positive integer ordered pairs $(x,y,z)$ in $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1$ Total no. of positive integer ordered pairs of $(x,y,z)$ that satisfy the equation $\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1$
$\bf{My\ Try:}$ Using Simple Guess $x=2\;,y=3\,z=6.$ satisfy $\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 1$
Now I have tried is there is any ordered pairs of $(x,y,z)$ that satisfy the above equation or not.
So Given $\displaystyle \frac{1}{z} = 1-\frac{1}{x}-\frac{1}{y} = 1-\left(\frac{x+y}{xy}\right)=\frac{xy-x-y}{xy}\Rightarrow z=\frac{xy}{xy-x-y} = 1+\frac{x+y}{xy-x-y}$
Now I did not understand How can i solve for Integer ordered pairs.
Help me
Thanks 

The only solutions are $(2,3,6); (2,4,4); (3,3,3)$ (and of course its permutations). This can be argued as follows (in fact you are almost there):
As you have it, we need $\dfrac{x+y}{xy-x-y}$ to be an integer. Without loss of generality we can take $x \leq y$. We have $x+y \leq 2y$ and $xy-x-y \geq (x-2)y$. Hence, for $x>2$, we have
$$\dfrac{x+y}{xy-x-y} \leq \dfrac2{x-2}$$ which is not an integer for $x>4$. Hence, this bounds $x$ between $2$ and $4$. Hence, only three cases to check:


*

*$x=2$. We need $\dfrac{y+2}{y-2}$ to be an integer and $y \geq x$. This gives us $y=3,4,6$.

*$x=3$. We need $\dfrac{y+3}{2y-3}$ to be an integer and $y \geq x$. This gives us $y=3,6$.

*$x=4$. We need $\dfrac{y+4}{3y-4}$ to be an integer and $y \geq x$. This gives us $y=4$.


Now put this together to get the set of solutions as $(2,3,6); (2,4,4); (3,3,3)$ and its permutations.
