4. 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 $\qquad$
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
4. C.

Solution: Let $3n+1=a^2$, $5n-1=b^2$, where $a, b$ are positive integers. Then, $7n+13=9(3n+1)-4(5n-1)=(3a)^2-(2b)^2=(3a-2b)(3a+2b)$.

From this, we know that $3a-2b$ is a positive integer, and $3a-2b \neq 1$. If $3a-2b=1$, then
$27n+9=(3a)^2=(2b+1)^2=4b^2+4b+1$, which implies $27n=4(n^2+n-2)$, so $4 \mid n$. Let
$n=4k$, then $5n-1=20k-1$ is not a perfect square, leading to a contradiction! Therefore, $3a-2b \geq 2$. Hence, from (1), we get that $7n+13$ is a composite number; also, because $8(17n^2+3n)=[(3n+1)+(5n-1)][4(3n+1)+(5n-1)]$
$$
=(a^2+b^2)[(2a)^2+b^2]=(2a^2+b^2)^2+(ab)^2,
$$

Therefore, the correct choice is C. (For example, 65 is one of the above $n$.)