Three, (25 points) Let $x, y, a, m, n$ be positive integers, and $x+y=a^{m}, x^{2}+y^{2}=a^{n}$. Find how many digits $a^{30}$ has.

保留源文本的换行和格式，直接输出翻译结果。
Three, from the known we get
$$
a^{2 m}=x^{2}+y^{2}+2 x y=a^{n}+2 x y \text {. }
$$

From the problem and equation (1), we know that $a^{2 m}>a^{n}$. Therefore, $2 m>n$.
Dividing both sides of equation (1) by $a^{n}$, we get
$$
a^{2 m-n}=1+\frac{2 x y}{a^{n}} \text {. }
$$

Since the left side of equation (2) is a positive integer, $a^{n}$ must divide $2 x y$.
Thus, $2 x y \geqslant a^{n}=x^{2}+y^{2}$, which means $(x-y)^{2} \leqslant 0$.
Solving this, we get $x=y$.
Therefore, $2 x=a^{m}, 2 x^{2}=a^{n}$. So, $a^{2 m-n}=2$.
Considering that $a$ and $2 m-n$ are both positive integers, we have $a=2$, at this time,
$$
2 m-n=1 \text {. }
$$

Next, we need to find how many digits $2^{30}$ has.
Since $2^{30}=\left(2^{10}\right)^{3} = 1024^{3} > \left(10^{3}\right)^{3} = 10^{9}$, so, $2^{30}$ is a number with no fewer than 10 digits.
$$
\begin{array}{l}
\text { Also, } \frac{2^{30}}{10^{9}}=\frac{1}{10} \times \frac{1024^{3}}{1000^{3}}<\frac{1}{10} \times\left[\frac{1025}{1000}\right]^{3} \\
=\frac{1}{10} \times\left[\frac{41}{40}\right]^{3}<\frac{1}{10} \times\left[\frac{41}{40} \times \frac{40}{39} \times \cdots \times \frac{12}{11}\right] \\
=\frac{1}{10} \times \frac{41}{11}=\frac{41}{110}<1,
\end{array}
$$

Therefore, $2^{30}<10^{9}$.
Hence, $2^{30}$ is a number with fewer than 10 digits.
Thus, we can conclude that $2^{30}$ is a 9-digit number.
(Li Ming, Wang Chengyong, Wuhu County No.3 Middle School, Anhui Province, 233300)