# Cyclic Function of Order 5

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

Question
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.