Given a six-digit number $\overline{a b c d e f}$, it is known that $\overline{a b c d e f}$, $\bar{a}$, $\overline{b c}$, and $\overline{d e f}$ are all non-zero perfect squares. Find all six-digit numbers that satisfy these conditions.
Solution: Let $M=\overline{a b c d e f}=n^{2}, a=x^{2}, \overline{b c}=y^{2}$, $\overline{d e f}=z^{2}$, where $1 \leqslant x \leqslant 3, 4 \leqslant y \leqslant 9, 10 \leqslant z \leqslant$ 31.
Then $M=n^{2}=10^{5} x^{2}+10^{3} y^{2}+z^{2}$.
(1) If $x=1$, when $y=6$,
$$
\begin{array}{l}
M=136000+z^{2}, \\
368^{2}<136100 \leqslant M \leqslant 136961<371^{2} .
\end{array}
$$

And $M=369^{2}=136161$ does not meet the condition, so the $M$ that meets the condition is $M=370^{2}=136900$.

When $y=4,5,7,8,9$, $z$ does not meet the condition, so $M$ does not exist.
(2) If $x=2$, when $y=4,5,6,7,8,9$, $z$ does not meet the condition, so $M$ does not exist.
(3) If $x=3$, when $y=8$,
$$
\begin{array}{l}
M=964000+z^{2}, \\
981^{2}<964100 \leqslant M \leqslant 964961<983^{2} .
\end{array}
$$

So the $M$ that meets the condition is $M=982^{2}=964324$.
When $y=4,5,6,7,9$, $z$ does not meet the condition, so $M$ does not exist.
Therefore, there are 2 six-digit numbers that meet the condition:
$$
136900,964324 .
$$