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.

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 $y=3$, $z=7$. 
