Example 12 Let $m>n \geqslant 1$, find the smallest $m+n$ such that
$$
1000 \mid 1978^{m}-1978^{n} .
$$
\begin{array}{l}\text { Solution: Let } k=m-n \text {, then } 1978^{m}-1978^{n} \equiv \\ 0(\bmod 1000) \Leftrightarrow 2^{n} \cdot 989^{n}\left(1978^{k}-1\right) \\ \equiv 0\left(\bmod 2^{3} \cdot 5^{3}\right) \Leftrightarrow\left\{\begin{array}{l}2^{n} \equiv 0\left(\bmod 2^{3}\right) \Rightarrow n \geqslant 3, \\ 1978^{k}-1 \equiv 0\left(\bmod 5^{3}\right) .\end{array}\right. \\ \because 1978 \equiv 3(\bmod 5), \therefore 1978^{4} \equiv 1(\bmod 5), \\ \because 1978 \equiv 3(\bmod 25), \therefore 1978^{20} \equiv 3^{20} \equiv \\ 1(\bmod 25), \\ \because 1978 \equiv-22\left(\bmod 5^{3}\right),(-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), \\ \therefore 1978^{20} \equiv 26\left(\bmod 5^{3}\right), \\ \therefore 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(25+1)(100+1) \equiv 1\left(\bmod 5^{3}\right), \\ \therefore 100 \mid k, \therefore k_{\text {min }}=100, m+n=m-n+ \\ 2 n_{\text {min }}=106. \\\end{array}