# Getting a bound via polynomial equations

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.

Question:

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.

Remarks:

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 $$(x_0,\zeta_m^tx_1,\zeta_m^{2t}x_2,\ldots,\zeta_m^{(m-1)t}x_{m-1},y),\quad t=2,4,\ldots,m-2,$$
and
$$(-x_0,-\zeta_m^tx_1,-\zeta_m^{2t}x_2,\ldots,-\zeta_m^{(m-1)t}x_{m-1},y),\quad t=1,3,\ldots,m-1,$$
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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