8-25 Let $p_{0}>2$ be a given prime number, and the sequence $a_{0}, a_{1}, a_{2}, \cdots$ satisfies $a_{0}=0$, $a_{1}=1$, and
$$
a_{n+1}=2 a_{n}+(a-1) a_{n-1}, n=1,2, \cdots
$$

where the natural number $a$ is a parameter. Find the smallest value of $a$ that satisfies the following two conditions.
(i) If $p$ is a prime and $p \leqslant p_{0}$, then $p \mid a_{p}$;
(ii) If $p$ is a prime and $p>p_{0}$, then $p \times a_{p}$.
［Solution］From the recurrence relation and $a_{0}=0, a_{1}=1$, we have
$$
a_{n}=\frac{1}{2 \sqrt{a}}\left[(1+\sqrt{a})^{n}-(1-\sqrt{a})^{n}\right] .
$$

Thus, for any odd prime $p$,
$$
\begin{aligned}
a_{p} & =\frac{1}{2 \sqrt{a}}\left[\sum_{k=0}^{p} C_{p}^{k} \frac{k}{2}-\sum_{k=0}^{p}(-1)^{k} C_{p}^{k} a^{\frac{k}{2}}\right] \\
& =\frac{1}{2 \sqrt{a}}\left(2 p \sqrt{a}+2 C_{p}^{3} a^{\frac{3}{2}}+\cdots+2 C_{p}^{p-2} a \frac{p-2}{2}+2 C_{p}^{p} a^{\frac{p}{2}}\right) \\
& =p+C_{p}^{3} a+\cdots+C_{p}^{p-2} a \frac{p-3}{2}+a^{\frac{p-1}{2}} .
\end{aligned}
$$

Since $C_{p}^{k}=\frac{p!}{k!(p-k)!}$, for $1 \leqslant k \leqslant p-1$ we have
$$
p \mid C_{p}^{k} \text {. }
$$

This implies
$$
\left.p\left|a_{p} \Leftrightarrow p\right| a \frac{p-1}{2} \Leftrightarrow p \right\rvert\, a .
$$

Given $a_{2}=2$, the smallest value of $a$ is the product of all primes in the interval $\left[3, p_{0}\right]$.