Example 22 (2003 China National Training Team Test Question) A positive integer cannot be divisible by $2$ or $3$, and there do not exist non-negative integers $a, b$, such that $\left|2^{a}-3^{b}\right|=n$, find the minimum value of $n$.

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Solve: Because $\left|2^{1}-3^{1}\right|=1, \left|2^{2}-3^{2}\right|=5, \left|2-3^{2}\right|=7, \left|2^{4}-3^{3}\right|=11, \mid 2^{4}-3^{1}|=13,| 2^{6}-3^{4}|=17,| 2^{3}-3^{3}|=19,| 2^{5}-3^{2}|=23,| 2^{1}-3^{3}|=25,| 2^{5}-3^{1}|=29,| 2^{5}-3^{0} \mid=31$, hence the minimum value of $n$ is not less than 35.

Secondly, we prove that there do not exist non-negative integers $a, b$ such that $\left|2^{a}-3^{b}\right|=35$. We use proof by contradiction. If not, then there exist non-negative integers $a, b$ such that $\left|2^{a}-3^{b}\right|=35$.
(1) If $2^{a}-3^{b}=35$, since $2^{a}>35$, hence $a \geqslant 6$. Taking the equation modulo 8, we get $-3^{b} \equiv 3(\bmod 8)$, i.e., $3^{b} \equiv 5(\bmod 8)$. But $3^{2 k} \equiv 1(\bmod 8), 3^{2 k+1} \equiv 3(\bmod 8)$, so this is impossible.
(2) If $3^{b}-2^{a}=35$, since $3^{b}>35$, hence $b \geqslant 4$. Taking the equation modulo 9, we get $-2^{a} \equiv-1$ (mod9), i.e., $2^{a} \equiv 1(\bmod 9)$, and
$$
\begin{array}{l}
2^{6 k} \equiv 1(\bmod 9), 2^{6 k+1} \equiv 2(\bmod 9), 2^{6 k+2} \equiv 4(\bmod 9), \\
2^{6 k+3} \equiv 8(\bmod 9), 2^{6 k+4} \equiv 7(\bmod 9), 2^{6 k+5} \equiv 5(\bmod 9) .
\end{array}
$$

Hence $a=6 k$ (where $k$ is a non-negative integer), so $3^{b}-8^{2 k}=35$. Taking both sides modulo 7, we get $3^{b} \equiv 1(\bmod 7)$, and
$$
\begin{array}{l}
3^{6 k^{\prime}} \equiv 729^{k^{\prime}} \equiv 1(\bmod 7), 3^{6 k^{\prime}+1} \equiv 3(\bmod 7), 3^{6 k^{\prime}+2}=2(\bmod 7), \\
3^{6 k^{\prime}+3} \equiv 6(\bmod 7), 3^{6 k^{\prime}+4} \equiv 4(\bmod 7), 3^{6 k^{\prime}+5} \equiv 5(\bmod 7),
\end{array}
$$

Hence $b=6 k^{\prime}$ (where $k^{\prime}$ is a positive integer), so $3^{6 k^{\prime}}-2^{6 k}=35$, i.e.,
$$
\left(3^{3 k^{\prime}}-2^{3 k}\right)\left(3^{3 k^{\prime}}+2^{3 k}\right)=35=5 \times 7 .
$$

Therefore $\left\{\begin{array}{l}3^{3 k^{\prime}}-2^{3 k}=1 \\ 3^{3 k^{\prime}}+2^{3 k}=35\end{array}\right.$ or $\left\{\begin{array}{l}3^{3 k^{\prime}}-2^{3 k}=5, \\ 3^{3 k^{\prime}}+2^{3 k}=7 .\end{array}\right.$
Solving the system of equations, we get $\left\{\begin{array}{l}3^{3 k^{\prime}}=18 \\ 2^{3 k}=17\end{array}\right.$ or $\left\{\begin{array}{l}3^{3 k^{\prime}}=6, \\ 2^{3 k}=1,\end{array}\right.$ which are both impossible.
In summary, $n_{\text {min }}=35$.