1993 Putnam A2

Let $(x_n)_{n \geq 0}$ be a sequence of nonzero real numbers

such that $x_n^2 - x_{n-1}x_{n+1} = 1$ for $n=1,2,3,\dots$. Prove there

exists a real number $a$ such that $x_{n+1} = ax_n - x_{n-1}$ for all $n

\geq 1$.

设 $(x_n)_{n \geq 0}$ 为非零实数序列

使得 $x_n^2 - x_{n-1}x_{n+1} = 1$ 对于 $n=1,2,3,\dots$。证明那里

存在一个实数 $a$，对于所有 $n，$x_{n+1} = ax_n - x_{n-1}$

\geq 1$。