# Solution to Diophantine equation $\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$

I have to prove the following, but I don't know how to start.

The only solutions in positive integers of the equation $$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2} \qquad \gcd(x,y,z)=1$$ are given by $$x=2st(s^2+t^2) \qquad y=s^4-t^4 \qquad z=2st(s^2-t^2)$$ where $s,t$ are relatively prime positive integers, one of which is even, with $s>t$.

I tried the following. After multiplying the equation by $x^2y^2z^2$, you get: $$(yz)^2+(xz)^2=(xy)^2,$$ which is a Pythagorean equation of the form $x^2+y^2=z^2$, which has solutions $x=2st$, $y=s^2-t^2$ and $z=s^2+t^2$ with some conditions. But I don't think this is a good approach. Any hints on how to start/continue with this problem are very much appreciated!

• maybe rewriting it as $z^4 = (z^2 - x^2)(z^2 - y^2)$ might help .. (just a suggestion) – r9m Aug 3 '15 at 17:55

## 1 Answer

Your instincts were good, this is just like the Pythagorean problem. Look at the unit circle $$X^2+Y^2=1$$. We know, from stereographic projection or other means, that the rational solutions are given by $$(X,Y)=\left(\frac {s^2-t^2}{s^2+t^2},\frac{2st}{s^2+t^2}\right)$$ For Pythagorus we clear the denominators but you would prefer that we clear the numerators, which we do by dividing by a common multiple, namely $(s^2-t^2)(2st)$. Inspection shows that we have achieved the desired form.

• Thanks for your answer. I'm still trying to solve it without using the unit circle. The proof for $x^2+y^2=z^2$ starts with that $x$ should be taken even and $y$ and $z$ odd. I tried a similar approach for $(yz)^2+(xz)^2=(xy)^2$, but that doesn't seem to work here. – user95864 Aug 4 '15 at 8:01
• @user95864 Well, all three problems are equivalent (rational points on circle, Pythagorean triples, your problem). If you have a solution you like for any of them then you get the others. You want to repeat Euclid's argument for your problem without going through Pythagorus? Should be possible (though the degree 4 bit worries me). – lulu Aug 4 '15 at 11:05
• Yes, that's where I got stuck. The book I'm working with does not use the circle, only Pythagorean triples and their properties. So I tried to use that. – user95864 Aug 4 '15 at 11:11
• @user95864 As I say, if you have the Pythagorean triples, then you can get the rational points on the circle (same parametrization that appears in my solution). I tend to think of it geometrically (via stereographic projection) but you can do it algebraically (as I expect your book does). Doesn't matter. Once you have that form, then you can proceed as I did. Going from scratch...well, as you've seen the factoring tricks are a lot harder to unwind. In theory, I suppose it has to work...but it sure doesn't have to be pleasant. Good luck! – lulu Aug 4 '15 at 11:16