If $f^m(x)=x$ then $f^2(x)=x$

Question

$f:\mathbb{R}\rightarrow\mathbb{R}$ and $f(x)$ is continuuos then if there is an $m$ which for that $m$, $f^m(x)=x$ then prove $f^2(x)=x$. ($f^m(x)$ means $m$ times $f\circ f\circ ...\circ f\circ f(x)$.)

Answer 1

Here's a proof sketch:

1. Since $f^m(x) = x$ is a bijection, $f(x)$ is also a bijection.

2. Since $f(x)$ is a continuous bijection $\mathbb{R}\to\mathbb{R}$, it is strictly monotonic.

3. Since $f(x)$ is strictly monotonic, $g(x) = f^2(x)$ is strictly increasing.

4. Since $g(x)$ is strictly increasing, then $g(x) < x \implies g^m(x) < x$ (similarly for $>$)

5. Since $g^m(x) = x,$ it follows that $f^2(x) = g(x) = x.$

You'll still need to justify each step (and in one step, you might need to briefly justify the assumption), but it should be straight-forward to do so.