8-35 The sequence of integers $u_{0}, u_{1}, u_{2}, u_{3}, \cdots$ satisfies $u_{0}=1$, and for each positive integer $n$
$$u_{n+1} u_{n-1}=k u_{n}$$

where $k$ is some fixed positive integer. If
$$u_{2000}=2000$$

find all possible values of $k$.
［Solution］ Let $u_{1}=u$. By the problem statement,
$$u_{2}=\frac{k u_{1}}{u_{0}}=k u$$

If $u=0$, then $u_{1}=0, u_{2}=0$. If $u_{t-1}=0$, then $k u_{t}=u_{t-1} \cdot$ $u_{t+1}=0$, hence $u_{t}=0$. Therefore, for any positive integer $n, u_{n}=0$. This contradicts $u_{2000}=2000$. Thus, $u \neq 0$.

By the problem statement, after calculation we get
$$\begin{array}{ll}
u_{3}=\frac{k u_{2}}{u_{1}}=k^{2}, & u_{4}=\frac{k u_{3}}{u_{2}}=\frac{k^{2}}{u}, \\
u_{5}=\frac{k u_{4}}{u_{3}}=\frac{k}{u}, & u_{6}=\frac{k u_{5}}{u_{4}}=1, \\
u_{7}=\frac{k u_{6}}{u_{5}}=u . &
\end{array}$$

From this, we can see that the given sequence is a periodic sequence with a period of 6. Therefore, by
$$2000=6 \times 333+2$$

we have
$$u_{2000}=u_{2}=k u,$$

Thus, $k$ should satisfy
$$\left\{\begin{array}{l}
k u=2000, \\
\frac{k}{u} \text { is an integer, }
\end{array}\right.$$