It is easy to prove, that $ a\le 4$ and chek, that $ a = 4$ give not solution. $ a = 2$ give $ (2,2,2),(2,2,4),(2,4,8)$. $ a = 3$ give $ (3,5,15)$. ModEdit: luckily there's a solution on http://www.kalva.demon.co.uk/imo/isoln/isoln921.html

Peter wrote: Find all integers $\,a,b,c\,$ with $\,1<a<b<c\,$ such that \[(a-1)(b-1)(c-1)\hspace{0.2in}\text{is a divisor of}\hspace{0.2in}abc-1.\] $x=a-1 \; , \; y=b-1 \; , \; z=c-1 \; \Longrightarrow \; xyz \mid (xy+yz+zx+x+y+z)$ $x \geq 3 \; \Longrightarrow \; y \geq 4 \; , \; z \geq 5 \; \Longrightarrow$ $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} \leq \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{12}+\frac{1}{20}+\frac{1}{15}=\frac{59}{60}<1$ $\Longrightarrow \; xyz>xy+yz+zx+x+y+z$. It is a contradiction. $\Longrightarrow \; x \leq 2$ The rest is easy and all solutions are: $(x,y,z)=\{(1,3,7),(2,4,14)\}$ $(a,b,c)=\{(2,4,8),(3,5,15)\}$

The problem is IMO 1992 Probrem 1. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=154338&#p154338