Lately, I've been toying with involutory functions (functions that are their own inverses), especially involutory rational functions with linear numerator and denominator. Then I noticed that some functions have the property that $f^n(x)=x$, where $n$ is greater than $2$. Call such a function an $n$-involution. So far I have proven the following (where $a$ represents some constant):
- If a function is of the form $f(x)=\frac{ax-a^2}{x}$, it is a $3$-involution.
- If a function is of the form $f(x)=\frac{ax-a^2}{x+a}$, it is a $4$-involution.
- If a function is of the form $f(x)=\frac{ax-a^2}{x+2a}$, it is a $6$-involution.
- If a function of the form $f(x)=\frac{ax+b}{cx+d}$ is an $n$-involution, then so is the function $g(x)=\frac{ax+c}{bx+d}$.
No matter how hard I try, I can't find a function of this form that is a $5$-involution. This problem essentially boils down to finding a $2$ x $2$ matrix whose fifth power is the $2$ x $2$ identity matrix, since the iterations of rational functions of that form can be mapped onto $2$ x $2$ matrices. How would I go about finding this? What about for a $7$-involution?