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I am solving an algebraic equation: $x+y+z=xyz$ for $x, y,z\in\mathbb{N}$
I want to find out what $x, y, z$ are and if there are more possibilities.
By guessing I found out that $x,y,z=1,2,3$ but I want to solve it solve it algebraically.
I solved the Equation for $x$:
$$ x+y+z=xyz $$ $$ \text{Subtract } xyz $$ $$ x+y+z-xyz=0 $$ $$ \text{Add }{-y-z} $$ $$ x-xyz=-y-z $$ $$ \text{Factor out }x $$ $$ x(-yz+1)=-y-z $$ $$ \text{Divide by }{-yz+1} $$ $$ \frac{x(-yz+1)}{-yz+1}=\frac{-y-z}{-yz+1} $$ $$ x=\frac{-y-z}{-yz+1} $$ $$ \text{Take Factor as }{-1} $$ $$ x=\frac{-1(-y-z)}{-1(-yz+1)} $$ $$ x=\frac{y+z}{yz-1} $$ But I can't go any further by replacing $x$: $$ \frac{y+z}{yz-1}+y+z=\left(\frac{y+z}{yz-1}\right)yz $$ Can please you help me?

Michael Hardy
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    You won't be able to "solve" this equation since there is no unique solution. Instead, note that if $xy\neq1$, then $$z=\frac{x+y}{xy-1}$$satisfies the equation. There are no other solutions, since if $xy=1$, then $x+y=0$, and that is clearly not possible to satisfy in the reals. – Rushabh Mehta Dec 23 '20 at 19:27
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    You need to specify what the values of $x,y,z$ are allowed to be. For example do you mean them to be positive integers? Or rationals? Or reals? Also do you want them all different? – coffeemath Dec 23 '20 at 19:30

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