$1 \cdot 83$ Find all integers $a, b, c$, where $1<a<b<c$, such that
$$
(a-1)(b-1)(c-1)
$$

is a divisor of $a b c-1$.
(33rd International Mathematical Olympiad, 1992)
$$
\begin{array}{l}
\text { [Solution 1] Given } a \geqslant 2, \quad b \geqslant 3, c \geqslant 4 \\
\text { Since } \\
\frac{a-1}{a}=1-\frac{1}{a} \geqslant \frac{1}{2} \text {, } \\
\frac{b-1}{b}=1-\frac{1}{b} \geqslant \frac{2}{3} \text {, } \\
\frac{c-1}{c}=1-\frac{1}{c} \geqslant \frac{3}{4} \text {. } \\
\end{array}
$$

Multiplying the three inequalities, we get
$$
\frac{(a-1)(b-1)(c-1)}{a b c} \geqslant \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4}=\frac{1}{4} .
$$

Let $s=\frac{a b c-1}{(a-1)(b-1)(c-1)}$, then
$$
\frac{a b c-1}{(a-1)(b-1)(c-1)}b>a>1 \text { implies } a \geqslant 3, \quad b \geqslant 5, \quad c \geqslant 7 . \\
\text { If } b \geqslant 7, \quad \text { then } c \geqslant 9 \text {, thus } \\
\frac{(a-1)(b-1)(c-1)}{a b c} \geqslant \frac{2}{3} \cdot \frac{6}{7} \cdot \frac{8}{9}=\frac{32}{63}>\frac{1}{2}, \\
2(a-1)(b-1)(c-1)>a b c .
\end{array}
$$

This contradicts (2). Therefore, we must have $b=5, a=3$, substituting into (2) gives
$$
\begin{array}{c}
16(c-1)=15 c-1, \\
c=15 .
\end{array}
$$

Thus, $a=3, b=5, c=15$.
(3) If $s=3$, i.e.,
$$
3(a-1)(b-1)(c-1)=a b c-1\frac{1}{3} \text {, }
$$
$$
3(a-1)(b-1)(c-1)>a b c .
$$

This contradicts (3). Therefore, we must have $a=2$. If $b \geqslant 5$, then
$$
\begin{array}{l}
\frac{(a-1)(b-1)(c-1)}{a b c} \geqslant \frac{1}{2} \cdot \frac{4}{5} \cdot \frac{5}{6}=\frac{1}{3} \text {. } \\
3(a-1)(b-1)(c-1) \geqslant a b c . \\
\end{array}
$$

This still contradicts (3). Therefore, $b$ can only be 3 or 4.
If $b=3$, from (3) we have
$$
6(c-1)=6 c-1 .
$$

This equation has no solution.
If $b=4$, from (3) we get
$$
\begin{array}{l}
9(c-1)=8 c-1, \\
c=8 .
\end{array}
$$

Thus, $a=2, b=4, c=8$.
In summary, we have two sets of solutions:
$$
\left\{\begin{array} { l } 
{ a = 3 , } \\
{ b = 5 , } \\
{ c = 1 5 . }
\end{array} \quad \left\{\begin{array}{l}
a=2, \\
b=4, \\
c=8 .
\end{array}\right.\right.
$$
[Solution 2] Let $x=a-1, y=b-1, z=c-1$, then from $1x=1$ and $y$ is odd, so $y=3$
At this point $\quad x y z=3 z$,
$$
(x+1)(y+1)(z+1)-1=8 z+7 .
$$

Thus, $3 z \mid 8 z+7$, i.e.,
$$
3 z \mid 9 z+(7-z)
$$

Since $3 z$ is a divisor of $7-z$ and $z>3$, we have $z=7$
Thus, $x=1, \quad y=3, \quad z=7$, i.e.,
$$
a=2, \quad b=4, \quad c=8 \text {. }
$$
(3) When $x=2$, the problem becomes
$$
2 y z \mid 3(y+z)+y z+2
$$

If $y$ and $z$ are both odd or one is odd and the other is even, then $3(y+z)+y z+2$ is odd, and cannot be a multiple of $2 y z$, so $y$ and $z$ must both be even.
Thus,
$$
\begin{aligned}
y \geqslant 4, z & \geqslant y+2 . \\
2 y z \leqslant & 3 y+3 z+y z+2 \\
& \leqslant y z+3 z+3 z-6+2 \\
& =y z+6 z-4 \\
& <y z+6 z,
\end{aligned}
$$

Thus, $y=4$. At this point we have
$$
8 z \mid 7 z+14,
$$

i.e., $8 z \mid 8 z+(14-z)$.
$\begin{array}{ll}\text { Thus } & z=14 . \\ \text { Therefore } x=2, & y=4, z=14,\end{array}$
$a=3, \quad b=5, \quad c=15$.
In summary, the positive integers $a, b, c$ have two sets of solutions:
$$
\left\{\begin{array} { l } 
{ a = 2 , } \\
{ b = 4 , } \\
{ c = 8 . }
\end{array} \quad \left\{\begin{array}{l}
a=3, \\
b=5, \\
c=15 .
\end{array}\right.\right.
$$