8.69 Find the number of ordered pairs of integers $(m, n)$ where $1 \leqslant m \leqslant 99,1 \leqslant n \leqslant$ $99,(m+n)^{2}+3 m+n$ is a perfect square.
(17th American Invitational Mathematics Examination, 1999)
[Solution] From $1 \leqslant m \leqslant 99,1 \leqslant n \leqslant 99$ we get
$$
\begin{array}{l}
(m+n)^{2}+3 m+n(m+n)^{2} .
\end{array}
$$

Thus, $(m+n)^{2}<(m+n)^{2}+3 m+n<(m+n+2)^{2}$.
If $(m+n)^{2}+3 m+n$ is a perfect square, then it must be true that
$$
(m+n)^{2}+3 m+n=(m+n+1)^{2} .
$$

However, $(m+n)^{2}+3 m+n=(m+n+1)^{2}+n-m-1$.
Therefore, it must be true that $n-m-1=0$.
That is, $m=n-1$.
At this point, $n=2,3, \cdots, 99, m=1,2, \cdots, 98$.
Therefore, the number of ordered integer pairs $(m, n)$ is 98:
$$
(m, n)=(1,2),(2,3), \cdots,(98,99) \text {. }
$$