7・67. Given that $m, n$ range over all positive integers, find the minimum value of $12^{m}-5^{n}$.
(China National Training Team Test Question, 1989)
[Solution] First, note that $12^{m}-5^{n}$ is an odd number, and since $112-5 \mid=7$. We prove that 7 is the smallest value that meets the condition.
Let
$$
s=\left|12^{m}-5^{n}\right| \text{. }
$$

Assume $s<7$, then $s=1,3,5$.
Since
$$
3 \nmid 5,5 \times 12 \text{, }
$$

$3=\left|12^{m}-5^{n}\right|, 5=\left|12^{m}-5^{n}\right|$ cannot hold.
If $s=1$, there are two possibilities.
(1) $12^{m}-5^{n}=1$.
$$
5^{n}=12^{m}-1=11 Q(m)
$$

where $Q(m)$ is an integer.
In this case, $11 \mid 5^{n}$, which is impossible.
(2) $12^{m}-5^{n}=-1$.
$$
5^{n}=12^{m}+1
$$

If $m$ is odd, then
$$
5^{n}=13 T(m),
$$

at this time, $13 \mid 5^{n}$, which is impossible.
If $m$ is even, let $m=2 t$, then
$$
5^{n}=144^{t}+1 \text{. }
$$

When $t$ is odd,
$$
\begin{array}{c}
5^{n}=145 \cdot K(m), \\
145 \mid 5^{n} \text{. }
\end{array}
$$

Since $145=5 \cdot 29$, then
$$
29 \mid 5^{n} \text{. }
$$

This is impossible.
When $t$ is even, let $t=2 q$, then
$$
5^{n}=\left(144^{2}\right)^{q}+1 \text{. }
$$

Since the unit digit of $144^{2}$ is 6, the unit digit of $\left(144^{2}\right)^{q}+1$ is 7, while the unit digit of $5^{n}$ is 5, which is also impossible.
Therefore,
$$
\left|12^{m}-5^{n}\right| \neq 1
$$

Thus, the smallest value of $112^{m}-5^{n} \mid$ is 7.