When studying the existence problem of power residue deference sets, I came across the following system of polynomial equations over $\mathbb{C}$, \begin{cases} &\sum\limits_{j=0}^{m-1}x_jx_{2k-j}=0,\ k=1,\ldots,\frac{m}{2}-1,\\ &x_kx_{m-k}=1,\ k=1,\ldots,\frac{m}{2}-1,\\ &x_\frac{m}{2}=1,\\ &x_kx_{\frac{m}{2}+k}=w^kx_{2k},\ k=1,\ldots,\frac{m}{2}-1,\\ &w^\frac{m}{2}=1, \end{cases} where $m\ge6$ is an even number and the indices of $x$'s are counted mod $m$. Denote by $V\subset\mathbb{C}^{m+1}$ the solution set of the system, and note that it is nonempty since $$ (x_0,x_1,\dots,x_{m-1},w)=(\frac{2-m}{2},1,\ldots,1,1) $$ is always a solution to the system.


What can we say about $A_m=\inf\{|x_0|^2\mid(x_0,x_1,\dots,x_{m-1},y)\in V,x_0\neq0\}$?

(If one replaces $x_j$ by $x_j/x_0$, the question is equivalent to asking for $\sup|x_0|^2$ in the new system.)

Computation results and conjectures:

By Groebner basis computation for $m$ up to $20$, I found that, except $m=8$ (this corresponds to existence of $8$th-power residue difference sets), the system is always zero-dimensional (i.e., has only finitely many solutions) and $x_0$ thus satisfies a univariate polynomial equation listed below:

when $m=6$, $$ (x-2)(x+2)(7x^2-1)=0; $$ when $m=10$, $$ x(x-4)(x+4)(11x^2-1)=0; $$ when $m=12$, $$ (x-3)(x+3)(x-5)(x+5)(5x-7)(5x+7)(13x^2-1)=0; $$ when $m=14$, $$ (x-6)(x+6)(4x^2+3)=0; $$ when $m=16$, $$ (x-7)(x+7)(7x-17)(7x+17)(17x^2-1)=0; $$ when $m=18$, $$ x(x-8)(x+8)(19x^2-1)=0; $$ when $m=20$, $$ (x-7)(x+7)(x-9)(x+9)(9x-31)(9x+31)(13x-67)(13x+67)=0. $$ From these equations we see $A_6=\frac{1}{7}$, $A_{10}=\frac{1}{11}$, $A_{12}=\frac{1}{13}$, $A_{14}=\frac{3}{4}$, $A_{16}=\frac{1}{17}$, $A_{18}=\frac{1}{19}$ and $A_{20}=\frac{961}{81}$.

The computation results suggest several conjectures when $m\neq8$, among which the following three are directly related to the question:

(a) the system is zero-dimensional;

(b) $A_m\ge\frac{1}{m+1}$ and $A_m=\frac{1}{m+1}$ if and only if $m+1$ is a prime power.


I'm not sure which subjects are this question related to, so feel free to change the tags.

For my purpose in the study of power residue difference sets, the conditions $x_0\in\mathbb{R}$ and $|x_1|=\dots=|x_{m-1}|=1$ can be imposed (that is to say, replace $V$ by $V\cap\mathbb{R}\times(\mathbb{S}^1)^m$) when considering the question.

Let $\zeta_m$ be any $m$-th root of unity. An observation (I don't know if it will useful or not...) is that if $(x_0,x_1,\ldots,x_{m-1},y)\in V$, then $$ (x_0,\zeta_m^tx_1,\zeta_m^{2t}x_2,\ldots,\zeta_m^{(m-1)t}x_{m-1},y),\quad t=2,4,\ldots,m-2, $$
$$ (-x_0,-\zeta_m^tx_1,-\zeta_m^{2t}x_2,\ldots,-\zeta_m^{(m-1)t}x_{m-1},y),\quad t=1,3,\ldots,m-1, $$
are all in $V$.

Updated: If we apply discrete Fourier transform to $(x_1,x_3,\dots,x_{m-1})$ and $(x_0,x_2,\dots,x_{m-2})$ respectively, the system takes a "dual" form. For example, suppose $l=\frac{m}{2}$ to be odd hereafter and let $$ w^{2s+1}x_{2s+1}=\sum\limits_{j=0}^{l-1}y_j\zeta_l^{(s-\frac{l-1}{2})j}, \quad x_{2s}=\sum\limits_{j=0}^{l-1}z_j\zeta_l^{sj} $$ for $s=0,\dots,l-1$, where $\zeta_l$ is a fixed primitive $l$th root of unity. Then the original system (which I will call the system on $x$-level) turns out to be the following system on $(y,z)$-level: \begin{cases} &\sum\limits_{j=0}^{l-1}z_j=x_0,\\ &y_s^2+z_{s-r}^2=\frac{x_0^2+m-1}{l^2},\ s=0,\ldots,l-1,\\ &\sum\limits_{j=0}^{l-1}y_jy_{j+s}=0,\ s=1,\ldots,l-1,\\ &\sum\limits_{j=0}^{l-1}y_j^2=1,\\ &\sum\limits_{j=0}^{l-1}z_jz_{j+s}=\frac{x_0^2-1}{l},\ s=1,\ldots,l-1,\\ &\sum\limits_{j=0}^{l-1}z_j^2=\frac{x_0^2+l-1}{l},\\ &y_s=\sum\limits_{j=0}^{l-1}z_jz_{2j-s}+\frac{x_0^2-1}{l},\ s=0,\ldots,l-1, \end{cases} where $w^2=\zeta_l^r$. There are mainly two benefits to consider the system on $(y,z)$-level instead of $x$-level. First, the variable $w$ in the system on $x$-level is treated as a parameter (corresponds to $r$) in the system on $(y,z)$-level, which enables my machine to compute for larger $m$ (computation result for $m=22$ confirms conjectures (a) and (b)). Second, when consider the solution of the $x$-level system in $\mathbb{R}\times(\mathbb{S}^1)^m$, the corresponding $(y,z)$-level system is then imposed the condition that all the variables are in $\mathbb{R}$.



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