Example 44 (2002-2003 Hungarian Mathematical Olympiad Final Question) 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.
(1) $a_{n}=10^{n}-2$.
First, prove the maximality.
Among $n$-digit decimal integers, only $10^{n}-1>10^{n}-2$.
But $10^{n}-1=9 \times \frac{10^{n}-1}{9}$
$$
\begin{array}{l}
=\left(\frac{9+\frac{10^{n}-1}{9}}{2}+\frac{\frac{10^{n}-1}{9}-9}{2}\right)\left(\frac{9+\frac{10^{n}-1}{9}}{2}-\frac{\frac{10^{n}-1}{9}-9}{2}\right) \\
=\left(\frac{9+\frac{10^{n}-1}{9}}{2}\right)^{2}-\left(\frac{\frac{10^{n}-1}{9}-9}{2}\right)^{2} .
\end{array}
$$

Since $\frac{10^{n}-1}{9}$ is odd, $10^{n}-1$ can be expressed as the difference of two perfect squares. This contradicts the problem's condition.
Next, prove that $10^{n}-2$ satisfies the condition.
If $10^{n}-2$ can be expressed as the difference of two perfect squares, then it must be congruent to 0, 1, or 3 modulo 4. But $10^{n}-2 \equiv 2 \pmod{4}$, so $10^{n}-2$ cannot be expressed as the difference of two perfect squares.
If $10^{n}-2$ can be expressed as the sum of two perfect squares, then it must be divisible by 4 or congruent to 2 modulo 8. But $10^{n}-2$ is not divisible by 4 and is congruent to 6 modulo 8 (since $n>2$).
Therefore, $10^{n}-2$ cannot be expressed as the sum of two perfect squares.
(2) From $9^{2}(n-1)+64=k^{2}$, we get
$9^{2}(n-1)=(k-8)(k+8)$.
Since $n \geqslant 3$ and $-8 \neq 8 \pmod{9}$, we have
$81 \mid (k-8)$ or $81 \mid (k+8)$.
If $81 \mid (k-8)$, then $k_{\text{min}}=89, n=98$.
If $81 \mid (k+8)$, then $k_{\text{min}}=73, n=66$.
Therefore, $n_{\min}=66$.