Are there any solutions to the equation $s^{2}(1+t^{2})^{2}+t^{2}(1+s^{2})^{2}=u^2$ where $s,t,u\in \mathbb{Q}$ and $0 < s,t<1$? If so, is there a simple way to parametrize them all?


If I am understanding the geometry behind this problem, even if we pick a specific value for $t$ we are left with an elliptic curve, and it is possible there could be infinitely many solutions.


This question arises from some related (and somewhat esoteric) questions I have about rational points on unit circles. But this question looked "pretty" enough, I thought I'd ask the experts here.



Does the following Diophantine equation have nontrivial rational solutions?
(*Edited with more details.*)


I know this is a *really* old question, but this has a nice connection to a problem considered by Euler, what is now known as *Euler bricks*, and I couldn't resist. The OP's equation is equivalent to finding three rationals $a,b,c$ such that,


$$\begin{aligned}
a^2+b^2\; &= u\_1^2\\
a^2-c^2\; &= u\_2^2\\
b^2-c^2\; &= u\_3^2
\end{aligned}\tag1$$


A solution (essentially by Euler) is,


$$a = \frac{s^2+1}{2s},\quad b = \frac{t^2+1}{2t},\quad c = 1\tag2$$


where $s,t$ must satisfy,


$$s^2(t^2+1)^2+t^2(s^2+1)^2 = w^2\tag3$$


and is the equation considered by the OP. For ***any*** solution to $(1)$ with $c\neq0$, then $s,t$ can be recovered as,


$$s = \frac{a\pm\sqrt{a^2-c^2}}{c} = \frac{a\pm u\_2}{c}$$


$$t = \frac{b\pm\sqrt{b^2-c^2}}{c} = \frac{b\pm u\_3}{c}$$


A small solution to $(3)$ is,


$$s = \frac{4p}{p^2-1},\quad t = \frac{3p^2+1}{p(p^2+3)}\tag4$$


This is the essentially the same one given by Euler for *[Euler bricks](http://mathworld.wolfram.com/EulerBrick.html)*,


$$\begin{aligned}
\alpha^2+\beta^2\; &= v\_1^2\\
\alpha^2+\gamma^2\; &= v\_2^2\\
\beta^2+\gamma^2\; &= v\_3^2
\end{aligned}$$


where we have used $p \to p\sqrt{-1}$, and have correspondingly tweaked $(1)-(4)$. From this initial rational point $(4)$, one can then generate an infinite more.


