Example 30 (1995 National Training Team Selection Exam Question) Find the smallest prime $p$ that cannot be expressed as $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers.
After verification, $2,3,5,7,11,13,17,19,23,29,31,37$ can all be written in the form $\mid 3^{a}-2^{b} \mid$, where $a, b$ are non-negative integers:
$$
\begin{array}{l}
2=3^{1}-2^{0}, 3=2^{2}-3^{0}, 5=2^{3}-3^{1}, \\
7=2^{3}-3^{0}, 11=3^{3}-2^{4}, \quad 13=2^{4}-3^{1}, \\
17=3^{4}-2^{6}, \quad 19=3^{3}-2^{3}, \quad 23=3^{3}-2^{2},
\end{array}
$$
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$29=2^{5}-3^{1}, 31=2^{5}-3^{0}, 37=2^{6}-3^{3}$.
It is conjectured that 41 is the smallest prime number that cannot be expressed in this form. To verify this conjecture, we use proof by contradiction.
If the equation
$$
2^{u}-3^{v}=41
$$

has non-negative integer solutions $(u, v)$, then
$2^{u}>41$, i.e., $u \geqslant 6$.
Therefore, $-3^{v} \equiv 1(\bmod 8)$, but the remainder of $3^{v}$ modulo 8 can only be 1 or 3, which is a contradiction. Hence, equation (1) has no non-negative integer solutions.
If the equation
$$
3^{x}-2^{y}=41
$$

has non-negative integer solutions $(x, y)$, then
$3^{x}>41$, i.e., $x \geqslant 4$.
Therefore, $2 y \equiv 1(\bmod 3)$, and $y$ must be even. Let $y=2 t$, then (2) becomes
$$
3^{x}-4^{t}=41 \text {. }
$$

Thus, we have
$3^{x} \equiv 1(\bmod 4)$.
Hence, $x$ is also even. Let $x=2 s$, then (2) becomes
$3^{2 s}-2^{2 t}=41$, i.e., $\left(3^{s}+2^{t}\right)\left(3^{x}-2^{t}\right)=41$.
Since 41 is a prime number, we have
$$
\left\{\begin{array}{l}
3^{s}+2^{t}=41, \\
3^{s}-2^{t}=1,
\end{array} \text { i.e., } \left\{\begin{array}{l}
3^{s}=21, \\
2^{t}=20 .
\end{array}\right.\right.
$$

This is impossible. Therefore, equation (2) has no non-negative integer solutions.
In conclusion, 41 cannot be expressed in the form $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers. Hence, the smallest prime number $p$ is 41.