$12 \cdot 71$ Find natural numbers $m$ and $n$, such that $m^{m}+(m n)^{n}=1984$.
(Kyiv Mathematical Olympiad, 1983)
［Solution］Since
$$
\begin{array}{c}
5^{5}=3125>1984 \\
m^{m}256 \cdot 16>1984 .
\end{array}
$$

Therefore, if $n=4$, only $m=1$ is possible, but in this case
$$
1^{1}+(1 \cdot 4)^{4} \neq 1984 .
$$

Thus,
$$
m \leqslant 4, n \leqslant 3 .
$$

Also, when $m<4, \quad n<3$,
$$
m^{m}+(m n)^{n}<4^{4}+12^{3}=1984 .
$$

Therefore, only $m=4, n=3$. In this case,
$$
m^{m}+(m n)^{n}=4^{4}+12^{3}=1984 .
$$

Hence, the equation has a unique solution in natural numbers: $m=4, \quad n=3$.