Number of ordered positive rationals (x,y,z) satisfying following conditions.

Question

How many ordered triples $(x,y,z)$ of positive rational numbers satisfy the conditions: $x+y+z$, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$, and $xyz$ are all integers.

Answer 1

Then $yz+xz+xy$ is also an integer. It follows from the Vieta relations that $x$, $y$, and $z$ are the roots of a monic cubic with integer coefficients, so they are integers.

Thus we want all triples $(x,y,z)$ of positive integers such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is an integer.

Now it is a short search, since the smallest of the integers must be $\le 3$, and indeed is $3$ only in the case $x=y=z=3$.

Smallest is $2$ yields $(2,3,6)$ and $(2,4,4)$ (and permutations). And smallest is $1$ again yields essentially two solutions.