Roots of a Cubic Polynomial with Elementary Symmetric Polynomial Coefficients Let $R_n$ be a set of $n$ distinct nonzero rational numbers. Let $e_k$ be elementary symmetric polynomials over $R_n$---i.e. $e_0=1$, $e_1 = \sum_{1\le i\le n} r_i$, $e_2 = \sum_{1\le i<j\le n} r_i r_j$, etc.
For all $n>3$, does there exits an $R_n$ such that $e_3 x^3 + e_2 x^2 + e_1 x + 1$ has distinct rational roots? ($e_3\ne0$)
For quadratic polynomial, it is easy to characterize $R_n$ using simple algebra. I don't know how to approach the cubic or higher order polynomials.

We first consider the case $n$ odd, since the solution is simpler.
Let $n = 2m + 1$.
Choose $a_i: 1 \le i \le m$ nonzero distinct rational values such that the sum of squares is itself a square:
$$a_1^2 + a_2^2 + \ldots + a_{m}^2 = A^2$$
(One way to generate such a list would be to iteratively use Pythagorean triples, ie something like $a_1 = 3, a_2 = 4, a_3 = 12, a_4 = 84, \ldots $)  
We define our set as follows:
$$
R_{n} = \{ \pm a_1, \pm a_2, \ldots, \pm a_m, B\} 
$$
It is now easy to check that for this choice we have $e_1 = B, e_2 = -A^2, e_3 = -A^2B$.  We can now factor
$$e_3x^3 + e_2x^2 + e_1x + 1
= -A^2Bx^3 - A^2x^2 + Bx + 1
= (Bx+1)(Ax+1)(-Ax+1)$$
Thus, the three roots are distinct.
I note that in the even case, if $0\in R_n$ were allowed then $R_{n} = \{ \pm a_1, \pm a_2, \ldots, \pm a_m, B,0\}$ would similarly work.
Now we consider $n \ge 4$ even.  Set $n = 2m + 2$.  Now choose $a_1, \ldots a_m$ as before, but rescale them so that their sum of squares is $9$.  Our set is now
$$
R_{n} = \{ \pm a_1, \pm a_2, \ldots, \pm a_m, 10, 4\} 
$$
The values of the symmetric polynomials can be checked directly: $$e_1 = 14\\ e_2 = 40 - \sum a_i^2 = 31\\ e_3 = 14\left(-\sum a_i^2\right) = -126$$  Our polynomial now factors as
$$
e_3x^3+e_2x^2+e_1x+1
= -126x^3+31x^2+14x+1
= -(2x-1)(9x+1)(7x+1)
$$
The even case is more difficult using this method since the more complicated example of $R_4 = \{-3,3, 10,4\}$ is necessary as a starting point.
