^ In decimal, the natural number $a$ is composed of $n$ identical digits $x$, the number $b$ is composed of $n$ identical digits $y$, and the number $c$ is composed of $2 n$ identical digits $z$. For any $n \geqslant 2$, find all numbers $x, y, z$ that satisfy $a^{2}+b=c$.
According to the problem,
$$
\left(\frac{x\left(10^{n}-1\right)}{9}\right)^{2}+\frac{y\left(10^{n}-1\right)}{9}=\frac{z\left(10^{2 n}-1\right)}{9} .
$$

Simplifying, we get
$$
\left(9 z-x^{2}\right) \cdot 10^{n}+9 z+x^{2}=9 y \text {. }
$$

Since $9 z+x^{2}<200, 9 y<100$, when $n \geqslant 3$, $9 z-x^{2}=0$. Thus, $x=3, z=1$ or $x=6, z=4$. Or $x=z=9$. In the first two cases, $y$ is 2, 8 respectively. The third case has no solution.

When $n=2$, it is also possible that $9 z-x^{2}=-1$, thus $x=8, z=7$, and the corresponding $y=3$. Therefore, the solutions are $x=3, y=2, z=1$ or $x=6, y=8, z=4$. When $n=2$, there is another solution: $x=8$, $y=3, z=7$.