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 $$\begin{array}{rl}{a}_1+a_2+\cdots +a_n& =a_1+(a_2+a_3+a_4)+(a_5+a_6+\cdots +a_{16})\\ & +\cdots +(a_{4^s-1}+1)+\cdots +a_{4^s}.\\ & =4^0\cdot 4^0+(4^1-4^0)\cdot 4^1+\cdots +(4^s-4^{s-1})\cdot 4^s\\ & =\frac{4^{2s+1}+1}{5}=\frac{4n^2+1}{5}.\end{array}$$ 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 $$\begin{array}{rl}{a}_1+a_2+\cdots +a_n& =a_1+(a_2+a_3+a_4)+\cdots +(a_{4^s-1}+1)+\cdots +a_{4^s}\\ & +(a_{4^{s+1}}+\cdots +a_n)\\ & =4^0\cdot 4^0+(4^1-4^0)\cdot 4^1+\cdots +(4^s-4^{s-1})\cdot 4^s\\ & +(n-4^s)\cdot 4^{s+1}\\ & =\frac{4^{2s+1}+1}{5}+(n-4^s)\cdot 4^{s+1}\\ & =\frac{5n\cdot 4^{s+1}-4^{2s+2}+1}{5}=\frac{5n{a}_n-a_n^2+1}{5}.\end{array}$$ This implies that $$\begin{array}{rl}{b}_n& =\frac{1}{n^2}\left(\frac{5n{a}_n-a_n^2+1}{5}-\frac{1}{5}\right)=\frac{5n{a}_n-a_n^2}{5n^2}\\ & =-\frac{1}{5}{\left(\frac{a_n}{n}\right)}^2+\frac{a_n}{n}=P\left(\frac{a_n}{n}\right).\end{array}$$ 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 $n^{\prime }=n-4^s$ ($1\le n^{\prime }<3\cdot 4^s$), we have $$b_n=\frac{4^{2s+1}+5n^{\prime }\cdot 4^{s+1}}{5(4^s+n^{\prime }{)}^2}=\frac{4+20\cdot \frac{n^{\prime }}{4^s}}{5{\left(1+\frac{n^{\prime }}{4^s}\right)}^2}.$$ 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 $n^{\prime }=\lfloor 4^s{x}_0\rfloor$, we have $\frac{n^{\prime }}{4^s}\le x_0$ and $\frac{n^{\prime }}{4^s}>\frac{4^s{x}_0-1}{4^s}=x_0-\frac{1}{4^s}$ so ${\lim }_{s\to \infty }\frac{n^{\prime }}{4^s}=x_0$. We deduce that $$\underset{s\to \infty }{\lim }{b}_{4^s+\lfloor 4^s{x}_0\rfloor }=f(x_0)=\frac{2024}{2025}.$$ The statement is proved. $\square$