20 Let $a, b, c$ be integers satisfying the inequality $1<a<b<c$, and $(a b-1)(b c-1)(c a-1)$ is divisible by $a b c$. Find the values of $a, b, c$.
20 Because
$$
(a b-1)(b c-1)(c a-1)=a b c(a b c-a-b-c)+(a b+b c+c a-1) .
$$

By the given condition
$$
a b c \mid(a b-1)(b c-1)(c a-1),
$$

so
$$
a b c \mid a b+b c+c a-1 \text {. }
$$

Also, because
$$
\frac{a b+b c+c a-1}{a b c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a b c},
$$

it should be a positive integer, so
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>1 \text {. }
$$

If $a \geqslant 3$, then
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{1}{3} \times 3=1 .
$$

This contradicts (1), so $a=2$. If $b \geqslant 4$, then
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{1}{2}+2 \times \frac{1}{4}=1,
$$

which also contradicts (1), so $b=3$. If $c \geqslant 6$, then
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1,
$$

which again contradicts (1), so $c=4,5$. If $c=4$, then
$$
\begin{array}{c}
(a b-1)(b c-1)(a a-1)=5 \times 11 \times 7, \\
a b c=2 \times 3 \times 4,
\end{array}
$$
(2) cannot be divided by (3), so it is discarded. Therefore, $c=5$, at this time
$$
\begin{array}{c}
(a b-1)(b c-1)(c a-1)=5 \times 14 \times 9, \\
a b c=2 \times 3 \times 5,
\end{array}
$$
(4) can be divided by (5), which meets the requirement. Hence
$$
a=2, b=3, c=5 .
$$