real-valued functional equations on a closed interval

Question

Let $a,b \in \mathbb{R}$ with $a<b$. Prove that there exists non-constant real-valued functions $f$ and $g$, both defined on $[a,b]$, satisfying $f(x+y) = f(x) + \frac{f(y)g(x)}{1-f(x)f(y)}$ and $g(x+y) = \frac{g(x)g(y)}{\left(1-f(x)f(y)\right)^2}$ whenever $x,y,x+y \in [a,b]$.

I've been trying to brush up on functional equations, and this came up and stumped me. I have no intuition on how to start this or if it's even true. What's the key observation that leads to a short self-contained proof of this statement?

Answer 1

Approach: Wishful thinking. Lots of it. Esp since this is likely an olympiad problem with a nice solution.

Observe that the first equation looks like something very familiar, especially that denominator.

It looks like $ \tan (x+y) = \frac{ \tan x + \tan y } { 1 - \tan x \tan y} $.

Now, guess what $f(x)$ could be.

Wishful thinking. Lets see what happens with $ f(x) = \tan x $.
If this gets ugly subsequantly, we can try to modify it, e.g. $f(x) = \tan kx, f(x) = k \tan x, \ldots$

Now, guess what $g(x)$ is.

$\tan (x + y ) = \tan x + \frac{ (\tan y ) ( 1 + \tan^2 x ) } { 1 - \tan x \tan y } $.
So $g(x) = 1 + \tan ^2 x $.

Now, verify that the second equation holds.

Just do it. Thank goodness it works out.
If it didn't hold, go back and edit the original guess for $f(x)$.