(50 points) Let $n \geqslant 3$ be a positive integer, and $a_{n}$ be the largest $n$-digit integer that is neither the sum of two perfect squares nor the difference of two perfect squares.
(1) Find the expression for $a_{n}$;
(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) First, prove the lemma: A positive integer $n$ is neither the sum of two perfect squares nor the difference of two perfect squares if and only if $n \equiv 2(\bmod 4)$.
$\because \quad 2 k+1=(k+1)^{2}-k^{2}, 4 k=(k+1)^{2}-(k-1)^{2}, \therefore \quad$ the lemma is proved.
$\because a_{n}$ is the largest $n$-digit integer, $\therefore \quad a_{n}<10^{n}-1$, where $n \geqslant 3$.
$\because \quad 10^{n}-2 \equiv-2 \equiv 2(\bmod 4), \therefore \quad a_{n}=10^{n}-2, n \geqslant 3$.
(2) Let the required positive integer be $n$, at this time the sum of the squares of the digits of $a_{n}$ is a perfect square $k^{2}$, then $81 n-17=k^{2}$, thus $81\left|k^{2}+17 \Leftrightarrow 81\right| k^{2}-64 \Leftrightarrow 81 \mid(k+8)(k-8)$.
$\because k+8 、 k-8$ are not congruent modulo 3, i.e., they are not both divisible by 3, $\therefore \quad 81 \mid k+8$, or $81 \mid k-8$. If $81 \mid k-8$, then $k_{\text {min }}=89$, at this time $n=98$; if $81 \mid k+8$, then $k_{\text {min }}=73$, at this time $n=66$. Therefore, $n_{\text {min }}=66$, at this time the sum of the squares of the digits of $a_{66}$ is $73^{2}=5329$.