Four. (20 points) It is known that the five-digit number $\overline{a b c d e}$ satisfies the following conditions:
(1) None of its digits are zero;
(2) It is a perfect square;
(3) The digit $a$ in the ten-thousands place is a perfect square, the two-digit number formed by the digits in the thousands and hundreds places, $bc$, and the two-digit number formed by the digits in the tens and units places, $de$, are also perfect squares.
Find all five-digit numbers that satisfy the above conditions.
Let $M^{2}=\overline{a b c d e}$, and $a=m^{2}$ (a single digit), $\overline{b c}=n^{2}$ (a two-digit number), $\overline{d e}=t^{2}$ (a two-digit number). Then
$$
M^{2}=m^{2} \times 10^{4}+n^{2} \times 10^{2}+t^{2} \text {. }
$$

From equation (1), we have
$$
\begin{array}{l}
M^{2}=\left(m \times 10^{2}+t\right)^{2} \\
=m^{2} \times 10^{4}+2 m t \times 10^{2}+t^{2} .
\end{array}
$$

Comparing equations (1) and (2), we get $n^{2}=2 m t$.
Since $n^{2}$ is a multiple of 2, $n$ is also a multiple of 2, so $n^{2}$ is a multiple of 4 and is a perfect square.
Thus, $n^{2}=16$ or 36 or 64.
When $n^{2}=16$, we get $m t=8$, then $m=1,2,4,8, t=8$, $4,2,1$. The last two solutions do not meet the conditions, so they are discarded.
Therefore, $M^{2}=11664$ or 41616.
When $n^{2}=36$, we get $m t=18$. Then $m=2,3,1, t=9,6$, 18. The last solution does not meet the conditions, so it is discarded.
Therefore, $M^{2}=43681$ or 93636.
When $n^{2}=64$, we get $m t=32$. Then $m=1,2,4,8, t=$ $32,16,8,4$ all do not meet the conditions, so they are discarded.

Therefore, the five-digit numbers that satisfy the conditions are only 4: 11 664, 41616, 43681, 93636.