6. Given $m, n \in \mathbf{Z}$, and $1 \leqslant m \leqslant 99,1 \leqslant n \leqslant 99$, then the number of ordered pairs $(m, n)$ such that $(m+n)^{2}+3 m+n$ is a perfect square is ( ) .
A. 98
B. 100
C. 96
D. 99
6. A. From $1 \leqslant m \leqslant 99,1 \leqslant n \leqslant 99$, we get
$$
\begin{array}{l}
(m+n)^{2}+3 m+n(m+n)^{2}, \text { thus } \\
(m+n)^{2}<(m+n)^{2}+3 m+n<(m+n+2)^{2} .
\end{array}
$$

If $(m+n)^{2}+3 m+n$ is a perfect square, then it must be that $(m+n)^{2}+3 m+n=(m+n+1)^{2}$, simplifying which gives $m=n+1$, at this point $n=1,2, \cdots, 98, m=2,3, \cdots, 99$. Therefore, the number of ordered integer pairs $(m, n)$ is 98: $(2,1),(3,2), \cdots$, $(99,98)$.