Vietnamese MO

a) We will prove that the polynomial $P(x) = -\frac{1}{5}x^2 + x$ satisfies the required properties. Obviously $b_1 = \frac{4}{5} = P(1) = P\left(\frac{1}{1}\right)$, so we only need to consider the case $n > 1$.
Notice that, for each non-integer real number $x$, then $[-x] = -[x] - 1$. Thus, for each positive integer $m > 1$, let $t$ be a natural number such that $4^t < m \le 4^{t+1}$, we have
* If $m = 4^{t+1}$, then $[-\log_4 m] = -t - 1$ and therefore $a_m = 4^{t+1}$.
* If $4^t < m < 4^{t+1}$, then we have $t < \log_4 m < t + 1$. Thus $\log_4 m$ not integer, and so
$$[-\log_4 m] = -[\log_4 m] - 1 = -t - 1 = 4^{t+1}.$$
In short, we always have $a_m = 4^{t+1}$, where $t$ is a natural number such that $4^t < m \le 4^{t+1}$. Now, we consider the following cases.

Case 1: $n$ is a power of 4. In this case, we have $n = 4^s$ where $s$ is some positive integer. Then $a_n = n$ and
$$\begin{aligned} a_1 + a_2 + \dots + a_n &= a_1 + (a_2 + a_3 + a_4) + (a_5 + a_6 + \dots + a_{16}) \\ &\quad + \dots + (a_{4^s-1} + 1) + \dots + a_{4^s}. \\ &= 4^0 \cdot 4^0 + (4^1 - 4^0) \cdot 4^1 + \dots + (4^s - 4^{s-1}) \cdot 4^s \\ &= \frac{4^{2s+1} + 1}{5} = \frac{4n^2 + 1}{5}. \end{aligned}$$
This implies that
$$b_n = \frac{1}{n^2} \left( \frac{4n^2 + 1}{5} - \frac{1}{5} \right) = \frac{4}{5} = -\frac{1}{5} \left( \frac{a_n}{n} \right)^2 + \frac{a_n}{n} = P \left( \frac{a_n}{n} \right).$$

Case 2: $n$ is not a power of 4. In this case, there exists a natural number $s$ such that $4^s < n < 4^{s+1}$. Then we deduce that $a_n = 4^{s+1}$ and
$$\begin{aligned} a_1 + a_2 + \dots + a_n &= a_1 + (a_2 + a_3 + a_4) + \dots + (a_{4^s-1} + 1) + \dots + a_{4^s} \\ &\quad + (a_{4^{s+1}} + \dots + a_n) \\ &= 4^0 \cdot 4^0 + (4^1 - 4^0) \cdot 4^1 + \dots + (4^s - 4^{s-1}) \cdot 4^s \\ &\quad + (n - 4^s) \cdot 4^{s+1} \\ &= \frac{4^{2s+1} + 1}{5} + (n - 4^s) \cdot 4^{s+1} \\ &= \frac{5n \cdot 4^{s+1} - 4^{2s+2} + 1}{5} = \frac{5na_n - a_n^2 + 1}{5}. \end{aligned}$$
This implies that
$$\begin{aligned} b_n &= \frac{1}{n^2} \left( \frac{5na_n - a_n^2 + 1}{5} - \frac{1}{5} \right) = \frac{5na_n - a_n^2}{5n^2} \\ &= -\frac{1}{5} \left( \frac{a_n}{n} \right)^2 + \frac{a_n}{n} = P \left( \frac{a_n}{n} \right). \end{aligned}$$
In short, the polynomial $P(x) = -\frac{1}{5}x^2 + x$ satisfies the requirements of the problem.

b) According to the result of part a), with $n$ not being a power of 4, we have
$$a_1 + a_2 + \cdots + a_n = \frac{4^{2s+1} + 1}{5} + (n - 4^s) \cdot 4^{s+1},$$
where $s$ is a natural number such that $4^s < n < 4^{s+1}$.
Let $n' = n - 4^s$ ($1 \le n' < 3 \cdot 4^s$), we have
$$b_n = \frac{4^{2s+1} + 5n' \cdot 4^{s+1}}{5(4^s + n')^2} = \frac{4 + 20 \cdot \frac{n'}{4^s}}{5 \left(1 + \frac{n'}{4^s}\right)^2}.$$
Note that the function $f(x) = \frac{4+20x}{5(1+x)^2}$ is continuous on the segment $[0, \frac{3}{4}]$ and $f(0) < \frac{2024}{2025} < f(\frac{3}{4})$ so there exists a real number $x_0 \in (0, \frac{3}{4})$ such that $f(x_0) = \frac{2024}{2025}$. In addition, it is easy to see that there exists a positive integer $s_0$ such that $1 \le \lfloor 4^s x_0 \rfloor < 3 \cdot 4^s$ for all positive integers $s \ge s_0$. Now, choosing $n' = \lfloor 4^s x_0 \rfloor$, we have $\frac{n'}{4^s} \le x_0$ and $\frac{n'}{4^s} > \frac{4^s x_0 - 1}{4^s} = x_0 - \frac{1}{4^s}$ so $\lim_{s \to \infty} \frac{n'}{4^s} = x_0$. We deduce that
$$\lim_{s \to \infty} b_{4^s + \lfloor 4^s x_0 \rfloor} = f(x_0) = \frac{2024}{2025}.$$
The statement is proved. $\square$