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 Let $3 n+1=a^{2}, 5 n-1=b^{2}, a, b$ be positive integers, then
$$
\begin{aligned}
7 n+13 & =9(3 n+1)-4(5 n-1) \\
& =(3 a)^{2}-(2 b)^{2} \\
& =(3 a-2 b)(3 a+2 b) .
\end{aligned}
$$

From this, we know that $3 a-2 b$ is a positive integer, and $3 a-2 b \neq 1$. Because if $3 a-2 b=1$, then
$$
27 n+9=(3 a)^{2}=(2 b+1)^{2}=4 b^{2}+4 b+1,
$$

which means
$$
27 n=4\left(b^{2}+b-2\right),
$$

then $4 \mid n$, let $n=4 k$, we get $5 n-1=20 k-1$ is not a perfect square, a contradiction! Therefore, $3 a-2 b \geqslant 2$. Hence, from (1), $7 n+13$ is a composite number; also because
$$
\begin{aligned}
8\left(17 n^{2}+3 n\right)= & {[(3 n+1)+(5 n-1)][4} \\
& =\left(a^{2}+b^{2}\right)\left[(2 a)^{2}+b^{2}\right] \\
& =\left(2 a^{2}+b^{2}\right)^{2}+(a b)^{2},
\end{aligned}
$$

so the answer is C. (For example, 65 can be a value of $n$ as described above)