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This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with functions and polynomials, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Find all functions of the form

$$f(x)=\frac{b}{cx+1}$$

where $b,c$ are integers and, for all real numbers $x$ such that $f(f(f(x)))$ is defined, the following equation holds:

$$f(f(f(x)))=x .$$

So I tried to make the substitutions yielding:

$$\frac{b}{c(\frac{b}{c(\frac{b}{cx+1})+1})+1}=x.$$

and then I simplified to:

$(bc+1)(cx^2+x-b)=0$.

However, I'm not sure if this is correct, and even so, I'm not sure how to approach the problem from here.

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    $\begingroup$ You can use $f(f(x))$ is inverse of $f(x)$ $\endgroup$
    – jonsno
    Jun 24 '17 at 7:53
  • $\begingroup$ and where is the denominator? $\endgroup$ Jun 24 '17 at 7:59
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If $b=0$ or $c=0$ the function $f(x):={b\over cx+1}$ is constant, hence cannot satisfy $f^{\circ3}={\rm id}$. If $bc\ne0$ then $f$ is a Moebius transformation with matrix $$\left[\matrix{0& b\cr c&1\cr}\right]\ .$$ The map $f^{\circ3}$ then has matrix $$\left[\matrix{0& b\cr c&1\cr}\right]^3=\left[\matrix{bc& b(bc+1)\cr c(bc+1)&2bc+1\cr}\right]\ .$$ The condition $f^{\circ3}={\rm id}$ means that the last matrix should be a nonzero multiple of the identity matrix. This enforces $bc=-1$, and it is easily seen that $bc=-1$ is also sufficient. When $b$ and $c$ have to be integers then only the cases $b=-c=\pm1$ remain.

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Plug $x=0:$

$$f(f(f(0)))=\frac{b(bc+1)}{2bc+1}=0 \Rightarrow b=0 \ \ or \ \ \ bc=-1.$$

However, $b=0$ does not suit.

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    $\begingroup$ This is interesting, gives possible solutions, but don't you have to also check that it is true for all $x$? $\endgroup$
    – enzotib
    Jun 24 '17 at 8:05
  • $\begingroup$ yes but once you have the guess it's easy to check. Alternatively, note that moebius functions are determined by their values at 0,1 and $\infty$ (this could break down the difficulty), or three points you like. $\endgroup$ Jan 31 at 2:07
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Note that the condition $f(f(f(x)))=x$ is equivalent to $f(f(x))=f^{-1}(x)$ (assuming of course that $f$ is invertible). Then you just calculate $f(f(x))$ and $f^{-1}(x)$ and then solve for $b$ and $c$.

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By the way, $$f(f(f(x)))=x$$ it's $$\frac{b(bc+1+cx)}{(c^2b+c)x+2bc+1}=x,$$ which gives $b(bc+1)=0$, $c(bc+1)=0$ and $bc+1=0$, which gives $bc+1=0$.

And indeed, $f(x)=\frac{b^2}{b-x}$ for $b\neq0$ works.

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  • $\begingroup$ Oh yeah, thanks for pointing that out! I edited my attempt to reflect this. $\endgroup$
    – anonymous
    Jun 24 '17 at 9:00
  • $\begingroup$ @anonymous Welcome! $\endgroup$ Jun 24 '17 at 9:07

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