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Let $p(x)$ be a polynomial of degree $n>2$, with roots $x_1,x_2,\dots,x_n$ (including multiplicities). Let $m$ be a positive even integer. Define the following mapping $$V_m(p)=\sum_{1\leq i<j\leq n}(x_i-x_j)^m.$$

QUESTION. For $\deg p(x)=n>2$ and $p'(x)$ its derivative, can you express $$\frac{V_m(p)}{V_m(p')}$$ as a function of $m$ and $n$ alone?

Remark. Prompted by Fedor's questions, as a showcase I just computed (not proved) that $$\frac{V_2(p)}{V_2(p')}=\frac{n^2}{(n-1)(n-2)}.$$

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Here a SageMath code that provides a function V(m) computing $V_m(p)$ in terms of elementary symmetric functions of $x_1,\dots,x_n$ (i.e. coefficients of $p$).

For example, if $p(x) = x^n - e_1 x^{n-1} + e_2 x^{n-2} + \dots$, then $$V_2(p) = (n-1)e_1^2 - 2n e_2,$$ $$V_4(p) = (n-1) e_1^4 - 4n e_2 e_1^2 + (2n+12)e_2^2 + (4n-12) e_3 e_1 - 4n e_4,$$ and so on.

From these expressions a proof for $m=2$ follows instantly. However, for larger $m$ the ratio $\frac{V_m(p)}{V_m(p')}$ does not appear to be a function of $n$, which I've tested computationally for $m$ up to $20$.

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    $\begingroup$ I find it rather interesting that it is a constant for $n=3$ and $m=4$. $\endgroup$
    – Fedor Petrov
    Commented Dec 31, 2020 at 13:38
  • $\begingroup$ @FedorPetrov: Indeed, but this does not extend to larger $m$. $\endgroup$
    – Max Alekseyev
    Commented Dec 31, 2020 at 14:36
  • $\begingroup$ @FedorPetrov: At dxdy.ru it was suggested that the case $n=3$ and $m=4$ reduces to $m=2$ via the identity: $$(a-b)^4+(a-c)^4+(b-c)^4 = \frac12\left((a-b)^2+(a-c)^2+(b-c)^2\right)^2.$$ $\endgroup$
    – Max Alekseyev
    Commented Jan 2, 2021 at 14:08
  • $\begingroup$ yes, it is equivalent, cf. my answer. $\endgroup$
    – Fedor Petrov
    Commented Jan 2, 2021 at 14:48
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If this were true, $V_m(p)/V_m(p'')=(V_m(p)/V_m(p'))\times(V_m(p')/V_m(p''))$ also would depend only on $m$ and $n=\deg p$, and so on, until we get $V_m(p)/V_m(p^{(n-2)})$. We have $$V_m(p^{(n-2)})=V_m\left(\frac{n!}2x^2-(n-1)!\left(\sum x_i\right)x+(n-2)!\sum_{i<j} x_ix_j\right)=c_{nm}\left((n-1)\left(\sum x_i\right)^2-2n\sum_{i<j}x_ix_j\right)^{m/2}=\tilde{c}_{nm} V_2(p)^{m/2}.$$ So if this were true, we would have $V_m(p)=C_{nm} (V_2(p))^{m/2}$. This is false already for $n=m=4$: if all the roots of $p$ are 0's and 1's, we have $V_4=V_2$, but $V_2^2/V_4=V_2$ is not fixed.

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