Let $p$ be an odd prime number and $\displaystyle E=\left(\mathbb{Z} / p \mathbb{Z} \right)\backslash \left\{ \overline{0}, \overline{1}\right\} $. Let $f : E \rightarrow E$ such that $$ f\left(\overline{x}\right)=\overline{1}-\overline{x}^{-1} $$ $\bullet$ I've to show that $f$ is a permutation and find its order.

$\bullet$ I've to show that the number of fixed point of $f$ is $\equiv \text{Card}\left(E\right) \ \text{mod} 3$.

$\bullet$ To conclude I need to find how many solutions they are to $\overline{x}^2-\overline{x}+1=0$ $, \ x \in \mathbb{Z}/ p \mathbb{Z}$

My idea was to show that $f$ was a bijection, then I think this proves it is a permutation of $E$. If $$ f\left(x\right)=y \text{ then } y=1-x^{-1} \Rightarrow xy=x-1 \Rightarrow x\left(y-1\right)=1 \Rightarrow y=x^{-1}+1 $$ So $f$ is a bijection and $f^{-1}: x \mapsto 1+x^{-1}$. Does this prove it is a permutation ? ( in fact I'm used to manipulate permutations on $\mathbb{N}$ so it disturbs me)

$$ f\circ f (x)=1-\left(1-x^{-1}\right)^{-1} $$ And I'm stuck here, but with the next question I guess $f\circ f \circ f \left(x\right)=x$.

  • $\begingroup$ This is not functional analysis. Functional analysis is the study of Banach spaces, Hilbert spaces, and (un)bounded operators. I removed the tag for you. $\endgroup$ Commented Jun 2, 2018 at 16:01
  • $\begingroup$ Does $1+\frac{1}{1-\frac{1}{x}} = x$? $\endgroup$ Commented Jun 2, 2018 at 16:23

1 Answer 1


The domain and range of the function $f$ are finite of the same size, so to show that it is a bijection, it suffices to show that it is surjective. Your work does this.

To find compositions, you can treat it as a normal rational function: $$ f(f(x)) = 1-\frac{1}{1-x^{-1}} = 1-\frac{x}{x-1} = \frac{1}{1-x}\\ f(f(f(x)) = 1 - (1-x) = x $$ So the order is three. Note that you could just do this work to show that it is a bijection, making sure to note that we're excluding $1$ and $0$ from the domain.

For the second problem, you just need to solve $1-\frac{1}{x} = x$, or $x^2-x+1=0$ mod $p$. The hint here is to use Quadratic Reciprocity. This will allow you to count the number of solutions that this equation has and, hence, the number of fixed points.


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