4. 229 A decimal natural number $a$ consists of $n$ identical digits $x$, and the number $b$ consists of $n$ identical digits $y$, while the 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.
Then
$$\begin{array}{l}
b=\underbrace{\frac{n \uparrow}{y w \cdots}}_{n \uparrow}=y \cdot \underbrace{11 \cdots 1}_{n \uparrow} \text {, } \\
c=\frac{n \uparrow}{z \approx \cdots}=z \cdot \frac{11 \cdots 1}{2 n \uparrow} \text {. }
\end{array}$$

That is
$$\begin{array}{l}
x^{2} \cdot \underbrace{11 \cdots 1^{2}}_{n \uparrow}+y \cdot \underset{n \uparrow}{11 \cdots 1}=\approx \underset{2 n \uparrow}{11 \cdots 1} \text {, } \\
x^{2} \cdot \underset{n \uparrow}{11 \cdots 1}+y=z\left(10^{n}+1\right), \\
\left(x^{2}-9 z\right) \underset{n \uparrow}{11 \cdots 1}=2 z-y \text {. }
\end{array}$$

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