# 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

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