Let us have $m,n$ positive integers, and suppose, that $ f o f ... f(m$ times$)$ and $f o f... f(n$ times$)$ have an $x$ fix point. For what $(m,n)$ positive integers will it be true, that $x$ is a fix point for $f$ as well? ($o$ means composition of functions, and an $x$ fix point for an $f$ function means, that $f(x) = x$ stands.
I don't have any ideas, how to answer this, any help will be appreciated! :)
The answer is that if $m$ and $n$ are relatively prime, that is, if their g.c.d. is $1$, then $f(x) = x$ stands.
To see this, let $m>n$ and consider $$f^m(x) = f(f(f(\cdots m \text{ times } \cdots (x)))\cdots) = f(f(f(\cdots m-n \text{ times } \cdots f(f( (\cdots n \text{ times } \cdots (x)))\cdots) = f(f(f(\cdots m-n \text{ times } \cdots (x)))\cdots) $$ because the $f^n(x)$ is just $x$. So $x$ is a fixed point of $f^{m-n}$. Now if $m-n > n$ repeat this, to subtract off another $n$, until we have that $x$ is a fixed point of $f^{m-kn}$ with $m-kn < n$.
Now flip the game, starting with $f^n(x)$ and showing that $x$ is a fixed point of $f^{n-\ell(m-kn)}(x)$. You can see that what we are doing here is isomorphic to doing Euclid's algorithm for finding the g.c.d of $m$ and $n$. If the answer is $1$ then in the end we have $f(x) = x$.
If g.c.d. $(m,n) = s > 1$ then that shows that $f^s(x) = x$ but it does not prove that $x$ is not a fixed point of $f$ as well. So $m$ and $n$ are not relatively prime, , then $f(x) = x$ might or might not stand. Consider for example, the identity function, and take $m=12, n=6$.