$30 \cdot 38$ If $4^{27}+4^{500}+4^{n}$ is a perfect square, then the largest integer $n$ is
(A) 1000.
(B) 900.
(C) 972.
(D) does not exist.
(1st "Hope Cup" National Mathematics Invitational Competition, 1990)
[Solution 1] Since $4^{27}+4^{500}+4^{n}=2^{54}\left(2^{2 n-54}+2^{946}+1\right)$, and
$$
2^{2 n-54}+2^{946}+1=2^{n-27}\left(2^{n-27}+2^{946-(n-27)}\right)+1 \text {, }
$$

Noticing that $(a-1)(a+1)+1=a^{2}$, considering $2^{n-27}$ as $a-1$, then
$$
2^{946-(n-27)}=2 \text {, or } 946-(n-27)=1
$$

In this case, the above expression is a perfect square, yielding $n_{\text {max }}=972$.
Therefore, the answer is $(C)$.
[Solution 2] Without loss of generality, assume $n \geqslant 27$, by
$$
4^{27}+4^{500}+4^{n}=2^{54}\left(2^{2 n-54}+2^{946}+1\right) \text {, }
$$

If the left side of the equation is a perfect square, then $2^{2 n-54}+2^{946}+1$ is a perfect square.
When $n=972$, we have
$$
2^{2 \cdot 972-54}+2^{946}+1=\left(2^{945}\right)^{2}+2 \cdot 2^{945}+1=\left(2^{945}+1\right)^{2} \text {, }
$$

But when $n>972$, because
$$
\left(2^{n-27}\right)^{2}<2^{2 n-54}+2^{946}+1<\left(2^{n-27}+1\right)^{2} \text {, }
$$

we know that $2^{2 n-54}+2^{946}+1$ is not a perfect square.
Thus, the largest $n$ sought is 972.
Therefore, the answer is $(C)$.