Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of  ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(2,6,3),(3,6,2),(3,2,6),(6,2,3),(6,3,2)}$.  Are these the only 10 solutions?
First, none of $x$, $y$ or $z$ can be $1$ ($x$, $y$ and $z$ are positive integers)
If I let $x=2$, then finding all solutions to $1/y+1/z = 1/2$    leads to $(4,4), (3,6)$ and $(6,3)$ which gives me $(x,y,z)$ as $(2,4,4), (2,3,6), (2,6,3)$ but this also means $(4,4,2), (4,2,4), (3,2,6), (3,6,2), (6,2,3), (6,3,2)$ are all valid triples for this equation.
If I let $x=3$, the only different values of $y$ and $z$ are $(3,3)$
How do I prove these are the only ten solutions? (without using any programming)
Known result:  If we denote $d(n^2)$ as the number of divisors of $n^2$, then the number of solutions of ${\frac {1} {x} }+{\frac {1} {y} } = {\frac {1} {n} }$   =  $d(n^2)$  (For positive $x$, $y$)
For ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$
$z = \frac{xy}{y(x-1)-x}$ where $xy \neq 0$
What happens after that?
Question is: how do we sho there are the only ten solutions?  I'm not asking for a solution.
Assuming $x \le y \le z$
$1 \le y \le \frac{xy}{y(x-1)-x}$ 
$\Longrightarrow 1 \le x \le y \le \frac{2x}{x-1} $ 
Got the answer.  I'll probably call @mathlove's answer. (Any additional answers I'll view later)
Liked @user44197 answer.

We may as well assume $x\le y\le z$ (and then count rearrangements of the variables as appropriate).  The smallest variable, $x$, cannot be greater than $3$ (or else $1/x+1/y+1/z\lt1/3+1/3+1/3=1$), nor can it be equal to $1$ (or else $1/x+1/y+1/z=1+1/y+1/z\gt1$).  So either $x=2$ or $x=3$.
If $x=3$, then $y=z=3$ as well (for the same reason as before), which gives the solution $(x,y,z)=(3,3,3)$.
If $x=2$, then $1/2+1/y+1/z=1$ implies
$${1\over2}={1\over y}+{1\over z}$$
Applying the inequality $y\le z$ to this equation, we see that $y$ must be greater than $2$ but cannot be greater than $4$, so $y=3$ or $y=4$.  Each of these gives a solution, $(x,y,z)=(2,3,6)$ and $(2,4,4)$.  
Counting rearrangements, we get the OP's $10$ solutions and no others.
