6. Let $n$ be a positive integer, and $3 n+1$ and $5 n-1$ are both perfect squares. For the following two propositions:
甲: $7 n+13$ must be a composite number;
乙: $8\left(17 n^{2}+3 n\right)$ must be the sum of two squares.
Your judgment is ( ).
(A) 甲 is correct, 乙 is incorrect
(B) 甲 is incorrect, 乙 is correct
(C) Both 甲 and 乙 are correct
(D) Neither 甲 nor 乙 is necessarily correct
6. C.

Let $3 n+1=a^{2}, 5 n-1=b^{2}\left(a, b \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
7 n+13=9(3 n+1)-4(5 n-1) \\
=(3 a)^{2}-(2 b)^{2} \\
=(3 a-2 b)(3 a+2 b) .
\end{array}
$$

Therefore, $3 a-2 b$ is a positive integer.
If $3 a-2 b=1$, then
$$
27 n+9=(3 a)^{2}=(2 b+1)^{2}=4 b^{2}+4 b+1 \text {, }
$$

which means $27 n=4\left(n^{2}+n-2\right)$.
Thus, 4|n.
Let $n=4 k$. Then $5 n-1=20 k-1$ is not a perfect square, leading to a contradiction.
Therefore, $3 a-2 b \geqslant 2$.
Hence, from equation (1), $7 n+13$ is a composite number.
$$
\begin{array}{l} 
\text { Also, } 8\left(17 n^{2}+3 n\right) \\
= {[(3 n+1)+(5 n-1)] . } \\
{[4(3 n+1)+(5 n-1)] } \\
=\left(a^{2}+b^{2}\right)\left[(2 a)^{2}+b^{2}\right] \\
=\left(2 a^{2}+b^{2}\right)^{2}+(a b)^{2},
\end{array}
$$

Therefore, the answer is $\mathrm{C}$ (for example, 65 is one of the values of $n$).