Find all integral solutions to $a+b+c=abc$. Find all integral solutions of the equation $a+b+c=abc$.  
Is $\{a,b,c\}=\{1,2,3\}$ the only solution? I've tried by taking $a,b,c=1,2,3$.

Of course, if one of $a,b,c$ are zero, then you can only have $\{a,b,c\}=\{0,k,-k\}$ for some integer $k$. So suppose none of $a,b,c$ are zero. Since $a,b,c$ satisfies $a+b+c=abc$ if and only if $-a,-b,-c$ satisfies $a+b+c=abc$, we can assume that either exactly one of $a,b,c$ is negative or none are negative.
Now dividing $a+b+c=abc$ by $abc$ results in the equation $$\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1.$$


*

*If one of $a,b,c$ are negative, so say $a<0$, (and so $b,c\geq 1$) then we have $\frac{1}{bc}\leq 1< 1-\frac{1}{ab}-\frac{1}{ac}=\frac{1}{bc}$, an impossibility. 

*If $a,b,c>0$, then  at least one of $ab, ac$ or $bc$ is at most $3$, and all are greater than $1$. Without loss of generality, we can say $1<ab\leq 3$. This can only be the case if $a,b\in \{1,2,3\}$, with exactly one of them equal to $1$. So say $a=1$. Hence $1+b+c=bc$ with $b=2$ or $b=3$. If $b=2$ we get $c=3$; if $b=3$, then $c=2$.
In summary, the integral solutions of $a+b+c=abc$ are $\{0,k,-k\}$ for any integer $k$, and $\{1,2,3\}$ and $\{-1,-2,-3\}$.
