6. Given that the natural number $n$ is not a multiple of 5. Then when 5 divides $n^{2}+1, n^{2}-1$, we have ( ).
(A) $5 \mid n^{2}+1$ or $5 \mid n^{2}-1$
(B) $5 \mid n^{2}+1$ and $5 \mid n^{2}-1$
(C) $5 \times n^{2}+1$ and $5 \times n^{2}-1$
(D) Cannot determine, any of the above may occur
6. (A).

Since $n$ is not a multiple of 5, we have $n \equiv \pm 1$ or $\pm 2(\bmod 5)$,
thus $n^{2} \equiv 1$ or $4(\bmod 5)$,
which gives $n^{2}-1 \equiv 0(\bmod 5)$ or $n^{2}+1 \equiv 0(\bmod 5)$. However, $\left(n^{2}-1\right)+\left(n^{2}+1\right) \neq 0(\bmod 5)$, so option $(B)$ cannot be selected.