CMO, 2024

Proof. First, since $\alpha$ is an irrational number, the integer part operation in the definition of $x_{n+1}$ always makes the corresponding number strictly smaller. For any integer $u$ satisfying $\frac{1}{\alpha-1} < u \le L$, we have $\lfloor \alpha u \rfloor > \alpha u - 1 > u$ and $\lfloor \alpha u \rfloor \le \alpha L$; while for any integer $v \ge L+1$, we have $\lfloor \frac{v}{\alpha} \rfloor > \frac{L}{\alpha} - 1 > \frac{1}{\alpha-1}$ and $\lfloor \frac{v}{\alpha} \rfloor < v$. Therefore, it is easy to verify by mathematical induction that for any positive integer $n$,
$$\frac{1}{\alpha - 1} < x_n \le \max\{x, \alpha L\},$$
which implies that the sequence $\{x_n\}$ is bounded. Since $\{x_n\}$ is a recursive sequence, it must eventually become periodic. Let $T$ denote the smallest positive period of $\{x_n\}$. By definition, there exists a positive integer $N$ such that for any integer $n \ge N$, we have $x_{n+N} = x_n$.
Since $\lfloor \alpha L \rfloor > \alpha L - 1 > L$, it follows that $\lfloor \alpha L \rfloor \ge L + 1$. Thus, for any integer $n \ge N$, we have $x_n \le \lfloor \alpha L \rfloor$. Consequently, for any integer $n \ge N$, if $x_n \ge L + 1$, then
$$x_{n+1} = \lfloor \frac{x_n}{\alpha} \rfloor < \frac{1}{\alpha} \cdot \alpha L = L,$$
and
$$x_{n+2} = \lfloor \alpha x_{n+1} \rfloor.$$
Clearly, there exists an integer $n \ge N$ such that $x_n > L$. Let $m \ge N + 1$ be the smallest integer such that
$$x_{m-1} = \min \{x_n \mid n \ge N \text{ and } x_n > L\}.$$
From the previous analysis, we know
$$x_m = \lfloor \frac{x_{m-1}}{\alpha} \rfloor < L \quad \text{and} \quad x_{m+1} = \lfloor \alpha x_m \rfloor > \alpha x_m - 1 > x_m.$$
Note that $x_{m+1} < \alpha x_m < \alpha \cdot \frac{x_{m-1}}{\alpha} = x_{m-1}$. By the minimality of $x_{m-1}$, we know $x_{m+1} \le L$. Since $\alpha x_m < x_{m+1} + 1 \le L + 1$, it follows that
$$x_m < \frac{L+1}{\alpha},$$

and hence $x_m \le \lfloor \frac{L+1}{\alpha} \rfloor$. Moreover, since $x_{m-1} \ge L+1$, we have
$$x_m = \lfloor \frac{x_{m-1}}{\alpha} \rfloor \ge \lfloor \frac{L+1}{\alpha} \rfloor.$$
Therefore, $x_m = \lfloor \frac{L+1}{\alpha} \rfloor$. By the minimality of $x_{m-1}$, we conclude that for any integer $n \ge n_0$, $x_n \ge x_m$.
For any integer $n \ge N$, if $x_m + 1 \le x_n \le L$, then
$$x_{n+1} = \lfloor \alpha x_n \rfloor > \alpha(x_m + 1) - 1 > \alpha \cdot \frac{L+1}{\alpha} - 1 = L,$$
which implies $x_{n+1} \ge L+1$. Moreover,
$$x_{n+2} = \lfloor \frac{x_{n+1}}{\alpha} \rfloor < \frac{x_{n+1}}{\alpha} < \frac{1}{\alpha} \cdot \alpha x_n = x_n.$$
Combining the above analysis, we have
$$x_m < x_{m+1} \le L < x_{m+2}.$$
Furthermore, for each $i = 1, 2, \dots, T-1$, when $i$ is odd, $x_{m+i} \le L$; when $i$ is even, $x_{m+i} > L$. Note also that $x_{m+1} > x_{m+3} > \dots > x_{m+T} = x_m$. Therefore, $T$ is an odd number.
Finally, note that $x_m = \lfloor \frac{L+1}{\alpha} \rfloor$ is independent of $x$. Thus, the minimal positive period $T$ is also independent of $x$.