China Mathematical Competition (Extra Test)

(1) We have $$a_n-a_{n+1}=a_n-\frac{a_n{a}_{n-1}+1}{a_n+a_{n-1}}=\frac{a_n^2-1}{a_n+a_{n-1}},n=1,2,\dots \stackrel{◯}{1}$$ If there exists a positive integer $n$ such that $a_{n+1}=a_n$, we get $$a_n^2=1\text{and}{a}_n+a_{n-1}\ne 0.$$ If $n=1$, we have $$|y|=1\text{ and }x\ne -y.\stackrel{◯}{2}$$ If $n>1$, then $$a_n-1=\frac{a_{n-1}{a}_{n-2}+1}{a_{n-1}+a_{n-2}}-1=\frac{(a_{n-1}-1)(a_{n-2}-1)}{a_{n-1}+a_{n-2}},n\ge 2,\stackrel{◯}{3}$$ and $$a_n+1=\frac{a_{n-1}{a}_{n-2}+1}{a_{n-1}+a_{n-2}}+1=\frac{(a_{n-1}+1)(a_{n-2}+1)}{a_{n-1}+a_{n-2}},n\ge 2.\stackrel{◯}{4}$$ Multiplying equations (3) and (4), we get $$a_n^2-1=\frac{a_{n-1}^2-1}{a_{n-1}+a_{n-2}}\cdot \frac{a_{n-2}^2-1}{a_{n-1}+a_{n-2}},n\ge 2.\stackrel{◯}{5}$$ From (5) we infer that $x$ and $y$ satisfy either (2) or $$|x|=1\text{ and }y\ne -x.\stackrel{◯}{6}$$ 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 $$\frac{a_n-1}{a_n+1}=\frac{a_{n-1}-1}{a_{n-1}+1}\cdot \frac{a_{n-2}-1}{a_{n-2}+1},n\ge 2.\stackrel{◯}{7}$$ Let $b_n=\frac{a_n-1}{a_n+1}$. Then, for $n\ge 2$, equation (7) becomes $$\begin{array}{rl}{b}_n& =b_{n-1}{b}_{n-2}=(b_{n-2}{b}_{n-3})b_{n-2}=b_{n-2}^2{b}_{n-3}\\ & =(b_{n-3}{b}_{n-4}{)}^2{b}_{n-3}=b_{n-3}^3{b}_{n-4}^2=\dots \end{array}$$ Then we get $$\frac{a_n-1}{a_n+1}={\left(\frac{y-1}{y+1}\right)}^{F_{n-1}}\cdot {\left(\frac{x-1}{x+1}\right)}^{F_{n-2}},n\ge 2,\stackrel{◯}{8}$$ here, $$F_n=F_{n-1}+F_{n-2},n\ge 2,F_0=F_1=1.\stackrel{◯}{9}$$ From (9) we get $$F_n=\frac{1}{\sqrt{5}}\left({\left(\frac{1+\sqrt{5}}{2}\right)}^{n+1}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{n+1}\right).\stackrel{◯}{10}$$ 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 $$a_n=\frac{(x+1{)}^{F_{n-2}}(y+1{)}^{F_{n-1}}+(x-1{)}^{F_{n-2}}(y-1{)}^{F_{n-1}}}{(x+1{)}^{F_{n-2}}(y+1{)}^{F_{n-1}}-(x-1{)}^{F_{n-2}}(y-1{)}^{F_{n-1}}},n\ge 0,\stackrel{◯}{11}$$ here $F_{n-1},F_{n-2}$ are determined by (10).