That's a strange problem: Peter wrote: Let $ p$ be an odd prime $ p$ such that $ 2h \neq 1 \; \pmod{p}$ for all $ h \in \mathbb{N}$ with $ h < p - 1$ Take $ h=\frac{p+1}{2}$ so that $ 2h = 1 \; \pmod{p}$. Then we must have $ h\geq p-1$, implying that $ p=3$. Peter wrote: and let $ a$ be an even integer with $ a \in] \tfrac{p}{2}, p [$. There is only one such integer: $ a=2$. Peter wrote: The sequence $ \{a_n\}_{n \ge 0}$ is defined by $ a_{0} = a$, $ a_{n + 1} = p - b_{n}$ \; $ (n \ge 0)$, where $ b_{n}$ is the greatest odd divisor of $ a_n$. Show that the sequence $ \{a_n\}_{n \ge 0}$ is periodic and find its minimal (positive) period. This sequence is $ 2,2,2,2,\dots$ and its period is 1.

True. Maybe they mean $ 2^h\neq 1$?

$ a_n$ is even number from interval $ (p/2,p)$, number of distinct means is $ N=[\frac{p+1}{4}]$. Because exist function f, suth that $ a_{n+1}=f(a_n)$, we get exist m and t, suth that $ m+t\le N, a_{i+t}=a_i \ \forall i>m$. We can found $ a_n=g(a_{n+1})=2^{l_n}(p-a_{n+1})$, were $ l_n$ defined unique $ l_n=[log_2(\frac{p}{p-a_{n+1}})]$. Therefore $ m=0$ and $ a_{i+t}=a_i \ \forall i$.

I'll try to look up what it should really be.

If $ a_{n+1}=f(a_n)$ and $ a_n\in S$, were $ S$ is finite set $ \#S=N$. Then in $ a_1,a_2,...,a_{N+1}$ exist at least 2 $ a_i=a_j \ (j>i)$, therefore $ a_{k+t}=a_k \ \forall k\ge i, \ t=j-i$.

Peter wrote: True. Maybe they mean $ 2^h\neq 1$? Yes. That makes more sense. And the length of the period in this case equals the number of even integers in the interval $ I = \left(\frac {p}{2},p\right)$, no matter what value of $ a$ was chosen. In other words, every even integer from the interval $ I$ appears in the period. The length of the period here can be also expressed as $ \frac {p\pm 1}{4}$ (whichever is integer). Proof. Inequality $ 2^h\neq 1$ for every $ h < p - 1$ simply means that the multiplicative order of $ 2$ modulo $ p$ equals $ p - 1$. Furthermore, in this case we have $ p\equiv 3, 5\pmod{8}$ and the smallest positive integer $ g$ such that $ 2^g\equiv - 1\pmod{p}$ is $ g = \frac {p - 1}{2}$. The greatest odd divisor $ b(x)$ of an integer $ x$ can be expressed as $ b(x) = x\cdot 2^{ - k(x)}$ for some non-negative integer $ k(x)$. Since the number of integers in $ I$ is $ \frac {p - 1}{2}$, we trivially have the following estimate: \[ \sum_{x\in I} k(x) \leq \left\lfloor \frac {(p - 1)/2 + 1}{2}\right\rfloor + \left\lfloor \frac {(p - 1)/2 + 1}{2^2}\right\rfloor + \dots \leq \frac {p - 1}{2}. \] Now suppose that $ a_m, a_{m + 1}, \dots, a_{m + t - 1}$ form the period, where $ t$ is the length of the period. Then all numbers $ a_m, a_{m + 1}, \dots, a_{m + t - 1}$ must be different (otherwise the period would be smaller). Furthermore, since $ a_{n + 1} = p - b(a_n) \equiv - a_n\cdot 2^{ - k(a_n)}\pmod{p}$ for every $ n$, we also have \[ a_m = a_{m + t} \equiv a_m ( - 1)^t 2^{ - k(a_m) - k(a_{m + 1}) - \dots - k(a_{m + t - 1})}\pmod{p}, \] implying that \[ 2^{k(a_m) + k(a_{m + 1}) + \dots + k(a_{m + t - 1})} \equiv ( - 1)^t\pmod{p} \] which can take place only if $ k(a_m) + k(a_{m + 1}) + \dots + k(a_{m + t - 1}) = \frac {p - 1}{2}$ and the integers $ a_{m}, a_{m + 1}, \dots, a_{m + t - 1}$ represent all even integers from $ I$. Q.E.D.