10.14 Find all such four-digit numbers, for which writing 400 to the left of them results in a perfect square.
(Kyiv Mathematical Olympiad, 1954)
[Solution] Let $\overline{a b c d}$ be the desired four-digit number.
Since the number $400 a b c d$ is a seven-digit number and is a perfect square, it can be expressed in the following form:
$$
\overline{400 a b c d}=(\overline{20 l m})^{2} .
$$

Rewriting this equation, we get

that is,
$$
\begin{aligned}
400 \cdot 10^{4}+\overline{a b c d} & =\left(20 \cdot 10^{2}+10 l+m\right)^{2}, \\
400 \cdot 10^{4}+\overline{a b c d} & =400 \cdot 10^{4}+4000(10 l+m)+(10 l+m)^{2}, \\
\overline{a b c d} & =4000(10 l+m)+(10 l+m)^{2} .
\end{aligned}
$$

Since $l, m \in\{0,1,2, \cdots, 9\}$.
If $l \geqslant 1$, then
$$
4000(10 l+m) \geqslant 4 \cdot 10^{4}
$$

In this case, equation (1) cannot hold.
Therefore,
$$
l=0 .
$$

Similarly, if $m \geqslant 3$, then $4000 m \geqslant 12000$, which also exceeds four digits, so $0<m \leqslant 2$.
When $l=0, m=1$, we have
$$
2001^{2}=4004001 \text {. }
$$

When $l=0, m=2$, we have
$$
2002^{2}=4008004 \text {. }
$$

Therefore, the desired four-digit numbers are 4001 and 8004.