Originally Posted by

**Mollier** **Problem:**

For which numbers $\displaystyle a,b,c,d$ will the function

$\displaystyle f(x)=\frac{ax+b}{cx+d} $

satisfy $\displaystyle f(f(x))=x$ for all $\displaystyle x$?

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**Attempt:**

$\displaystyle

f(f(x))=\frac{a\left[\frac{ax+b}{cx+d}\right]+b}{c\left[\frac{ax+b}{cx+d}\right]+d}\\

= \frac{\left[\frac{a^2x+ab}{cx+d}\right]+b}{\left[\frac{acx+bc}{cx+d}\right]+d}\\

=\frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)}

$ Correct up to here.

By using polynomial division I get the following result:

$\displaystyle

\left(\frac{a^2+bc}{ac+cd}\right)+\frac{(ab+bd)-(bc+d^2)\left(\frac{a^2+bc}{ac+cd}\right)}{(ac+cd) x+(bc+d^2)}

$ The trouble with this is that you are about to find that a = –d, so the denominator of the fraction in parentheses is zero.

By doing some hopefully correct algebra I get:

$\displaystyle

(a^2+bc)x+(ab+bd)=x

$

1. $\displaystyle (ab+bd)=0 $

So, $\displaystyle a=-d$

2. $\displaystyle (a^2+bc)=1 $

So, $\displaystyle a=\sqrt{1-bc} $

If $\displaystyle b$ and $\displaystyle c$ have opposite signs, their product is negative, and they can be any number.

If on the other hand, their signs are the same, their product is positive and:

$\displaystyle -1 \leq b \leq 1 $

$\displaystyle -1 \leq c \leq 1 $

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Am I making this more complicated than it is?

How can I better express my results regarding $\displaystyle b$, and $\displaystyle c$?