Solve the equation $a+b+c=abc$ for $a,b,c\in\mathbb{Z}$

Question

Solve for $a,b,c$ (where $a$, $b$, and $c$ are integers) the equation $$a+b+c=abc.$$ I would prefer a solution using trigonometry and I think that it might use the formula $\tan A + \tan B + \tan C=\tan A\tan B\tan C$ where $A$, $B$, and $C$ are angles of a triangle. The solutions might be $(a,b,c)$=$(1,2,3)$, $(-1,-2,-3)$ ,$(0,0,0)$ and their corresponding ordered pairs. Please see that ive asked for integral solution.

Answer 1

First, let's consider the solutions under the condition that $|a|\le |b|\le |c|$. Then, we have $$|ab||c|=|abc|=|a+b+c|\le |a|+|b|+|c|\le 3|c|$$$$\Rightarrow |c|(3-|ab|)\ge 0\Rightarrow c=0\ \text{or}\ |ab|\le 3.$$

$c=0$ leads $a=b=0$.

$|ab|\le 3$ leads $|a|^2\le |a||b|\le 3\Rightarrow a^2\le 3\Rightarrow a=0,\pm 1.$

1) $a=0$ leads $0=b+c\iff c=-b$, so $(a,b,c)=(0,b,-b).$

2) $a=-1$ leads $$-1+b+c=-bc\iff (b+1)(c+1)=2$$$$\iff (b+1,c+1)=(1,2),(2,1),(-1,-2),(-2,-1)$$$$\iff (b,c)=(0,1),(1,0),(-2,-3),(-3,-2).$$

3) $a=1$ leads $$1+b+c=bc\iff (b-1)(c-1)=2$$$$\iff (b-1,c-1)=(1,2),(2,1),(-1,-2),(-2,-1)$$$$\iff (b,c)=(2,3),(3,2),(0,-1),(-2,0).$$

Hence, when $|a|\le |b|\le |c|$, we have, for any $k\in\mathbb Z$ $$(a,b,c)=(0,k,-k),(-1,-2,-3),(1,2,3).$$

Hence, in general, the answer for your question is the followings : $$(a,b,c)=(0,k,-k),(k,0,-k),(k,-k,0),$$$$(-1,-2,-3),(-1,-3,-2),(-2,-1,-3),(-2,-3,-1),(-3,-1,-2),(-3,-2,-1),$$$$(1,2,3),(1,3,2),(2,3,1),(2,1,3),(3,2,1),(3,1,2)$$ where $k\in\mathbb Z$.

Answer 2

For $\mathbb{Z}$, a simple solution is $(1+2+3)=6=(1\times2\times3)$