Example 33 (2002-2003 Hungarian Mathematical Olympiad) Let $n$ be an integer greater than 2, and $a_{n}$ be the largest $n$-digit number that is neither the sum of two perfect squares nor the difference of two perfect squares.
(1) Find $a_{n}$ (expressed as a function of $n$);
(2) Find the smallest value of $n$ such that the sum of the squares of the digits of $a_{n}$ is a perfect square.
Solve: Since $10^{n}-1$ is the largest $n$-digit number, and
$$
\begin{aligned}
10^{n}-1 & =\frac{10^{n}-1}{9} \times 9 \\
& =\left[\frac{\left(9+\frac{10^{n}-1}{9}\right)-\left(9-\frac{10^{n}-1}{9}\right)}{2}\right]\left[\frac{\left(9+\frac{10^{n}-1}{9}\right)+\left(9-\frac{10^{n}-1}{9}\right)}{2}\right] \\
& =\left[\frac{9+\frac{10^{n}-1}{9}}{2}\right)^{2}-\left(\frac{9-\frac{10^{n}-1}{9}}{2}\right)^{2} .
\end{aligned}
$$

Since 9 and $\frac{1}{9}\left(10^{n}-1\right)$ are both odd, $10^{n}-1$ can be expressed as the difference of two squares of integers, i.e., $10^{n}-1$ does not satisfy the condition.
Next, we prove that $10^{n}-2$ satisfies the condition.
If $10^{n}-2 \equiv a^{2}-b^{2}, a, b$ are integers, then $a^{2} \equiv 0$ or $1(\bmod 4), b^{2} \equiv 0$ or $1(\bmod 4)$, hence $a^{2}-b^{2} \equiv 0,1$ or $3(\bmod 4)$, but $10^{n}-2 \equiv 2(\bmod 4)$, which is a contradiction.

If $10^{n}-2=a^{2}+b^{2}$, then $a^{2} \equiv 0(\bmod 4)$ or $a^{2} \equiv 1(\bmod 8), b^{2} \equiv 0(\bmod 4)$, or $b^{2} \equiv$ $1(\bmod 8)$, hence $a^{2}+b^{2} \equiv 0(\bmod 4)$ or $a^{2}+b^{2} \equiv 1$ or $2(\bmod 8)$, but $10^{n}-2 \equiv 2(\bmod 4)$ or $10^{n} \equiv 6(\bmod 8)$ (since $\left.n \geqslant 3\right)$, which is a contradiction.
Therefore, $10^{n}-2$ satisfies the problem's condition, so $a_{n}=10^{n}-2$.
(2) In $a_{n}=10^{n}-2$, there are $n-1$ digits that are 9, and the units digit is 8. From $9^{2}(n-1)+8^{2}=k^{2}$, we get $8 \mid(n-1)=(k-8)(k+8)$.

Since $n \geqslant 3$ and $k-8 \not \equiv k+8(\bmod 9)$, we have $8 \mid(k-8)$ or $8 \mid(k+8)$.
If $8 \mid(k-8)$, then $k_{\text {min }}=89$, if $8 \mid(k+8)$, then $k_{\text {min }}=66$.
Therefore, $n_{\text {min }}=66$.