$1 \cdot 36$ Find all such four-digit numbers, when 400 is written to their left, the result is a perfect square.

Find all such four-digit numbers, when 400 is written to their left, the result is a perfect square.
[Solution] Let $\overline{a b c d}$ be the required four-digit number. Since $\overline{400 a b c d}$ is a seven-digit number and a perfect square, we have
$$
\overline{400 a b c d}=(20 \mathrm{~lm})^{2}
$$

Rewriting (1) as
$$
400 \times 10^{4}+\overline{a b c d}=\left(20 \times 10^{2}+10 l+m\right)^{2} .
$$

Expanding and simplifying the right side of the above equation, we get
$$
\overline{a b c d}=4000(10 l+m)+(10 l+m)^{2} \text {. }
$$

Since $l$ and $m$ are the tens and units digits of a number, from (2) we have
$$
l=0,0<m \leqslant 2,
$$

which means $m=1$ or $m=2$.
Thus, the required numbers are 4001 or 8004.