6. If for a natural number $n$ not less than 8, when $3n+1$ is a perfect square, $n+1$ can always be expressed as the sum of $k$ perfect squares, then the minimum value of $k$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
6. (C).

From $3 n+1=a^{2}$, we know $3 \nmid a$. Therefore, $a=3 t \pm 1$. Thus, $3 n+1=9 t^{2} \pm 6 t+1 \Rightarrow n=3 t^{2} \pm 2 t$. Hence, $n+1=t^{2}+t^{2}+(t \pm 1)^{2}$.