Example 10 (1994 National Training Team Selection Test) Find all quadruples of positive integers $a, b, c, d$ such that the product of any three numbers divided by the remaining one leaves a remainder of 1.
The
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problem
solution First, prove that $a, b, c, d$ are all not less than 2, and are pairwise coprime.
This is because, according to the problem, we have $b c d = k a + 1$. Clearly, $a \geqslant 2$, otherwise, if $a=1$, then $b c d$ divided by $a$ would have a remainder of 0, which contradicts the problem statement.
At the same time, $a$ is coprime with any of $b, c, d$.
Similarly, $b \geqslant 2, c \geqslant 2, d \geqslant 2$, and $a, b, c, d$ are pairwise coprime.
Without loss of generality, assume $2 \leqslant a \leqslant 5, d \geqslant 7$.
Equation (1) can be rewritten as
$1 + \frac{1}{6 c d} = \frac{1}{2} + \frac{1}{3} + \frac{1}{c} + \frac{1}{d}$, thus
$\frac{1}{c} + \frac{1}{d} = \frac{1}{6} + \frac{1}{6 c d} < \frac{1}{6}$, so $c < 12$.
From $a=2$, we know $c$ is an odd number, hence $c \leqslant 11$.
From equation (3), we can derive $d = 6 + \frac{35}{c-6}$.
Since $d \geqslant 7$ and $d$ is a positive integer, $\frac{35}{c-6}$ must be a positive integer. Considering $c \leqslant 11$, we have $1 \leqslant c-6 \leqslant 5$, thus $c-6=1$ or $c-6=5$, i.e.,
$c=7$ or $c=11$.
When $c=7$, from equation (4) we get $d=41$,
When $c=11$, from equation (4) we get $d=13$.
Therefore, the four numbers $(a, b, c, d) = (2,3,7,11)$ or $(2,3,11,13)$.