12.121 Given that $a, b, c, d$ are pairwise distinct positive integers, and satisfy $a+b=$ $c d, a b=c+d$. Find the quadruple $(a, b, c, d)$ that meets the above requirements.
(Hungarian Mathematical Olympiad, 1987)
［Solution］If $a+b \geqslant a b$, then
$$
\begin{array}{c}
a b-a-b=(a-1)(b-1)-1 \leqslant 0, \\
(a-1)(b-1) \leqslant 1 .
\end{array}
$$

Since $a, b$ are positive integers, and $a \neq b$, it must be that
$$
a=1 \text { or } b=1 .
$$

If $a b>a+b$, then
$$
c+d=a b>a+b=c d .
$$

And we have
$$
(c-1)(d-1)<1 \text {. }
$$

In this case, $c=1$ or $d=1$.
Therefore, one and only one of $a, b, c, d$ is 1.
If $a=1$, then we have
$$
\left\{\begin{array}{l}
1+b=c d \\
b=c+d
\end{array}\right.
$$

This gives us
$$
\frac{1}{c d}+\frac{b}{c d}=\frac{1}{c d}+\frac{c+d}{c d}=\frac{1}{c d}+\frac{1}{c}+\frac{1}{d}=1 .
$$

So
$$
c=2, d=3 \text { or } c=3, d=2 \text {. }
$$

Substituting into the known system of equations, we get $b=5$.
Considering the cases where $b, c, d$ equal 1, we have the following eight solutions:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline$a$ & 1 & 1 & 5 & 5 & 2 & 3 & 2 & 3 \\
\hline$b$ & 5 & 5 & 1 & 1 & 3 & 2 & 3 & 2 \\
\hline$c$ & 3 & 2 & 3 & 2 & 1 & 1 & 5 & 5 \\
\hline$d$ & 2 & 3 & 2 & 3 & 5 & 5 & 1 & 1 \\
\hline
\end{tabular}