$x\_{n+1}=4x\_n(1-x\_n)$ I already proved that for $x\_n\subset [0,1]$, $x\_n=sin^2(2\pi y\_n)$


with $y\_{n+1}=\begin{cases}2y\_n & 0 \le y\_n < 0.5 \\vee 2y\_n -1 & 0.5 \le y\_n < 1 \end{cases}$


Now I would like to prove that for an arbitrary number $m\in\mathbb N$ there exists an $x\in [0,1]$ of the recursion with period lenght m


I think it can be shown using the fact that, if I write $y\_n$ in the binary system as $y\_n=\sum\_{k=1}^\infty a\_{k,n}2^{-k}$ the recursion for $y\_n$ equivalent to $a\_{k,n+1}=a\_{k+1,n}$ is, but I dont know how. 



Logistic map periodic point
Essentially you have proved that the logistic map is conjugate to the doubling map $Tx=2x\bmod 1$. Now, $T$ is in turn conjugate to the shift map $\sigma:\Sigma\to\Sigma$, where $\Sigma$ is the space of infinite 0-1 words. 


More precisely, if 
$$
\pi(w\_1,w\_2,\dots)=\sum\_{n=1}^\infty w\_n2^{-n},
$$
then you have
$$
\pi \sigma = T\pi.
$$
Since $\pi$ is 1-1, except for a countable set of finite words, you can just take any word $u$ of length $m$ and the corresponding infinite word $w=uuuu\dots$. Clearly, $\sigma^m(w)=w$, i.e., $w$ is $\sigma$-periodic of period $m$. (With a little effort you can make this the *smallest* period.) 


For instance, $w=001001001\dots$ is of period 3. 


Now take $x:=\pi(w)$; it is $T$-periodic of period $m$. Finally, use you conjugate map to turn $x$ into an $m$-periodic point for the logistic map. 


