636*. Find all five-digit numbers that are perfect squares and remain perfect squares when the first, the first two, or the first three digits are erased.
$\triangle$ Let the desired number be denoted as $\overline{a b c d e}$. Then each of the numbers

$$
\overline{d e}, \overline{c d e}, \overline{b c d e}, \overline{a b c d e}
$$

is a perfect square.

The numbers $\overline{d e}$ and $\overline{c d e}$ can only be perfect squares simultaneously if $\overline{d e}=25$. Then the digit $c$ is 2 or 6.

The number $\overline{b c d e}=\overline{b c 25}$, according to the rule for squaring natural numbers ending in the digit 5 (see statement of problem 563b in § 13), is a perfect square if and only if the number $\overline{b c}$ is the product of two consecutive natural numbers.

Let $c=2$ and consider the number $\overline{b 2}$. By checking the digit $b$, we establish that the number $\overline{b 2}$ is the product of two consecutive natural numbers when $b=1(12=3 \cdot 4)$, $b=4(42=6 \cdot 7)$, and $b=7(72=8 \cdot 9)$.

Let $c=6$. Similarly, we establish that the number $\overline{b 6}$ is such a product only when $b=5(56=7 \cdot 8)$.

We obtain four four-digit numbers:

$$
1225, \quad 4225, \quad 7225, \quad 5625
$$

which are perfect squares.

Similarly, but more tediously, we find the digit $a$. Each of the five-digit numbers

$$
\overline{a 1225}, \quad \overline{a 4225}, \quad \overline{a 7225}, \quad \overline{a 5625}
$$

must be a perfect square. For this, we need to find the digit $a$ such that each of the three-digit numbers

$$
\overline{a 12}, \quad \overline{a 42}, \quad \overline{a 72}, \quad \overline{a 56}
$$

is the product of two consecutive natural numbers. In the first case, only $a=8(812=28 \cdot 29)$ fits, in the second $a=3(342=18 \cdot 19)$, in the third $a=2$, in the fourth $a=1$ and $a=7$.

Answer: $81225, 34225, 27225, 15625, 75625$.