For a given rational number $0<q<1$ let $r(q)$ be $$ r(q) = \frac{2q(1-q^2)}{(1+q^2)^2}, \quad q \in (0, 1) $$ Is there a triple $(a, b, c)$ of rational numbers such that $$ r(a)+r(b)=r(c), \quad a,b,c \in (0,1)? $$
(Edit 1: This question arose when I was playing with magic squares of squares, which, to my knowledge, is an unsolved problem. I am not very familiar with elliptic curves and such, that's why I tried to brute force search a solution and didn't find any after trying about a $1000$ rationals ($10^9$ triples). Some insight on the problem could be very helpful.)
Edit 2: As @ThomasAndrews correctly mentioned in the comments, the range of $r$ is in fact the set of values $uv$ such that $u^2+v^2=1$. The problem above actually follows from the system $$ \begin{cases} s_1^2+r_1^2=1,\\ s_2^2+r_2^2=1,\\ s_3^2+r_3^2=1,\\ s_3r_3=s_1r_1+s_2r_2.\\ \end{cases} $$ I found that all rational points on the unit circle can be parameterized as $$ s = \frac{2q}{1+q^2}, \quad r = \frac{1-q^2}{1+q^2}, \quad q \in \mathbb{Q}, $$ hence the last equation is the real problem in the system.
Edit 3: The fact that the set $r([0,1])$ is dense in the interval $\left(0, \frac{1}{2}\right)$ (since $r$ is a continious function) also might help.