Find all integral solutions to $a+b+c=abc$. [duplicate]

Solution 1:

If at least one of $a,b,c$ is $0$ (i.e. $abc=0$), then $(a,b,c)$ satisfies the equation iff $\{a,b,c\}=\{t,-t,0\}$ for some $t\in\mathbb Z$.

Let $abc\neq 0$.

If WLOG $a=b$, then $(ca-1)^2=c^2+1\iff c=0$, but $abc\neq 0$, contradiction.
(or simply observe: $2a+c=a^2c\iff a^2c-2a-c=0, \Delta=4+4c^2=k^2, k\in\mathbb Z$
$\,\Rightarrow\, c^2+1=l^2, l\in\mathbb Z\iff c=0$, contradicting $abc\neq 0$).

If WLOG $a=-b$, then $c=-a^2c\iff c=0$, contradicting $abc\neq 0$.

Let WLOG $|a|>|b|>|c|>0$.

$$|abc|=|a||bc|=|a+b+c|\stackrel{\text{Triangle ineq.}}\le |a|+|b|+|c|< 3|a|$$

$$\iff |a|(|bc|-3)< 0$$

$|a|>0$, so $|bc|< 3$, and $(b,c)\in\{(2,1),(2,-1),(-2,1),(-2,-1)\}$.

Checking all of these only gives the solutions $(a,b,c)=(1,2,3),(-3,-2,-1)$.

Answer: $\{a,b,c\}=\{1,2,3\},\{-1,-2,-3\},\{t,-t,0\}$ for any $t\in\mathbb Z$.

Solution 2:

Note that if $a\ge b \ge c\ge 2$ you have $$3a\ge a+b+c=abc \ge 4a$$So for solutions with $a\ge b \ge c \ge 0$ you have $c=0$ or $c=1$

$c=0$ gives $a+b=0$, or $a=-b$. A non-negative solution has $a=b=0$. Note also the general solution $r, -r, 0$

$c=1$ gives $a+b+1=ab$ which can be rewritten $(a-1)(b-1)=2$ so $a=3, b=2$. We see also the solution $a=0, b=-1, c=1$ given by the factorisation of $2$ into negative factors.

Changing all the signs in a known solution gives another solution, so we can assume $abc\ge 0$, and we can also now assume that none of the numbers is $0$ or $1$.

Our computation for $1$ therefore gives the solutions for $-1$ too just by changing signs.

To test for further solutions we can change signs and choose $abc\gt 0$ with $a\gt 0\gt-2\ge b\ge c$. But then the inequalities at the top of this answer both hold, and there are no additional solutions.

We find all solutions by applying the symmetries of permuting values or sign change to the solutions already found.