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.
 A: They are the only possible solution. Proof is as follows:
Suppose that $d = gcd(x,y)$ and $x=d ~ x0$, $y=d ~y_0$ where $x_0$ and $y_0$ are co-prime.
Substituting in the original equation we get
$$ 1/x + 1/y + 1/z=1 \Rightarrow -d\,x_0\,y_0\,z+y_0\,z+x_0\,z+d\,x_0\,y_0 =0$$
Solving for $d$:
$$d=\frac{\left( y_0+x_0\right) \,z}{x_0\,y_0\,\left( z-1\right) }$$
Since $x_0$ and $y_0$ are co-prime, for $d$ to be an integer, $x_0\,y_0$ should divide $z$.
Hence we require
$$ z= k ~x_0 ~y_0$$
Substituting in the equation for $d$ and solving for $k$:
$$ k=\frac{d}{d\,x_0\,y_0-y_0-x_0}$$
This shows that $d$ is a multiple of $k$. Let
$$ d= \mu k$$. Then
$$k=\frac{k\,\mu}{k\,\mu\,x_0\,y_0-y_0-x_0}$$
Solving for $k$:
$$k=\frac{1}{x_0\,y_0}+\frac{1}{\mu\,y_0}+\frac{1}{\mu\,x_0}$$
This implies that $1 \le x_0 \le 3$, $1 \le y_0 \le 3$, $1 \le \mu\le 3$
It is possible to eliminate some of the 27 possible values since $k$ has to be an integers this will result in 12 possible values for $(x_0,y_0,\mu)$ and two of the solutions are repetitions giving the 10 solutions mentioned in the problem. 
A: 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.
A: HINT : You may suppose that $1\le x\le y\le z.$ This will make it easier to find the solutions.
A: Hint: Deduce that none of $x, y, z \in \mathbb{N}$ exceeds $7$. This can be done by mathlove's hint above. 
Additionally, you can show that there is only one $(a, 2, 2)$-tuple and thus use $(a, 2, 3)$ to bound the solutions.
A: There are 14 possible answers up to order for the 4-term analog of this problem.
2,3,7,42
2,3,8,24
2,3,9,18
2,3,10,15
2,3,12,12
2,4,5,20
2,4,6,12
2,4,8,8
2,5,5,10
2,6,6,6
3,3,4,12
3,3,6,6
3,4,4,6
4,4,4,4
