25. Among the positive integers $n$ between $1 \sim 100$, the number of $n$ for which $n^{2}+4$ and $n+3$ have a common divisor greater than 1 is ( ).
(A) 0
(B) 3
(C) 5
(D) 7
(E) 11
25.D.

Assume $n^{2}+4$ and $n+3$ can both be divided by $d$. Since $n^{2}+4-(n+3)(n-3)=13$, if $n^{2}+4$ and $n+3$ have a common divisor greater than 1, it must be 13. Therefore, there must exist an integer $k$ such that $n=13 k+10$ for $n+3$ to be divisible by 13. At this time,
$$
\begin{array}{l}
(13 k+10)^{2}+4=169 k^{2}+260 k+104 \\
=13\left(13 k^{2}+20 k+8\right)
\end{array}
$$

can also be divided by 13. It is easy to see that when $k \in\{0,1,2,3,4,5,6\}$, $n=$ $13 k+10$ is between 1 and 100.