Triples of positive integers $a,b,c$ with rational $\sqrt{\frac{c-a}{c+b}},\sqrt{\frac{c+a}{c-b}}$ While working on a physics problem, I came up with a certain question in number theory:

For positive integers $c>b>a$, can $\dfrac{c-a}{c+b}$ and $\dfrac{c+a}{c-b}$ both be rational squares?

I asked this question on MSE chat (link) and a number of small solutions were quickly found, e.g.  $$(a,b,c)=(2,13,14), (11,13,14), (9,23,27), (16,56,65).$$ A list of other such examples was provided by DHMO during that conversation.
These solutions are moreover primitive, in that further solutions can be generated from them by multiplying $(a,b,c)$ through by a common integer square. In that respect, the problem is analogous to that of finding Pythagorean triples. In that case, there is a well-known formula for generating a (primitive) Pythagorean triple from a pair of (coprime) integers.
What I want to know for this question: Is there an analogous formula that generates primitive triples satisfying the above condition?
 A: In order to get all solutions, it helps to write the equations with a term that cancels between numerator and denominator (and thus doesn't need to be a square like the rest of the numerators and denominators):
$$\left\{\begin{aligned}\frac{c-a}{c+b}&=\frac{q\times r^2}{q\times s^2}\\\frac{c+a}{c-b}&=\frac{t\times u^2}{t\times v^2}\end{aligned}\right.$$
Breaking that into separate numerators and denominators, SageMathCell gives:
$$\begin{align}
a &= -\frac{s^{2} t u^{2} - {\left(2 \, t u^{2} - t v^{2}\right)} r^{2}}{2 \, {\left(r^{2} - s^{2}\right)}}\\
b &= -\frac{r^{2} t v^{2} + {\left(t u^{2} - 2 \, t v^{2}\right)} s^{2}}{2 \, {\left(r^{2} - s^{2}\right)}}\\
c &= -\frac{s^{2} t u^{2} - r^{2} t v^{2}}{2 \, {\left(r^{2} - s^{2}\right)}}\\
q &= -\frac{t u^{2} - t v^{2}}{r^{2} - s^{2}}
\end{align}$$
Sage returns no solutions when solving for just $a, b, c$, but I'm not sure why. I'm also not enough of a mathematician to know if all of the above is remotely helpful.

Edit: As an example of where the extra variables make things nicer:
solve([q*r^2 == (14-2), q*s^2 == (14+13), t*u^2 == (14+2), t*v^2 == (14-13)], (q,s,t,v))
$$\left[q = \frac{12}{r^{2}}, s = \frac{3}{2} \, r, t = \frac{16}{u^{2}}, v = \frac{1}{4} \, u\right]$$
vs solving for (r,s,u,v), which gives
$$\left[r = \frac{2 \, \sqrt{3}}{\sqrt{q}}, s = -\frac{3 \, \sqrt{3}}{\sqrt{q}}, u = -\frac{4}{\sqrt{t}}, v = \frac{1}{\sqrt{t}}\right]$$
A: Solve the system.
$$\left\{\begin{aligned}&\sqrt{\frac{c-a}{c+b}}=\frac{k}{t}\\&\sqrt{\frac{c+a}{c-b}}=\frac{p}{s}\end{aligned}\right.$$
The solution can be written as.
$$c=t^2p^2-k^2s^2$$
$$a=t^2p^2+k^2s^2-2k^2p^2$$
$$b=t^2p^2+k^2s^2-2t^2s^2$$
