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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?

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    $\begingroup$ For an answer see here. $\endgroup$ – Dietrich Burde Jan 31 '16 at 19:20
  • $\begingroup$ Your equations are not wrong, but they are a little misleading: we know $z^2 < x^2$ and $z^2 < y^2$, so $(x^2-z^2)(y^2-z^2)$ is more natural than $(z^2-x^2)(z^2-y^2)$. $\endgroup$ – TonyK Jan 31 '16 at 19:24
  • $\begingroup$ @DietrichBurde: this answer is quite partial. $\endgroup$ – Yves Daoust Jan 31 '16 at 19:24
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    $\begingroup$ @SubhadeepDey Just coprime or pairwise coprime? Because $15,20,12$ are coprime... $\endgroup$ – Bart Michels Jan 31 '16 at 19:39
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    $\begingroup$ Yes, but that parametrisation gives us $(x,y,w)$ [apart from a typo, you wrote $m^2+n^2$ twice, the first should be $m^2 - n^2$]. We want $w \mid xy$, so that $z = \frac{xy}{w}$ is an integer. That gives us $(m^2+n^2) \mid k(m^2-n^2)2mn$. Since $m,n$ are coprime and of opposite parity, that only holds if $(m^2+n^2) \mid k$. And we find that $\gcd(x,y,z) = 1$ only if $k = m^2+n^2$. $\endgroup$ – Daniel Fischer Jan 31 '16 at 23:24
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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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  • $\begingroup$ If the OP meant $\gcd(x,y,z)=1$, then this is a good, short answer. $\endgroup$ – Tito Piezas III Feb 2 '16 at 15:35
  • $\begingroup$ If not, the answer is still short and good (+1). It seems to be a good idea to suppose that $gcd(x,y,z)=1$. $\endgroup$ – Dietrich Burde Feb 2 '16 at 16:36
  • $\begingroup$ Nice answer. +1 $\endgroup$ – Bumblebee Sep 14 '16 at 19:23

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