China Mathematical Competition (Extra Test)

(1) We have
$$a_n - a_{n+1} = a_n - \frac{a_n a_{n-1} + 1}{a_n + a_{n-1}} = \frac{a_n^2 - 1}{a_n + a_{n-1}}, \quad n = 1, 2, \dots \quad \textcircled{1}$$
If there exists a positive integer $n$ such that $a_{n+1} = a_n$, we get
$$a_n^2 = 1 \quad \text{and} \quad a_n + a_{n-1} \neq 0.$$
If $n=1$, we have
$$|y|=1 \text{ and } x \neq -y. \qquad \textcircled{2}$$
If $n > 1$, then
$$a_n - 1 = \frac{a_{n-1}a_{n-2} + 1}{a_{n-1} + a_{n-2}} - 1 = \frac{(a_{n-1}-1)(a_{n-2}-1)}{a_{n-1} + a_{n-2}}, \quad n \ge 2, \qquad \textcircled{3}$$
and
$$a_n + 1 = \frac{a_{n-1}a_{n-2} + 1}{a_{n-1} + a_{n-2}} + 1 = \frac{(a_{n-1}+1)(a_{n-2}+1)}{a_{n-1} + a_{n-2}}, \quad n \ge 2. \qquad \textcircled{4}$$
Multiplying equations (3) and (4), we get
$$a_n^2 - 1 = \frac{a_{n-1}^2 - 1}{a_{n-1} + a_{n-2}} \cdot \frac{a_{n-2}^2 - 1}{a_{n-1} + a_{n-2}}, \quad n \ge 2. \qquad \textcircled{5}$$
From (5) we infer that $x$ and $y$ satisfy either (2) or
$$|x|=1 \text{ and } y \neq -x. \qquad \textcircled{6}$$
Conversely, if $x$ and $y$ satisfy either (2) or (6), then $a_n =$ constant when $n \ge 2$ and the constant can only be either $1$ or $-1$.

(2) From (3) and (4), we get
$$\frac{a_n - 1}{a_n + 1} = \frac{a_{n-1} - 1}{a_{n-1} + 1} \cdot \frac{a_{n-2} - 1}{a_{n-2} + 1}, \quad n \ge 2. \qquad \textcircled{7}$$
Let $b_n = \frac{a_n - 1}{a_n + 1}$. Then, for $n \ge 2$, equation (7) becomes
$$\begin{aligned} b_n &= b_{n-1}b_{n-2} = (b_{n-2}b_{n-3})b_{n-2} = b_{n-2}^2 b_{n-3} \\ &= (b_{n-3}b_{n-4})^2 b_{n-3} = b_{n-3}^3 b_{n-4}^2 = \dots \end{aligned}$$
Then we get
$$\frac{a_n - 1}{a_n + 1} = \left(\frac{y-1}{y+1}\right)^{F_{n-1}} \cdot \left(\frac{x-1}{x+1}\right)^{F_{n-2}}, \quad n \ge 2, \qquad \textcircled{8}$$
here,
$$F_n = F_{n-1} + F_{n-2}, \quad n \ge 2, \quad F_0 = F_1 = 1. \qquad \textcircled{9}$$
From (9) we get
$$F_n = \frac{1}{\sqrt{5}} \left( \left( \frac{1+\sqrt{5}}{2} \right)^{n+1} - \left( \frac{1-\sqrt{5}}{2} \right)^{n+1} \right). \qquad \textcircled{10}$$
The range of $n$ in (10) can be extended to negative integers. For example, $F_{-1} = 0, F_{-2} = 1$. Since (8) holds for any $n \ge 0$, we get
$$a_n = \frac{(x+1)^{F_{n-2}} (y+1)^{F_{n-1}} + (x-1)^{F_{n-2}} (y-1)^{F_{n-1}}}{(x+1)^{F_{n-2}} (y+1)^{F_{n-1}} - (x-1)^{F_{n-2}} (y-1)^{F_{n-1}}}, \quad n \ge 0, \qquad \textcircled{11}$$
here $F_{n-1}, F_{n-2}$ are determined by (10).