10.19 The digits $x (x \neq 0)$ and $y$ are such that for any $n \geqslant 1$, the number $x_{n}^{x x \cdots x} 6 y y \cdots y$ is a perfect square. Find all $x$ and $y$.
(18th All-Soviet Union Mathematical Olympiad, 1984)
［Solution］First, consider the case when $n=1$.
If
$$
\overline{x 6 y 4}=A^{2},
$$

then $A$ is a two-digit number with 2 or 8 as its unit digit, and the tens digit is no less than 3.
By examining $32^{2}, 38^{2}, 42^{2}, 48^{2}, 52^{2}, 58^{2}, 62^{2}, 68^{2}, 72^{2}, 78^{2}, 82^{2}, 88^{2}, 92^{2}, 98^{2}$, we find that only
$$
68^{2}=4624, \quad 98^{2}=9604
$$

have 6 as the hundreds digit.
In this case,
$x=4, y=2$, or $x=9, y=0$.
Furthermore, by
$$
\begin{array}{l}
(\underbrace{6 \cdots 6}_{n \uparrow} 8)^{2}=\underbrace{4 \cdots 4}_{n \uparrow} 6 \underbrace{2 \cdots 2}_{n \uparrow} 4 . \\
(\underbrace{2 \cdots 9}_{n \uparrow} 8)^{2}=\underbrace{9 \cdots 9}_{n \uparrow} 6 \underbrace{0 \cdots 0}_{n \uparrow} 4 \text {. } \\
\end{array}
$$

we know that the only solutions are
$$
\left\{\begin{array} { l } 
{ x = 4 , } \\
{ y = 2 , }
\end{array} \quad \left\{\begin{array}{l}
x=9, \\
y=0 .
\end{array}\right.\right.
$$