Rightarrow (z^2-x^2)(z^2-y^2)=z^4 \\ & \Rightarrow (x+z)(x-z)(y+z)(y-z)=z^4\end{split}$$
How should I go on?
 [/INST] For the equation.
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$$
Use a Pythagorean triple.
$$a^2+b^2=c^2$$
Obtained solutions.
$$x=ac$$
$$y=bc$$
$$z=ab$$
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