14. Let $a \in \mathbf{N}_{+}, a \geqslant 2$, set $A=\left\{y \mid y=a^{x}, x \in \mathbf{N}_{1}\right\}, B=|y| y=(a+1)$ $x+b, x \in \mathbf{N}_{+}$| Does there exist $b$ in the closed interval $[1, a]$ such that $A \cap B \neq \phi$? If it exists, find one possible value of $b$ and the corresponding $A \cap B$. If it does not exist, explain the reason.
14. Let $b \in [1, a]$ such that $A \cap B \neq \phi$, i.e., there exists $y_{0} \in A$ and $y_{0} \in B$, such that $y_{0} = a^{m}, (m \in \mathbf{N}_{+})$ and $y_{0} = (a+1)n + b (n \in \mathbf{N}_{+})$.
Then there should exist $m, n \in \mathbf{N}_{+}$, such that $a^{n} = (a+1)n + b \cdot (1 \leqslant b \leqslant a)$
That is, $n = \frac{a^{m} - b}{a+1} (m, n \in \mathbf{N}_{+}, 1 \leqslant b \leqslant a)$
Since $a^{m} - b = [(a+1) - 1]^{m} - b = (a+1)^{m} - C_{m}^{1}(a+1)^{m-1}$
$$
+ \cdots + C_{m}^{m-1}(-1)^{m-1}(a+1) + (-1)^{m} - b
$$

Therefore, for $n \in \mathbf{N}_{+}$, it should make $(-1)^{m} - b$ divisible by $a+1$.
(1) When $m$ is a positive even number, $1 - b$ should be divisible by $a+1$.
Since $a \geqslant 2, 1 \leqslant b \leqslant 2$, hence only when $b = 1$ it meets the requirement.
(2) When $m$ is a positive odd number, $-1 - b$ should be divisible by $a+1$.
Since $2 \leqslant b+1 \leqslant a+1$, hence only when $b = a$ it meets the requirement.
In summary, the $b$ that satisfies the condition exists, and its values are $b = 1$ or $b = a$.
When $b = 1$,
$A \cap B = \{y \mid y = a^{2K}, K \in \mathbf{N}_{+}\}$

When $b = a$,
$A \cap B = \{y \mid y = a^{2K-1}, K \in \mathbf{N}_{+}\}$