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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?

Tito Piezas III
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user302454
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    For an answer see here. – Dietrich Burde Jan 31 '16 at 19:20
  • 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 $(z^2-x^2)(z^2-y^2)$. – TonyK Jan 31 '16 at 19:24
  • @DietrichBurde: this answer is quite partial. –  Jan 31 '16 at 19:24
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    @SubhadeepDey Just coprime or pairwise coprime? Because $15,20,12$ are coprime... – Bart Michels Jan 31 '16 at 19:39
  • It's $gcd(x_0, y_0, z_0) = 1 $. I think there are 3 satisfying pairs including $(15, 20, 12)$. But I want to check if the factorization is helpful... – user302454 Jan 31 '16 at 21:21
  • @SubhadeepDey yes, $x,y,z$ are coprime integers. – Dietrich Burde Jan 31 '16 at 21:28
  • It's easier, I think, to multiply with $x^2y^2$, set $w = \frac{xy}{z}$ - which must be an integer, since it's a rational number whose square is an integer, and then use the parametrisation of Pythagorean triples. $w$ must divide $\bigl(\gcd(x,y)\bigr)^2$, and $\gcd\bigl(x,y,\frac{xy}{w}\bigr) = 1$ then determines $\gcd(x,y)$ in terms of the parametrisation. That gives you a parametrisation of the (infinitely many) coprime solutions. – Daniel Fischer Jan 31 '16 at 21:49
  • you mean that $\Big(x, y, \frac{xy}{w}\Big) = \Big((m^2-n^2), (2mn), (m^2+n^2)\Big)$ ? – user302454 Jan 31 '16 at 22:06
  • Not quite, it's $\bigl((m^2-n^2)(m^2+n^2), 2mn(m^2+n^2), 2mn(m^2-n^2)\bigr)$. – Daniel Fischer Jan 31 '16 at 22:08
  • How so? I thought that all the pythagorean triplets are given by $\Big(k(m^2+n^2), 2kmn, k(m^2+n^2)\Big)$ and we get all the primitive triplets for $k=1$ – user302454 Jan 31 '16 at 22:12
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    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 $k = m^2+n^2$. – Daniel Fischer Jan 31 '16 at 23:24
  • @DietrichBurde, oh, I thought pairwise coprime. –  Feb 01 '16 at 10:56

1 Answers1

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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.

$$x=ac$$

$$y=bc$$

$$z=ab$$

individ
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