8. 57 Given the sequence $\left\{a_{n}\right\}$ with the general term
$$a_{n}=15 n+2+(15 n-32) \cdot 16^{n-1}, n=0,1,2, \cdots$$
(1) Prove that for each non-negative integer $n, 15^{3} \mid a_{n}$.
(2) Find all $n$ such that
$$1991\left|a_{n}, 1991\right| a_{n+1}, 1991 \mid a_{n+2} \text {. }$$
[Solution] (1) Since
$$\begin{aligned}
2 a_{n+1}-a_{n}= & 2\left[15(n+1)+2+(15 n-17) \cdot 16^{n}\right]-[15 n+2+(15 n \\
& \left.-32) \cdot 16^{n-1}\right] \\
= & 15 n+30+2+(31 \times 15 n-32 \times 16) \cdot 16^{n-1} \\
= & a_{n+2}+(31 \times 15 n-32 \times 16) \cdot 16^{n-1}-(15 n-2) \cdot \\
& 16^{n+1} \\
= & a_{n+2}-15^{3} n \cdot 16^{n-1},
\end{aligned}$$

Therefore, $a_{n+2}=2 a_{n+1}-a_{n}+15^{3} n \cdot 16^{n-1}, n=0,1,2, \cdots$
Given $a_{0}=0, a_{1}=0$, it is easy to prove by induction that $15^{3} \mid a_{n}$.
(2) Let $p$ be a prime number greater than 5. If
$$p\left|a_{n}, p\right| a_{n+1}, p \mid a_{n+2}$$

then from the recurrence formula in (1), we can deduce that $p \mid n$. Using the general term formula, we can deduce that $16^{n} \equiv 1(\bmod p)$. Conversely, if $p \mid n, 16^{n} \equiv 1(\bmod p)$, it is easy to see from the general term formula that $p \mid a_{n}$. Also,
$$a_{n+1}=15 n\left(1+16^{n}\right)+17\left(1-16^{n}\right)$$

so $p \mid a_{n+1}$. Using the recurrence formula in (1), we can see that $p \mid a_{n+2}$. The above proves that $p \mid a_{n}$, $p\left|a_{n+1}, p\right| a_{n+2}$ if and only if $p \mid n$ and $16^{n} \equiv 1(\bmod p)$. If $n_{0}$ is the smallest natural number such that $16^{n} \equiv 1(\bmod p)$, then
$$\left\{n ; p\left|a_{n}, p\right| a_{n+1}, p \mid a_{n+2}\right\}=\left\{k\left[p, n_{0}\right] ; k=0,1,2, \cdots\right]$$

where $\left[p, n_{0}\right]$ denotes the least common multiple of $p$ and $n_{0}$.
Clearly, $1991=11 \times 181$, and both 11 and 181 are prime numbers. From $16^{n} \equiv 5^{n}(\bmod 11)$, it is easy to calculate that the smallest natural number $n$ such that $16^{n} \equiv 1(\bmod 11)$ is $n=5$. By Fermat's theorem, we know that $2^{180} \equiv 1(\bmod 181)$, i.e., $16^{45} \equiv 1(\bmod 181)$. If a natural number $n<45$ and
$$16^{n} \equiv 1(\bmod 181)$$

then $n \mid 45$, but $16^{1} \equiv 16(\bmod 181)$, and by calculation, we get
$$16^{5} \equiv 43(\bmod 181), 16^{9} \equiv 59(\bmod 181)$$

Thus, the smallest natural number $n$ such that $16^{n} \equiv 1(\bmod 181)$ is $n=45$. Therefore, the set of all $n$ such that $1991 \mid a_{n}$, $1991\left|a_{n+1}, 1991\right| a_{n+2}$ is
$$[5,11,45,181] k=11 \times 45 \times 181 k=89595 k$$

where $k=0,1,2, \cdots$