3. When $n$ takes all natural numbers greater than 1, among the following four expressions, the one that never takes a perfect square value is ( ).
(A) $5 n^{2}-5 n-5$
(B) $3 n^{2}-3 n+3$
(C) $9 n^{2}-9 n+9$
(D) $11 n^{2}-11 n-11$
3.(C).
$\because$ When $n$ is a natural number greater than 1, we have $(n-1)^{2}<n^{2}-n+1<n^{2}$,
i.e., $(3 n-3)^{2}<9 n^{2}-9 n+9<\left(3 n^{2}\right)$, $\therefore 9 n^{2}-9 n+9=(3 n-2)^{2}$ or $(3 n-1)^{2}$.
However, there are no positive integer solutions in these cases. But when $n$ is equal to $3, 2, \mathrm{~A}$, $(A)(B)(D)$ are all perfect squares.