I am seeking to understand the structure of solutions to the diophantine equation $$\tan^2(\pi r) + \tan^2(\pi s) = 1.$$ I am conjecturing that there are no rational solutions $r, s \in \mathbb{Q}$ to the equation satisfying both $\tan(\pi r) >0$ and $\tan(\pi s) >0$. (This is so we exclude "trivial" solutions such as $r = 1$ and $s = \frac{1}{4}$, which imply $\tan(\pi r) = 0$ and $\tan(\pi s) = 1$).
On the other hand, if there are "non-trivial" rational solutions, can the form of the solutions be characterized?
I have been unable to make progress on this, so thought I'd ask for any thoughts. Thank you!
Hint: in picture:
Blue is $y_1=tan^2(\pi x)$
Red is : $y=1- tan^2(\pi x)$
The soluyions are $x=0.2, 0.8, 1.2, 1.8,\cdot\cdot\cdot$
for graphing I let $\pi=3.14$ and $3.14\times x$ is in radian, if you want to calculate $y$ by hand using a calculator , you need to convert it to degrees:
$x=0.2\Rightarrow \pi x=3.14\times 0.2\approx 0.2\times 180^o=36^o$
$tan^2 36*o\approx 0.5$
$tan^2 36+tan^2 36=1$
I will work on a synthetic solution and will include the result if I find something.
Update: For synthetic solutions we may write:
$tan^2 u=1-\frac{cos 2u}{cos^2 u}$ ; $u= \pi r$
$v=\pi s$
$$tan^2 (\pi r)=1- tan^2 (\pi s)$$
$$1-\frac{cos 2\pi r}{cos^2 (\pi r)}=1-\frac {sin^2(\pi s)}{cos^2 (\pi s)}$$
OR:
$$\frac{cos (2\pi r)}{cos^2 (\pi r)}=\frac {sin^2(\pi s)}{cos^2 (\pi s)}$$
The following system of equation is consistent:
$\begin{cases}cos (2\pi r)=sin^2(\pi s)\\cos^2(\pi r)=cos^2 (\pi s)\end{cases}$
which must be solved. $u=v\approx 36^o$ meets the condition of the original equation, other solutions comes from equating the numerators:
$cos u=cos v\Rightarrow u+v=2k\pi$ or $u-v=2k\pi$
Which gives :
$u=0\rightarrow v=2k\pi\pm\frac {\pi}4$
$v=0\rightarrow u=2k\pi\pm\frac {\pi}4$
Also if $u=v$, as picture shows, then:
$u=v\approx 2k\pi\pm \frac {\pi}{5}$