$1 \cdot 43$ Try to find all integers $a, b, c$, where $1<a<b<c$, such that $(a-1)(b-1)(c-1)$ is a divisor of $a b c-1$.

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[Solution] Given $a \geqslant 2, b \geqslant 3, c \geqslant 4$, hence
$$
\frac{a-1}{a}=1-\frac{1}{a} \geqslant \frac{1}{2}, \frac{b-1}{b} \geqslant \frac{2}{3}, \frac{c-1}{c} \geqslant \frac{3}{4} .
$$

Therefore, $a b c \leqslant 4(a-1)(b-1)(c-1)$,
$$
S=\frac{a b c-1}{(a-1)(b-1)(c-1)}b>a>1$, we get $a \geqslant 3, b \geqslant 5, c \geqslant 7$.
If $b \geqslant 7$, then $c \geqslant 9$, thus
$$
\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} \text {. }
$$

This contradicts (2), so we must have $b=5, a=3$, substituting into (2) gives
$$
16(c-1)=15 c-1,
$$

Solving this, we get $c=15$.
(C) If $S=3$, i.e.,
$$
3(a-1)(b-1)(c-1)=a b c-1\frac{1}{3},
$$

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

This still leads to a contradiction, so we must have $b=3,4$.
If $b=3$, from (3) we get
$$
6(c-1)=6 c-1,
$$

This equation has no solution.
If $b=4$, from (3) we get
$$
9(c-1)=8 c-1,
$$

Solving this, we get $c=8$.
In summary, the solution to this problem is