$3 \cdot 68$ For any positive integer $q_{0}$, consider the sequence $q_{1}, q_{2}, \cdots, q_{n}$ defined by $q_{i}=\left(q_{i-1}-1\right)^{3}+3 \quad(i=1$, $2, \cdots, n)$. If each $q_{i}(i=1,2, \cdots, n)$ is a power of a prime. Find the largest possible value of $n$.
（Hungarian Mathematical Olympiad, 1990)
［Solution］Since $m^{3}-m=m(m-1)(m+1) \equiv 0(\bmod 3)$,
thus
$$
\begin{aligned}
q_{i} & =\left(q_{i-1}-1\right)^{3}+3 \\
& \equiv\left(q_{i-1}-1\right)^{3} \\
& \equiv q_{i-1}-1 \quad(\bmod 3) .
\end{aligned}
$$

Therefore, among $q_{1}, q_{2}, q_{3}$, there must be one that is divisible by 3, and this number should be a power of 3.
If $3 \mid(q-1)^{3}+3$, then

Thus
$$
3 \mid(q-1)^{3}
$$
$3 \mid q-1$.
$$
3^{3} \mid(q-1)^{3} \text {, }
$$

and
$$
3 \mid(q-1)^{3}+3 \text {, }
$$

Therefore, only when $q_{i}=1$, $\left(q_{i}-1\right)^{3}+3$ is a power of 3. This happens only when $i=0$. But when $q_{0}=1$, it leads to
$$
q_{1}=3, \quad q_{2}=11, \quad q_{3}=1003=17 \cdot 59 .
$$

Thus, the maximum value of $n$ is 2.