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 $a,b,c=1,2,3$.

The solutions with $abc=0$ obviously are of form $0,b,-b$.  And since $a,b,c$ is a solution iff $-a,-b,-c$ is, we find all solutions with $abc\neq 0$ by finding all with $abc>0$. We show the only such solution is $1,2,3$.  
If $abc>0$ then at least one of $a,b,c$ is $>0$ and the other two are alike in sign.  But if that sign was negative then the sum would be less than the greatest of $a,b,c$ while the product would be greater, so this would not give a solution. So if $abc>0$ then $a,b,c$ are each $>0$.
If $a=b$, apply the quadratic formula to find $a$ in the equation $2a+c=a^2c$.  You find $a$ is a positive integer iff $c=0$ which is a case we already know. 
So we can assume $0<a<b<c$.  
For fixed $0<a<b$ consider 
  $$f_{ab}(c) = abc - a-b-c=(ab-1)c-(a+b)$$ 
as a linear function of $c$. We want an integer solution $f_{ab}(c)=0$ with $c>2$.  The slope of $f_{ab}$ is $ab-1$ which is $>0$, so there is at most one $c$, and indeed for $a=1,b=2$ there is the known solution $c=3$.  When $a=1$ and $b>2$  the smallest value of $c$ we need consider is $c=b+1$. For that value of $c$ the function is already strictly positive and since it has positive slope there is no solution with any greater value of $c$.
When $a>1$ then again  $b>2$,  the smallest value of $c$ we need consider is $c=b+1$ and for that value of $c$ the function is already strictly positive and since it has positive slope there is no solution.
