$4 \cdot 229$ A decimal natural number $a$ consists of $n$ identical digits $x$, and number $b$ consists of $n$ identical digits $y$, number $c$ consists of $2n$ identical digits $z$. For any $n \geqslant 2$, find the digits $x, y, z$ that make $a^{2}+b=c$ true.
[Solution] Let $a=\overline{x x_{n} \cdots x}=x \cdot \underbrace{11 \cdots 1}_{n \uparrow}$,

then we have
$$
x^{2} \cdot \underbrace{11 \cdots 1^{2}}_{n \uparrow}+y \cdot \underbrace{11 \cdots 1}_{n \uparrow}=\approx \cdot \underbrace{11 \cdots 1}_{2 n \uparrow},
$$
$$
\begin{aligned}
x^{2} \cdot \underline{11 \cdots 1}+y & =z\left(10^{n}+1\right), \\
\left(x^{2}-9 \approx\right) & \frac{11 \cdots 1}{n \uparrow}=2 z-y .
\end{aligned}
$$

Thus $\quad\left\{\begin{array}{l}x^{2}-9 z=0, \\ 2 z-y=0 .\end{array}\right.$ or $\quad\left\{\begin{array}{l}n=2, \\ 2 z-y=11, \\ x^{2}-9 z=1 .\end{array}\right.$
Solving these, we get
$$
\begin{array}{l}
x=3, y=2, z=1, n \geqslant 2 ; x=6, y=8, z=4, n \geqslant 2 \\
x=8, y=3, \approx=7, n=2
\end{array}
$$