6. Let $n$ be a positive integer, and $3n+1$, $5n-1$ are both perfect squares. Consider the following two propositions:
(1) $7n+13$ must be a composite number;
(2) $8(17n^2+3n)$ must be the sum of two perfect squares.
Which of the following is correct?
(A)(1)
(B)(2)
(C)(1)(2)
(D) None
6. C.

Let $3 n+1=a^{2}, 5 n-1=b^{2} (a, b$ are positive integers). 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
$$
27 n+9=(3 a)^{2}=(2 b+1)^{2},
$$

which means $27 n=4\left(b^{2}+b-2\right)$.
Thus, $41 n$.
Assume $n=4 k$.
Then $5 n-1=20 k-1$ is not a perfect square, leading to a contradiction.
Therefore, $3 a-2 b \geqslant 2$.
Hence, $7 n+13$ is a composite number, so conclusion (1) is correct.
$$
\begin{array}{l} 
\text { Also, } 8\left(17 n^{2}+3 n\right) \\
= {[(3 n+1)+(5 n-1)] . } \\
{[4(3 n+1)+(5 n-1)] } \\
=\left(a^{2}+b^{2}\right)\left[(2 a)^{2}+b^{2}\right] \\
=\left(2 a^{2}+b^{2}\right)^{2}+(a b)^{2} .
\end{array}
$$

Therefore, conclusion (2) is correct.