BMO2 1995 Question 1 - Solve the equation for positive integers (a,b,c) Find all triples of positive integers (a, b, c) such that
$$ \left( 1 + \frac{1}{a} \right)\left( 1 + \frac{1}{b} \right)\left( 1 + \frac{1}{c} \right) = 2 $$
I tried to expand, yielding $ a+b+c+ab+bc+ca+1 = abc $, but I couldn't go anywhere from here. 
This problem clearly involves using the fact that $a, b, c$ are positive integers, since there are three variables and only one equation. It would be great if someone also addressed this condition and maybe tried to solve the problem for other requirements, like when $a, b, c$ are positive or negative integers.
Finally - can anyone post or link other such problems which involves a mix of algebra manipulation and number theory. I would really appreciate it.
(Here is the full paper: https://bmos.ukmt.org.uk/home/bmo2-1995.pdf)

Considering cases works pretty well. If the product is to be $2$, then at least one of the numbers must be less than $4$, and all numbers are clearly $>1$.
Order the numbers so that $a\le b\le c$. Then $a = 2$ or $a=3$. If $a=2$, then $4\le b\le 6$ (since $b=2$ or $b=3$ produces a product too large, and if $b=7$ then $c\ge 7$ and the product is too small). This gives the three solutions $(2,4,15)$, $(2,5,9)$, and $(2,6,7)$. Alternatively, if $a=3$, a similar analysis yields the solutions $(3,3,8)$ and $(3,4,5)$.
So there are five essentially different solutions: $(2,4,15)$, $(2,5,9)$, $(2,6,7)$, $(3,3,8)$, and $(3,4,5)$.
