33. Let $m \geqslant 3$. Prove the following: the arithmetic sequence $1+\operatorname{lm}(l=0,1, \cdots)$ must contain infinitely many primes. (i) The original proposition is equivalent to the statement that the arithmetic sequence must contain at least one prime. (ii) Let $q$ be a prime, $\mathrm{q} \mid m^{m}-1$ and $\delta_{q}(m)=h$, then $q^{r} \| m^{h}-1$ if and only if $q^{r} \| m^{m}-1$. (iii) If $q$ satisfies the conditions in (ii) and $m \nmid q-1$, then $h < m$. (iv) Let the distinct prime factors of $m$ be $p_{1}, p_{2}, \cdots, p_{n}$; and let the sets be
$$\begin{array}{l} S_{1}=\left\{s=m /\left(p_{i_{1}} \cdots p_{i_{t}}\right): 1 \leqslant i_{1}<\cdots<i_{t} \leqslant n, 2 \nmid t\right\} \\ S_{2}=\left\{s=m /\left(p_{i_{1}} \cdots p_{i_{t}}\right): 1 \leqslant i_{1}<\cdots<i_{t} \leqslant n, 2 \mid t\right\} \end{array}$$

and
$$A_{1}=\prod_{s \in S_{1}}\left(m^{s}-1\right), \quad A_{2}=\prod_{s \in S_{2}}\left(m^{s}-1\right)$$

Prove: If all prime factors $q$ of $m^{m}-1$ are not congruent to $1 \pmod{m}$, then we must have $A_{1} = \left(m^{m}-1\right) A_{2}$. (v) When $m \geqslant 3$, the equation in (iv) cannot hold. Therefore, the arithmetic sequence must contain at least one prime.