A common example of a cyclic function of order $3$ is $$g(x) = \frac 1{1-x} $$ because $$g^3(x)=x.$$

Is there a similar type (i.e. rational function) of cyclic function which is of order $5$ instead?


Let $\phi=\frac{1+\sqrt{5}}{2}$, then consider $g(x)=\frac{1}{\phi-x}$.

It’s not possible if $g$ is a rational fraction with rational coefficients and $g$ isn’t just the identity. Indeed, this implies that $f \in \mathbb{C}(x) \longmapsto f\circ g \in \mathbb{C}(x)$ is surjective and thus that $g$ is injective from $\mathbb{C}$ to itself. It’s not hard to deduce that $g$ must be of “degree” one, that is, with both numerator and denominator of degree at most one. Now write $g(x)=\frac{ax+b}{cx+d}$, and let $A=\begin{bmatrix}a &b\\c&d\end{bmatrix}$.

You can easily compute that the conditions on $g$ force $A$ to be invertible, with rational entries and such that $A^5$ is scalar, say, $A^5=\alpha I_2$, $\alpha \neq 0$.

But then, this means that $\alpha$ has a fifth root (say, $\beta$) such that the characteristic polynomial of $A$ vanishes at $\beta$ – in particular $\beta$ is a element of a quadratic number field.

Write $\beta=u+v\sqrt{C}$, with $C$ being a square-free integer, as $\beta^5$ is rational, it follows that $C^2v^5+10Cv^3u^2+5vu^4=0$.

If $u=0$ or $v=0$, then $v=0$ or $C=0$, so $\beta$ is rational (and this implies that $A$ is diagonalizable in $\mathbb{Q}$ and then that $A$ is scalar and thus that $g(x)=x$).

So let $w=v/u \in \mathbb{Q}^*$ so that $P(w):=C^2w^4+10Cw^2+5=0$. If $5$ doesn’t divide $C$, then $P$ is irreducible by a slight elaboration on Eisenstein. If $C=5D$, then $D$ isn’t divisible by $5$ and $1/w$ is a root of $Q(x)=x^4+10Dx^2+5D^2$, which is irreducible by the same argument.


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