Example 1. Numbers $x\left(x \neq 0^{\circ}\right)$ and $y$ are such that for any $n \geqslant 1$, the number $\overline{x \cdots x 6 y \cdots y}$ is a perfect square. Find such $x$ and $y$. (18th All-Union Mathematical Olympiad for Secondary Schools)
Given the condition $\overline{x 6 y 4}=m^{2}$, since the hundreds digit is 6 and the units digit is 4, we have $40<n<100$, and the last digit of this two-digit number must be 2 or 8. Since $68^{2}=4624$ and $98^{2}=9604$, $m$ can only be 68 or 98.
Thus, we conjecture:
(1) $4 \cdots 462 \cdots 24=\overline{6 \cdots 68^{2}}$,
(2) $9 \cdots 960 \cdots 04=\overline{9 \cdots 98^{2}}$.
$$
\begin{array}{l}
\text { Solution (1) } \because \underbrace{4 \cdots 4}_{n \uparrow} 6 \underbrace{2 \cdots 24}_{n \uparrow} \\
=4 \cdot 10^{n+1}\left(10^{n}+\cdots+10+1\right) \\
+2 \cdot 10^{n+1}+2\left(10^{n}+\cdots+10+1\right)+2 \\
=4 \cdot 10^{n+1} \cdot \frac{10^{n+1}-1}{9}+2 \cdot 10^{n+1} \\
+2 \cdot \frac{10^{n+1}-1}{9}+2 \\
=\frac{\left[4 \cdot 10^{2(n+1)}+16 \cdot 10^{n+1}+16\right]}{9} \\
=\left[\frac{2}{3}\left(10^{n+1}+2\right)\right]^{2} \\
=(\frac{1}{3} \cdot 2 \underbrace{0 \cdots 0}_{n \uparrow} 4)^{2} \\
=\overline{6 \cdots 68^{2}} \text {. } \\
\end{array}
$$
$$
\begin{array}{l}
\text { (2) } \because \underbrace{9 \cdots 9}_{n \uparrow} 6 \underbrace{0 \cdots 04}_{n \uparrow} \\
=\left(10^{n}-1\right) \cdot 10^{n+2}+6 \cdot 10^{n+1}+4 \\
=10^{2(n+1)}-10^{n+2}+(10-4) \cdot 10^{n+1}+4 \\
=10^{2(n+1)}-10^{n+2}+10^{n+2}-4 \cdot 10^{n+1}+4 \\
=10^{2(n+1)}-4 \cdot 10+4 \\
=\left(10^{n+1}-2\right)^{2}=\overline{9 \cdots} \overline{8^{2}} . \\
\therefore x=4, y=2 \text { or } x=9, y=0 .
\end{array}
$$