8.57 Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=15 n+2+(15 n-.32) \cdot 16^{n-1}, n=0,1,2, \cdots \text {. }
$$
(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, \( a_{n+2}=2 a_{n+1}-a_{n}+15^{3} n \cdot 16^{n-1}, \) for \( n=0,1,2, \cdots \).
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 relation in (1), we get \( p \mid n \). Using the general term formula, we can deduce that \( 16^{n} \equiv 1 \pmod{p} \). Conversely, if \( p \mid n \) and \( 16^{n} \equiv 1 \pmod{p} \), 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 relation 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 \( 16^{n} \equiv 1 \pmod{p} \). If \( n_{0} \) is the smallest natural number such that \( 16^{n_{0}} \equiv 1 \pmod{p} \), 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 \( 16^{n} \equiv 5^{n} \pmod{11} \), it is easy to calculate that the smallest natural number \( n \) such that \( 16^{n} \equiv 1 \pmod{11} \) is \( n=5 \). By Fermat's Little Theorem, we know \( 2^{180} \equiv 1 \pmod{181} \), i.e., \( 16^{45} \equiv 1 \pmod{181} \). If a natural number \( n<45 \) and
$$
16^{n} \equiv 1 \pmod{181},
$$

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

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

where \( k=0,1,2, \cdots \).