$2 \cdot 84$ 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 $n$.
［Solution］Since
$$m^{3}-m=m(m-1)(m+1) \equiv 0(\bmod 3)$$

Therefore $\square$
$$q_{i}=\left(q_{i-1}-1\right)^{3}+3 \equiv\left(q_{i-1}-1\right)^{3} \equiv q_{i-1}-1(\bmod 3)$$

Thus, among $q_{1}, q_{2}, q_{3}$, there must be one divisible by 3, and it should be a power of 3. However, when 3 । $((q-$ $\left.1)^{3}+3\right)$, 3 । $(q-1)^{3}$. Therefore, 3 । $(q-1), 3^{3}$ । $(q-1)^{3},(q-1)^{3}+$ 3 can only be divisible by 3, not by $3^{2}$. Thus, only when $q_{i}=1$, $\left(q_{i}-1\right)^{3}+3$ is a power of 3. This must occur when $i=0$, but $q_{0}=1$ leads to
$$q_{1}=3, q_{2}=11, q_{3}=1003=17 \times 59$$

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