Let $a_{n}=n^{4}+64$, then in the sequence $\left\{a_{n}\right\}$
(A) there are no prime numbers
(B) there are only a finite number of prime numbers
(C) there are only a finite number of perfect squares
(D) there are infinitely many perfect squares
6. (A).
$$
n^{4}+64=n^{4}+16 n^{2}+64-16 n^{2}=\left(n^{2}+8\right)^{2}-(4 n)^{2}=\left(n^{2}+4 n+8\right) \times\left(n^{2}-4 n+8\right) \text {, }
$$

When $n \geqslant 1$, this is the product of two positive integers greater than 1, so $n^{4}+64$ is not a prime number. Also, when $n \geqslant 6$, $\left(n^{2}\right)^{2}<n^{4}+64<\left(n^{2}+1\right)^{2}$, i.e., $n^{4}+64$ lies between two consecutive perfect squares, and thus cannot be a perfect square. When $n=1$, $2,3,4,5$, it is easy to verify that $n^{4}+64$ is not a perfect square.