30 Given a natural number $a \geqslant 2$, the set $A=\left\{y \mid y=a^{x}, x \in \mathbf{N}\right\}$. Does there exist $b$ in the interval $[1, a]$ such that $C=A \cap B \neq \emptyset$? Here $B=\{y \mid y=(a+1) x+b, x \in \mathbf{N}\}$. If it exists, find all possible values of $b$ and the corresponding set $C$; if it does not exist, explain the reason.
30 Suppose there exist $b \in [1, a], s \in \mathbf{N}, t \in \mathbf{N}$, such that
$$
a^{s}=(a+1) t+b,
$$

then
$$
t=\frac{a^{s}-b}{a+1} .
$$

When $s=2 m$, $\frac{a^{2 m}-1}{a+1}=(a-1)\left(a^{2(m-1)}+a^{2(m-2)}+\cdots+1\right)$ is an integer, so
$$
t=\frac{a^{2 m}-b}{a+1}=\frac{a^{2 m}-1}{a+1}-\frac{b-1}{a+1}
$$

is a natural number only when $b=1$.
When $s=2 m+1(m \in \mathbf{N})$, $\frac{a^{2 m+1}+1}{a+1}=a^{2 m}-a^{2 m-1}+\cdots+1$ is an integer, so
$$
t=\frac{a^{2 m+1}-b}{a+1}=\frac{a^{2 m+1}+1}{a+1}-\frac{b+1}{a+1}
$$

is a natural number only when $b=a$.
When $s=1$, $t=\frac{a-b}{a+1} \notin \mathbf{N}$.
In summary, when $b=1$ or $b=a$, $C=A \cap B \neq \emptyset$, and
1) when $b=1$, $C=\left\{y \mid y=a^{2 m}, m \in \mathbf{N}, m \geqslant 1\right\}$,
2) when $b=a$, $C=\left\{y \mid y=a^{2 m+1}, m \in \mathbf{N}, m \geqslant 1\right\}$.