$x_{n+1}=4x_n(1-x_n)$ 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.