3 . Find natural numbers $m, n$, such that $m^{m}+(m n)^{n}$ $=1984$.
Solution: When $m=4, n=3$,
$$
\begin{array}{l}
m^{m}+(m n)^{n}=4^{4}+12^{3} \\
=256+1728=1984 .
\end{array}
$$

And $5^{5}=3125>1984$, so $m \leqslant 4, n \leqslant 4$. When $n=4$, since
$$
1984>(4 m)=256 m^{4},
$$

$m=1$, but
$$
1^{1}+(1 \cdot 4)^{4}-1984 \text {. }
$$

Thus, $m \leq 4, n \leqslant 3$.
But if $m<4$ or $n<3$,
$$
n_{n}{ }^{7}+(m n)^{n}<4^{4}+12^{3}=1984 .
$$

Therefore, the only solution to this problem is $m=4, n=3$.