Are there nontrivial rational solutions to $\sqrt{1-x^2} + \sqrt{1-y^2} = \sqrt{1-z^2}$? Obviously $(a,0,a)$,$(-a,0,a)$,$(0,a,a)$,$(0,-a,a)$ are solutions.
I tried finding solutions brute-forcing this problem, but I discovered there are no solutions with numerator and denominator smaller than 100'000.
I can prove that the question is equivalent to finding a bunch of natural numbers that satisfy:
$x'^2+ka^2 = w^2\\y'^2+kb^2 = w^2\\z'^2+kc^2 = w^2\\a+b=c$
(here is $x=x'/w$, $y=y'/w$, $z=z'/w$)
It can be proven they are equivalent as follows:
To go from the title to the 4 equations:

*

*Call $w$ the lcm of the denominators of x, y, z

*Define $x'=xw$, $y'=yw$, $z'=zw$

*The equation is now $\sqrt{w^2-x^2} + \sqrt{w^2-y^2} = \sqrt{w^2-z^2}$

*This equation can only hold if all sqrt are natural numbers, up to a common factor $k$ by this theorem

*Now call $a=\sqrt{w^2-x^2}/\sqrt{k}$, $b=\sqrt{w^2-y^2}/\sqrt{k}$, $c=\sqrt{w^2-z^2}/\sqrt{k}$

*The original equation becomes $a+b=c$, the other 3 can be calculated by the definitions of the previous step.

To go from the 4 equations to the title:

*

*Rewrite the first 3 equations to $a=\sqrt{w^2-x^2}/\sqrt{k}$, $b=\sqrt{w^2-y^2}/\sqrt{k}$, $c=\sqrt{w^2-z^2}/\sqrt{k}$

*Plug in in equation 4.

*Substitude $x=x'/w$, $y=y'/w$, $z=z'/w$
But apparently being able to split a number into two squares in at least three different ways, is quite a rare property, so I would suspect that finding three with the extra property that $a+b=c$ is probably impossible, which makes me believe that there are no solutions, or that solutions are very rare.
But is there a way to find a solution, or prove that non exists except for the trivial case?
Edit: Added proof of equivalence.

Here is a way to generate solutions for $k=3$.
It is well known that every number whose prime factors are all of the form $6m+1$, is also of the form $n=a^2+3b^2$.  Then
$$(2n)^2=4a^4+24a^2b^2+36b^4\\
=(2a^2-6b^2)^2+3(4ab)^2\\
=(a^2+6ab-3b^2)^2+3(a^2-2ab-3b^2)^2\\
=(a^2-6ab-3b^2)^2+3(a^2+2ab-3b^2)^2$$
Note two of the unsquared left-hand terms add up to the third, and two of the unsquared right-hand terms add up to the third.
There are also solutions for $k$ not equal to 3.
