10 . 18 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 $x, y, z$ such that $a^{2}+b=c$ holds.
（20th All-Soviet Union Mathematical Olympiad, 1986）
［Solution］From the given, we have
$$
\left(\frac{\overline{x x \cdots x}}{n \uparrow}\right)^{2}+\frac{\overline{y y \cdots y}}{n \uparrow}=\overline{\overline{z z \cdots z}} .
$$

That is, \quad(x \cdot \underbrace{11 \cdots 1}_{n \uparrow})^{2}+y \cdot \underbrace{11 \cdots 1}_{n \uparrow}=z \cdot \frac{11 \cdots 1}{2 n \uparrow}.
Therefore, we have
$$
\begin{array}{l}
x^{2}(\underbrace{11 \cdots 1}_{n \uparrow})^{2}+y(\underbrace{11 \cdots 1}_{n \uparrow})=z \cdot(1 \underbrace{00 \cdots 01}_{n-1 \uparrow})(\underbrace{11 \cdots 1}_{n \uparrow}), \\
x^{2} \cdot \underbrace{11 \cdots 1}_{n \uparrow}+y=z \cdot 1 \underbrace{00 \cdots 01}_{n-1 \uparrow} \\
\end{array}
$$

Let $x^{2}=\overline{u v}=10 u+v, \quad u, v \in\{0,1,2, \cdots, 9\}$.
At this point, equation (1) becomes
$$
u \cdot \frac{11 \cdots 1}{n \uparrow} 0+v \cdot \underbrace{11 \cdots 1}_{n \uparrow}+y=z \cdot 1 \underbrace{00 \cdots 0}_{n-1 \uparrow} 1
$$

Comparing the tens digit in equation (2).
The tens digit on the right side of equation (2) is 0, and the tens digit on the left side is $u+v+1-10$ (when $v+$ $y \geqslant 10$) or $u+v-10$ (when $v+y \leqslant 9$).
Thus,
$$
u+v=9 \text { or } u+v=10 \text {. }
$$

Since $10 u+v=x^{2}$, $x^{2}$ can only be
$$
09,36,81,64 \text {. }
$$

That is,
$$
x=3,6,9,8 \text {. }
$$

When $x=3$, we can solve for $y=2, z=1,(n \geqslant 2)$;
When $x=6$, we can solve for $y=8, z=4,(n \geqslant 2)$;
When $x=8$, we can solve for $y=3, \quad z=7,(n=2)$;
When $x=9$, there is no solution.