Question:
(Spivak Calculus 3rd, Chapter 3, Problem 8)
For which numbers $a,b,c,d$ will the function
$$f(x) = \frac{ax + d}{cx + b}$$
satisfy $f(f(x)) = x$ for all $x$?
Attempt at an answer:
I tried to substitute $f(x)$ into itself, resulting in
$$f(f(x)) = \frac{a(\frac{ax + d}{cx + b}) + d}{c(\frac{ax + d}{cx + b}) + b} = \frac{(a^2 + cd)x + (a+b)d}{[(a+b)c]x+(b^2+cd)}$$
When this expression is set equal to $x$, I get an equation
$$[(a+b)c]x^2 + (b^2-a^2)x - (a+b)d = 0$$
Now, since this equation must hold for all $x$, I can select several arbitrary values, creating a system of as many equation as I want. This is where I am a little stuck; it seems that picking the correct values of $x$ is a matter of guessing and a little luck. Is there a better way to go about doing this?