Find $a,b,c$ where $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$ Find integers $a,b,c$ such that $1<a<b<c$ and $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$ .I tried it solve using elementary number theory but I can't proceed . Somebody help me.

Denote $\bullet = \dfrac{abc-1}{(a-1)(b-1)(c-1)}$
Note that $$
\bullet = \dfrac{a}{a-1}\cdot \dfrac{bc-1}{(b-1)(c-1)} + \dfrac{1}{(b-1)(c-1)} \\
=\dfrac{a}{a-1} \cdot \left(1 + \dfrac{1}{b-1} + \dfrac{1}{c-1}\right) + \dfrac{1}{(b-1)(c-1)}.\tag{1}
$$
$(1)\Rightarrow$ (lower bound)
$$
\dfrac{a}{a-1}<\bullet.\tag{2}
$$

If $a=2$, then (since $c-1\ge b$) from $(1), (2)$ we get
$$2<\bullet \leq 2\cdot \left(1+\dfrac{1}{b-1}+\dfrac{1}{b}\right) + \dfrac{1}{(b-1)b};\tag{3}$$
and if (in addition) $b\ge 5$, then $(3)\Rightarrow$
$$2<\bullet \leq 2\cdot \left(1+\dfrac{1}{4}+\dfrac{1}{5}\right) + \dfrac{1}{20}=\dfrac{59}{20}<3,\tag{4}$$
so expression $\bullet$ cannot be integer in this case.
Considering cases $(a,b,c)=(2,3,c)$ and $(a,b,c)=(2,4,c)$, we find first  solution$$(a,b,c)=(2,4,8).\tag{*}$$

If $a=3$, then from $(1), (2)$ we get
$$\dfrac{3}{2}<\bullet \leq \dfrac{3}{2}\cdot \left(1+\dfrac{1}{b-1}+\dfrac{1}{b}\right) + \dfrac{1}{(b-1)b};\tag{5}$$
and if (in addition) $b\ge 7$, then $(5)\Rightarrow$
$$\dfrac{3}{2}<\bullet \leq \dfrac{3}{2}\cdot \left(1+\dfrac{1}{6}+\dfrac{1}{7}\right) + \dfrac{1}{42}=\dfrac{167}{84}<2,\tag{6}$$
so expression $\bullet$ cannot be integer in this case.
Considering cases $(a,b,c)=(3,4,c)$, $(a,b,c)=(3,5,c)$ and $(a,b,c)=(3,6,c)$, we find second solution $$(a,b,c)=(3,5,15).\tag{**}$$

If $a\ge 4$, then $b\ge 5$, $c\ge 6$, and  $(1), (2) \Rightarrow$
$$1<\bullet \le \dfrac{4}{3} \cdot \left(1 + \dfrac{1}{4} + \dfrac{1}{5}\right) + \dfrac{1}{20} = \dfrac{119}{60}<2.\tag{7}$$
Therefore there are no solutions for $a\ge 4$ (since value of expression $\bullet$ is between $1$ and $2$).
