Example 6 Let $m>n \geqslant 1$. Find the smallest $m+n$ such that
$$\text {1000| } 1978^{m}-1978^{n} \text {. }$$
Solve: Using the congruence symbol, the problem is to find the smallest $m+n$ such that
$$1978^{m}-1978^{n} \equiv 0(\bmod 1000)$$

is satisfied. First, let's discuss what conditions $m, n$ must meet for the above equation to hold. Let $k=m-n$. Equation (9) becomes
$$2^{n} \cdot 989^{n}\left(1978^{k}-1\right) \equiv 0\left(\bmod 2^{3} \cdot 5^{3}\right)$$

By Property VII and Property IX, it is equivalent to
$$\left\{\begin{array}{l}
2^{n} \equiv 0\left(\bmod 2^{3}\right) \\
1978^{k}-1 \equiv 0\left(\bmod 5^{3}\right)
\end{array}\right.$$

From (10), we know $n \geqslant 3$. Next, we find the $k$ that satisfies (11). First, find the smallest $l$ such that
$$1978^{t}-1 \equiv 0(\bmod 5)$$

holds, denoted as $d_{1}$. Since
$$1978 \equiv 3(\bmod 5)$$

we have $d_{1}=4$. Next, find the smallest $h$ such that
$$1978^{h}-1 \equiv 0\left(\bmod 5^{2}\right)$$

holds, denoted as $d_{2}$. From Chapter 1, §4, Example 5, we know $4 \mid d_{2}$. Noting that
$$1978 \equiv 3\left(\bmod 5^{2}\right)$$

From the calculation in Example 2, we have $d_{2}=20$. Finally, find the smallest $k$ such that
$$1978^{k}-1 \equiv 0\left(\bmod 5^{3}\right)$$

holds, denoted as $d_{3}$. From Chapter 1, §4, Example 5, we know $20 \mid d_{3}$. Noting that
$$\begin{array}{l}
1978 \equiv-22\left(\bmod 5^{3}\right) \\
\begin{aligned}
(-22)^{20} & \equiv(25-3)^{20} \equiv 3^{20} \equiv(243)^{4} \\
& \equiv 7^{4} \equiv(50-1)^{2} \equiv 26\left(\bmod 5^{3}\right)
\end{aligned}
\end{array}$$

Through calculation, we get
$$\begin{array}{c}
1978^{20} \equiv 26\left(\bmod 5^{3}\right), \quad 1978^{40} \equiv(25+1)^{2} \equiv 51\left(\bmod 5^{3}\right) \\
1978^{60} \equiv(25+1)(50+1) \equiv 76\left(\bmod 5^{3}\right) \\
1978^{80} \equiv(50+1)^{2} \equiv 101\left(\bmod 5^{3}\right) \\
1978^{100} \equiv(100+1)(25+1) \equiv 1\left(\bmod 5^{3}\right)
\end{array}$$

Therefore, $d_{3}=100$. So, from Chapter 1, §4, Example 5, we know $100 \mid k$, and the smallest $k=100$. Thus, the necessary and sufficient conditions for (10) and (11), i.e., (9) to hold are
$$n \geqslant 3, \quad 100 \mid m-n$$

Therefore, the smallest $m+n=(m-n)+2 n=106$. If the congruence concept and its properties are not used to solve this problem, it would be very complicated.