8.13 Does a perfect square exist whose sum of digits is (1)1983; (2)1984?
(Kiev Mathematical Olympiad, 1983)
[Solution] (1) If the sum of the digits of $n^{2}$ is 1983. Since 1983 is divisible by 3, then $n^{2}$ is also divisible by 3, which implies that $n$ is also divisible by 3, and thus $n^{2}$ is divisible by 9. Therefore, the sum of the digits of $n^{2}$ must be divisible by 9. However, 1983 is not divisible by 9, leading to a contradiction. Hence, there is no perfect square whose sum of digits is 1983.
(2) A perfect square whose sum of digits is 1984 does exist.
For example, $n=99 \cdots 97=10^{220}-3$, then
$$
\begin{array}{l}
n^{2}=10^{440}-6 \cdot 10^{220}+9 \\
=1 \underbrace{00 \cdots 0}_{440 \uparrow}-6 \cdot 1 \underbrace{00 \cdots 0}_{220 \uparrow}+9 \\
=\underbrace{99 \cdots 9}_{219 \uparrow} 4 \underset{219 \uparrow}{\underline{0 \cdots 09}} \\
\end{array}
$$

Thus, the sum of the digits of $n^{2}$ is
$$
219 \cdot 9+4+219 \cdot 0+9=1984 \text {. }
$$