12.65 How many ordered quadruples of integers $(a, b, c, d)$ satisfy:
$$
0<a<b<c<d<500, a+d=b+c \text { and } b c-a d=93 \text {. }
$$
(11th American Invitational Mathematics Examination, 1993)
[Solution] From $a+d=b+c$ and $0<a<b<c<d$, we know that there exists $\delta \in$ $N$, such that
$$
\left\{\begin{array}{l}
b=a+\delta \\
c=d-\delta .
\end{array}\right.
$$

Substituting (1) and (2) into $b c-a d=93$ yields
$$
\begin{array}{c}
(a+\delta)(d-\delta)-a d=93 . \\
\delta(d-a-\delta)=1 \cdot 93=3 \cdot 31 .
\end{array}
$$

This leads to four systems of equations:
I. $\left\{\begin{array}{l}\delta=1, \\ d-a-\delta=93 .\end{array}\right.$
II. $\left\{\begin{array}{l}\delta=3 \\ d-a-\delta=31 .\end{array}\right.$
III. $\left\{\begin{array}{l}\delta=93, \\ d-a-\delta=1 .\end{array}\right.$
IV. $\left\{\begin{array}{l}\delta=31, \\ d-a-\delta=3 .\end{array}\right.$

From (1), (2), and system I, we get
$$
\left\{\begin{array}{l}
b=a+1, \\
c=d-1=93+a+1-1=93+a, \\
d=c+1=94+a .
\end{array}\right.
$$

Since
$$
a+94<500 \text {, }
$$

we have
$$
0<a<406 \text {. }
$$

Thus, there are 405 sets of $(a, b, c, d)$.
From (1), (2), and system II, we get
$$
\left\{\begin{array}{l}
b=a+3 \\
c=d-3=a+31 \\
d=c+3=a+34 .
\end{array}\right.
$$

So
$$
\begin{array}{l}
a+34<500 \\
0<a<466
\end{array}
$$

Thus, there are 465 sets of $(a, b, c, d)$.
From (1), (2), and system III, we get
$$
\left\{\begin{array}{l}
b=a+93 \\
c=a+1 \\
d=a+94
\end{array}\right.
$$

This does not satisfy $a<b<c<d$.
From (1), (2), and system IV, we get
$$
\left\{\begin{array}{l}
b=a+31 \\
c=a+31 \\
d=a+62
\end{array}\right.
$$

This does not satisfy $a<b<c<d$.
Therefore, there are a total of $405+465=870$ sets of $(a, b, c, d)$.