Positive Integer solutions to $x+y+z = xyz$

Question

I just thought of the following Diophantine equation and want to know how many positive integer solutions exist for it ($x,y,z$ are positive integers)

$x+y+z = xyz$

Answer 1

One trivial solution is (1,2,3) Now to prove that there are no more, I will prove the following lemma first if a and b are positive integers such that $a,b>1$ then $ab>a+b$

Proof:

$ab>a+b$

$ab-a-b+1>1$

$(a-1)(b-1)>0$ Therefore all that we need to prove is that $(a-1)(b-1)$ is positive which follows from our assmption that $a>1$ and $b>1$

Now coming back to original question. It is clear that all of $x,y,z$ cannot be 1 if any two (say x,y) are equal to 1 then

$1+1+z = z$ which is not possible

If only one (say z) is equal to one then

$x+y+1=xy$

$xy-x-y+1=2$

$(x-1)(y-1)=2$

$(x,y) = (3,2) or (2,3)$

Now if neither of x,y,z is equal to 1

From our lemma

$xyz>xy+z>x+y+Z$

So equality cannot hold

So only solutions are $(1,2,3)$ and its permutations