8. If the set
$$
\begin{aligned}
A= & \{(m, n) \mid(m+1)+(m+2)+\cdots+ \\
& \left.(m+n)=10^{2015}, m \in \mathbf{Z}, n \in \mathbf{Z}_{+}\right\},
\end{aligned}
$$

then the number of elements in set $A$ is ().
(A) 4030
(B) 4032
(C) $2015^{2}$
(D) $2016^{2}$
8. B.

From the given, we have
$$
n(n+2 m+1)=2^{2016} \times 5^{2015} \text {. }
$$

Thus, one of $n$ and $n+2 m+1$ is odd and the other is even.
Therefore, one of $n$ and $n+2 m+1$ is even, with the possibilities being $2^{2016}, 2^{2016} \times 5, 2^{2016} \times 5^{2}, \cdots, 2^{2016} \times 5^{2015}$, for a total of 2016 cases. Swapping the order also yields 2016 cases. Therefore, set $A$ has a total of 4032 elements.