So I read in another thread about involutory functions, he claims for any real numbers $a$ and $b$, the function: $$f(x) = a + \frac{b}{x-a} = \frac{ax + (b-a^2)}{x-a}$$ satisfies $$f(f(x)) = a + \frac{b}{a + \frac{b}{x-a} - a} = a + (x-a) = x$$

And well, I thought I'd give it a try. So I want to find values of $a, b$ and $c$ at which the function below is involutory:

$$f(x) = \frac{x-a}{bx-c}$$

And I of course want to solve for $a, b$ and $c$. Now, I don't consider myself good enough to know in what direction I need to go to solve this.

My attempt was $$f(f(x)) = \frac{\frac{x-a}{bx-c}-a}{b\cdot\frac{x-a}{bx-c}-c} = x$$

$$\iff\frac{\frac{x-a-abx+ac}{bx-c}}{\frac{bx-ba-cbx+c^2}{bx-c}} = \color{red}{\frac{x-a-abx+ac}{bx-ba-cbx+c^2} = x}$$

Now, how would I continue on to find values for $a,b$ and $c$, which satisfy the equation I colored red? By glance I happened to notice that if $a=b=0$ and $c=-1$, the equation is satisfied; but I want to be able to do it algebraically.

Any help is appreciated, thanks.

Edit 1. I was thinking about using a matrix; however I haven't learned about them yet, but I'd be happy to study a solution using them to see if I can get a hang of it.


  • $\begingroup$ FYI: Still looking for an answer that's a little more aligned to my level. $\endgroup$ – B. Lee Oct 6 '14 at 17:24

Here's a proof from first principles. Suppose $f(x)$ is a Mobius transform i.e. $f(x)=\dfrac{ax+b}{cx+d}$ with $ad\neq bc$ (this prevents $f(x)$ from being a constant map.)

Suppose we want $f(f(x))=x$ for all $x$. In particular, we have $$-\frac{b}{a}=f(f(-b/a))=f(0)=\frac{b}{d}\implies b=0 \text{ or }a=-d$$ and

$$-\frac{d}{c}=f(f(-d/c))=f(\infty)=\frac{a}{c}\implies c=0 \text{ or }a=-d$$

Thus we have two relevant cases:

  • If $a\neq -d$ then $b=c=0\implies f(x)=\dfrac{ax}{d}$ and $1=f(f(1))=f(a/d)=a^2/d^2\implies a=d$. So $f(x)=x$ is just the identity map.

  • If $a=-d$ then $f(x)=\dfrac{ax+b}{cx-a}$. Then $$f(f(x))=f(\dfrac{ax+b}{cx-a})=\dfrac{a(ax+b)+b(cx-a)}{c(ax+b)-a(cx-a)}=\dfrac{a^2+bc}{a^2+bc}x=x$$ is satisfied identically. (Note that $a^2+bc=bc-ad\neq 0$ by assumption, so the cancellation is always valid.) Thus any nontrivial $f(x)=\dfrac{ax+b}{cx-a}$ is an involution.

  • $\begingroup$ Ah, you nailed it pretty good. There's few things on my mind now. Most notably, I wonder, how come for $x=0$ we get $ac-a=0$ and not $\frac{ac-a}{c^2-ba-cbx}$. I also tried entering the equation in Wolfram, and they only gave the solution $a=b=0$ and $c=0$, I don't know. I link the Wolfram input in my post. $\endgroup$ – B. Lee Oct 6 '14 at 16:54
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    $\begingroup$ @XMLParsing: Well, it'd have to be $\dfrac{ac-a}{c^2-ba}=0$ since there's still an $x$ on the bottom :). (That does raise the specter of $c^2=ab$, but that'd imply that the LHS diverges as $x\to 0$ rather than going to zero.) But I think I might add a second proof that goes directly from $f(x)=\dfrac{ax+b}{cx+d}$. $\endgroup$ – Semiclassical Oct 6 '14 at 16:59
  • $\begingroup$ This was a very though problem for me; too many new words in one day. I think I understood about half of what you said, hah. Anyway, thank you for your answer. Edit 1. Saw your edit, I would LOVE another approch, put it in another answer post if you will. That way users won't be confused later on. $\endgroup$ – B. Lee Oct 6 '14 at 17:02
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    $\begingroup$ We're not assuming that $x=0$ always. We're just checking whether it works for that particular choice, and in that chosen case the $cbx$ term in the denominator is zero as well. If we were doing $x=1$, then the $cbx$ term would matter. @XMLParsing $\endgroup$ – Semiclassical Oct 6 '14 at 19:20
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    $\begingroup$ @XMLParsing: Oof, your point just penetrated my thick skull. You're right, there is a hole in the first one; I could repair it via an extra case, but that's pretty tedious. The second proof should be waterproof in any regard, so I'm inclined to remove the first attempt altogether. $\endgroup$ – Semiclassical Oct 6 '14 at 19:31

factorizing your equation we obtain $(c-1)(bx^2-x(c+1)+a)=0$


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