China Girls' Mathematical Olympiad

Let $b_0=0$, $b_1=1$, $b_n=4b_{n-2}+b_{n-1}$ ($n\ge 2$). Then $$b_n=\frac{(2+\sqrt{5}{)}^n-(2-\sqrt{5}{)}^n}{2\sqrt{5}}.$$ In particular, $b_6=1292$, $b_7=5473$. For every $k=1,2,\dots ,5473$, there are unique integers $x_k,y_k$ such that $1292k=x_k+5473y_k$, and $1\le x_k\le 5473$. Since $(1292,5473)=1$, $x_1,x_2,\dots ,x_{5473}$ is a permutation of $\{1,2,\dots ,5473\}$, and it is clear that $\{y_k\}$ is nondecreasing: $$y_1\le y_2\le \cdots \le y_{5473}=1291.$$ For our convenience, let $f(x)=x-\lfloor x\rfloor$. We have $$\begin{array}{rl}f(x_k\sqrt{5})& =f(1292k\sqrt{5}-5473y_k\sqrt{5})\\ & =f\left(\frac{(2+\sqrt{5}{)}^6-(2-\sqrt{5}{)}^6}{2}k-\frac{(2+\sqrt{5}{)}^7-(2-\sqrt{5}{)}^7}{2}{y}_k\right)\\ & =f(-(2-\sqrt{5}{)}^{6}k+(2-\sqrt{5}{)}^7{y}_k).\end{array}$$ Since $$\begin{array}{rl}& 0<(2-\sqrt{5}{)}^{6}k-(2-\sqrt{5}{)}^7{y}_k\\ & \le 5473(2-\sqrt{5}{)}^6-1291(2-\sqrt{5}{)}^7<1,\end{array}$$ it follows that $$f(x_k\sqrt{5})=1-(2-\sqrt{5}{)}^k+(2-\sqrt{5}{)}^7{y}_k,$$ and this is strictly decreasing. Now, since $x_1=1292$, $x_{5473}=5473$, $x_{5472}=4181$, $x_{5471}=2889$, $x_{5470}=1597$, we see that $a_{1292}$ attains the maximum and $a_{1597}$ the minimum.