Sum and product of rational numbers is unity Consider the system of  equations:
$$\sum_{i=1}^n X_i = 1$$
$$\prod_{i=1}^n X_i = 1$$
It is reasonably simple to show that for $n\ge 4$, this system admits a rational solution $(x_1, \dots, x_n) \in \mathbb{Q}^n$. 
It suffices to find a solution for $n=4,5,6,7$ and for all larger values of $n$ solutions can be obtained by appending $(-1,-1,1,1)$ to a solution for $n-4$.  
For $n=2$, there are no rational solutions. I'm pretty certain the existence of rational solutions for $n=3$ is an open problem, but I cannot find any discussion of it, largely because its hard to search for. 
Is this a named problem? A solved problem? Can anyone direct me to any work that has been done on it?
 A: For $n=3$, taking $x_3 = 1 - x_1 - x_2$ we get
$-1 + x_1 x_2  - x_1 x_2^2 - x_1^2 x_2 = 0$.  This is an elliptic curve with Weierstrass form $s^3+(23/48)s-181/864+t^2$, according to Maple.  I think it ought to be a finite computation (not one that I know how to do) to find generators for the  rational points.
EDIT: Well, apparently this is an unsolved problem in general.  But unless I've made a mistake, Sage tells me the only rational points in the Weierstrass form are $(s,t) = (-1/12, \pm 1/2)$, which correspond to points at $\infty$ in the $x_1,x_2$ form), so there are no rational solutions in the case $n=3$. 
sage:  E = EllipticCurve([23/48,181/864])
sage:  E.torsion_subgroup()
 Torsion Subgroup isomorphic to Z/3 associated to the Elliptic Curve
 defined by y^2 = x^3 + 23/48*x + 181/864 over Rational Field
sage:  E.torsion_subgroup().gens()
 ((1/12 : 1/2 : 1),)
sage:  E.rank()
  0

EDIT: For $n=4$, taking $x_4 = 1 - x_1 - x_2 - x_3$ and $x_3 = 3/2$, the elliptic curve for $x_1$ and $x_2$ has rank $1$.  Thus there are infinitely many rational solutions for $n=4$ with $x_3 = 3/2$, including
$ \left[-\frac13,\frac43,\frac32,-\frac32\right]$ and 
$ \left[-\frac{289}{210}, \frac{150}{119}, \frac{3}{2}, -\frac{98}{255}\right]$.
I'd appreciate confirmation from elliptic curve experts that I'm using Sage correctly.
