Bulgarian Mathematical Olympiad Round 4 problem 1. Problem: Prove that there exists a unique triple of positive integers greater than 1 such that product of any two increased by one is divisible by the third number.
I found $(2,3,7)$ to satisfy this but cant prove this. Please help.

Clearly all numbers must be pairwise coprime, let them be $a<b<c$
Notice that $abc| ab+ac+bc+1$ and so $abc\leq ab+ac+bc+1$.
Notice that if $a\geq 3$ then $abc\geq 3bc> ab+ac+bc+1$ and so $a=2$.
It follows that $b|2c+1$ and $c|2b+1$ and so $bc|2b+2c+1$ In particular $bc\leq 2b+2c+1$.
It $b\geq 4$ then  $bc\geq 4c>2b+2c+1$ and so $b=3$.
Hence $a=2,b=3$. And finally we must have $c|2\times 3+1$. So $c=7$.
