Solve for $a+b+c$: Equation System ($a,b,c\in N$)

[Albert1]

$a,b,c \in N$, and the following equation system is given :

$\left\{\begin{matrix}
ab+bc+ca+2(a+b+c)=8045-----(1) & & & & \\
abc-a-b-c=-2-----(2) & & & &
\end{matrix}\right.$

find the value of $a+b+c$

 

[anemone]

My solution:

Spoiler

Note that $(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1$. Now, substitute what we're given into it, we get $(a+1)(b+1)(c+1)=-2+8045+1=8044=2(2)(2011)$, this implies $(a,\,b,\,c)=(1,\,1,\,2010)$ (up to permutations) and hence $a+b+c=2012$.