$30 \cdot 29$ Among the following numbers, the one that definitely cannot be the square of some natural number (where $n$ is a natural number) is
(A) $3 n^{2}-3 n+3$.
(B) $4 n^{2}+4 n+4$.
(C) $5 n^{2}-5 n-5$.
(D) $7 n^{2}-7 n+7$.
(E) $11 n^{2}+11 n-11$.
(Tianjin Junior High School Mathematics Competition; China Junior High School Mathematics League, 1984)
[Solution 1] $\because 3 n^{2}-3 n+3=3\left(n^{2}-n+1\right)$,
$\therefore$ If we let $n^{2}-n+1=3$, i.e., when $n=2$, $3 n^{2}-3 n+3$ is the square of 3.
Similarly, when $n=3$, $5 n^{2}-5 n-5$ is the square of 5.
When $n=3$, $7 n^{2}-7 n+7$ is the square of 7.
When $n=3$, $11 n^{2}+11 n-11$ is the square of 11.
From this, we can rule out $(A),(C),(D),(E)$.
Therefore, the answer is $(B)$.
[Solution 2] $\because 4 n^{2}+4 n+4=2^{2}\left(n^{2}+n+1\right)$,

and
$$
n^{2}<n^{2}+n+1<(n+1)^{2},
$$

we know
$n^{2}+n+1$ is not a perfect square.
$\therefore 4 n^{2}+4 n+4$ is not a perfect square.
Therefore, the answer is $(B)$.