find all triplets of positive rationals satisfying this property Find all $(x, y, z) \in Q^{+} | (a, b, c) \in Z^{+} $ for $ a = x + y + z, b = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}, c = xyz$.
I have tried using Viete's formulas, setting $x, y, z$ as the roots of some polynomial $p(t) = a_3t^3 + a_2t^2 + a_1t^2 + a_0 = a_3(t-x)(t-y)(t-z)$, but made little progress... one idea I had was to, since $(x, y, z) \in Q^{+}$, using the rational roots theorem (assuming that we can write p(t) with integer coefficients):
Let $x = \frac{s_1}{t_1}, y = \frac{s_2}{t_2}, z = \frac{s_3}{t_3}$ where $(s_1, s_2, s_3, t_1, t_2, t_3) \in Z$. This gives that $s | a_0 \ \forall s \in \{s_1, s_2, s_3\}$ and $t | a_3 \ \forall t \in \{t_1, t_2, t_3\}$. Does anybody have any ideas?
 A: You've made good progress. To finish, first use Vieta's formulas or just expand to get
$$\begin{equation}\begin{aligned}
p(t) & = (t-x)(t-y)(t-z) \\
& = t^3 - (x+y+z)t^2 + (yz+xz+xy)t - xyz
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note $bc = yz+xz+xy$ is an integer. Thus, all of the coefficients in \eqref{eq1A} are integers, with the coefficient of $t^3$ (i.e., $1$) and $xyz$ being non-zero, so the Rational root theorem states all of the rational roots of \eqref{eq1A} are of the form $\frac{p}{q}$ where $q$ is a factor of the coefficient of $t^3$, i.e., $1$, so $q=1$. Thus, $x$, $y$ and $z$ are all positive integers.
This automatically means that $a$ and $c$ are integers, so they don't impose any constraints. For dealing with $b$, due to symmetry, WLOG consider $x \le y \le z$.
With $x = 1$, we can have $y = 1$, which means that $z = 1$. If $y = 2$, then $b = 1 + \frac{1}{2} + \frac{1}{z}$, so if $z \ge 3$, this is not an integer, which means $z = 2$. Finally, if $y \ge 3$, then we have $\frac{1}{y}+\frac{1}{z} \le \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$, so $b$ can't be an integer.
With $x = 2$, if $y = 2$, then $\frac{1}{z}$ must be an integer, which is not possible since $z \ge 2$. If $y = 3$, then $z = 6$ is required. Next, if $y = 4$, then only $z = 4$ works. Finally, if $y \ge 5$, then $b \le \frac{1}{2} + 2\left(\frac{1}{5}\right) = \frac{9}{10}$, which is not permitted.
With $x = 3$, only $y = z = 3$ allows $b$ to be an integer (i.e., $1$ in this case).
Finally, $x \ge 4$ means that $b \le 3\left(\frac{1}{4}\right)$, which is not allowed.
Thus, the $(x,y,z)$ are, in some order of vales, the elements of $\{(1,1,1),(1,2,2),(2,3,6),(2,4,4),(3,3,3)\}$.
