I need to determine all positive integer solutions for the equation:  $$\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1.$$  This is how I have tried to do it:   Mulitiplied both sides by $xyz$ to get $$yz+xz+xy+z+x+y+1=xyz.$$  Factor it: \begin{align} x(y+z+1-yz)+yz+y+z &amp;= -1   \\ x(y(1-z)+z+1)+y(1+z)+z &amp;= -1 \end{align}  If $z=0$, we get \begin{align} x(y+1)+y=-1 &amp;\iff xy+x+y=-1   \\ &amp;\iff (x+1)(y+1)=0, \end{align} which gives us $x=y=-1$.  Is this all positive integer solutions? Or have I missed something?  EDIT: I am stupid.   New attempt.   If $z=1$, I get $2x+2y=-2 \iff x+y=-1$.  Then there is no solutions of positive integers for both $x$ and $y$ at the same time.    If I try for $z=2$, I get \begin{align} x(y(1-2)+2+1)+y(1+2)+2=-1 &amp;\iff x(3-y)+3y+2=-1   \\ &amp;\iff 3x+3y+3-xy=0 \end{align} and I won't get a solution where all the variables are positive integers.
Can any one help me with this?  Determine all integer triples $(x,y,z)$ such that $1 ≤ x ≤ y ≤ z$ and $x + y + z + xy + yz + xz = xyz − 1$.  I thought of Vieta's formula but don't let me lead you into a dead end.
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