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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.

Vlodya
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  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Jul 05 '23 at 19:44
  • What's the motivation for this question? – Derivative Jul 05 '23 at 19:46
  • @Derivative I have edited the post to better express it, but basically it is related to the possible differences between squares in a magic square of squares (if there exists one) – Vlodya Jul 05 '23 at 20:04
  • @AnotherUser thanks a lot! – Vlodya Jul 05 '23 at 20:05
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    The range of $r$ is the set of values $uv,$ where $u,v$ are positive rationals and $u^2+v^2=1.$ Not sure how that helps. – Thomas Andrews Jul 05 '23 at 20:11
  • @ThomasAndrews that is actually kind of where the problem is coming from :) But great observation and thanks a lot! It could be really helpful in case I'm juat overcomplicating things – Vlodya Jul 05 '23 at 20:26
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    Well, give us everything you know. Don't make us guess. – Thomas Andrews Jul 05 '23 at 20:28
  • @ThomasAndrews sorry if the first version of the comment sounded a little bit rude, I misclicked and sent the unfinished comment. Thanks a lot again, I'll add more context. – Vlodya Jul 05 '23 at 20:33

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