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.

  • $\begingroup$ Try expanding $(r_1x+1)(r_2x+1)...(r_nx+1)$. $\endgroup$
    – Nate
    Commented May 16, 2015 at 21:22
  • $\begingroup$ @Nate I'd appreciate if you'd explain how that would help. $\endgroup$
    – mhp
    Commented May 21, 2015 at 6:59
  • $\begingroup$ Hmm... I'm not sure what I was thinking now either, I probably just misunderstood what you were asking. Sorry. $\endgroup$
    – Nate
    Commented May 21, 2015 at 23:55

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .