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.
 A: 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 $2$).
A: Not completely different from Oleg517's. write $x=a-1$, $y=b-1$, $z=c-1$, we have $1\le x < y <z$ and $$\frac{abc-1}{(a-1)(b-1)(z-1)} = 1 + \frac{x+y+z + xy + yz + zx}{xyz}$$ is an integer.
On the other hand, $$0< x+y+z + xy + yz + zx \le xyz + (xy+yz+zx) = xyz + x(y+z) + yz < xyz + xyz + yz < 3xyz,$$ so $x+y+z + xy + yz + xz = xyz$ or $2xyz$. 
First, solve $x+y+z + xy + yz + zx = xyz$: $x>1$ is easy to see; on the other hand, if $x\ge 3$, then the left side is bounded by $$xyz(1/20+1/15+1/12+1/3+1/4+1/5) <xyz$$. So $x=2$, and we get $y=4$, $z=14$ by solving $(y-3)(z-3)=11$. 
Second, solve $x+y+z+xy+yz+zx=2xyz$: it is easy to see $x=1$ by arguments similar to the above, then $y=3$, $z=7$. 
