12・147 Find the smallest prime that cannot be expressed as $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers.
(China National Team Selection Exam, 1995)
[Solution] It has been verified that $2,3,5,7,11,13,17,19,23,29,31,37$ can all be expressed in the form $\left|3^{a}-2^{b}\right|$, where $a$ and $b$ are non-negative integers.
$$
\begin{array}{ll}
2=3^{1}-2^{0}, & 3=2^{2}-3^{0}, \\
5=2^{3}-3^{1}, & 7=2^{3}-3^{0}, \\
11=3^{3}-2^{4}, & 13=2^{4}-3^{1}, \\
17=3^{4}-2^{6}, & 19=3^{3}-2^{3}, \\
23=3^{3}-2^{2}, & 29=2^{5}-3^{1}, \\
31=2^{5}-3^{0}, & 37=2^{6}-3^{3} .
\end{array}
$$

We prove that 41 is the smallest prime that cannot be expressed as $\left|3^{a}-2^{b}\right|$.
This is equivalent to proving that the indeterminate equations
$$
\begin{array}{l}
2^{u}-3^{v}=41 \\
3^{x}-2^{y}=41
\end{array}
$$
have no non-negative integer solutions.
Suppose $(u, v)$ is a non-negative integer solution to equation (1), then
$$
2^{u}>41, u \geqslant 6 \text{. }
$$

Therefore, $2^{u} \equiv 0(\bmod 8)$,
then $-3^{v} \equiv 1(\bmod 8)$,
or $3^{v} \equiv-1(\bmod 8)$,
when $v$ is a non-negative even number, $3^{v} \equiv 1(\bmod 8)$
when $v$ is a positive odd number, $3^{v} \equiv 3(\bmod 8)$
(4), (5) contradict (3). Therefore, equation (1) has no non-negative integer solutions.

Suppose $(x, y)$ is a non-negative integer solution to equation (2), then
$$
3^{x}>41 \text{, so } x \geqslant 4 \text{. }
$$

Therefore, $3^{x} \equiv 0(\bmod 3), 2^{y} \equiv 1(\bmod 3)$,
thus $y$ can only be even, let $y=2 t$. Then $2^{y} \equiv 0(\bmod 4)$,
thus $3^{x} \equiv 1(\bmod 4)$,
from this we know that $x$ can only be even, let $x=2 s$, then
$$
41=3^{x}-2^{y}=3^{2 s}-2^{2 t}=\left(3^{s}+2^{t}\right)\left(3^{s}-2^{t}\right) \text{. }
$$

Thus, $\left\{\begin{array}{l}3^{s}+2^{t}=41, \\ 3^{s}-2^{t}=1 .\end{array}\right.$
Solving gives $3^{s}=21,2^{t}=20$ which has no non-negative integer solutions for $s, t$.
Therefore, equation (2) has no non-negative integer solutions. Hence, the smallest prime is 41.