The generalized k-resultant modulus set problem in finite fields

Abstract

Let ${\mathbb{F}}_q^d$ be the $d$-dimensional vector space over the finite field ${\mathbb{F}}_q$ with $q$ elements. Given $k$ sets $E_j\subset {\mathbb{F}}_q^d$ for $j=1,2,\dots ,k$, the generalized $k$-resultant modulus set, denoted by ${\Delta }_k(E_1,E_2,\dots ,E_k)$, is defined by $${\Delta }_k(E_1,E_2,\dots ,E_k)=\left\{\Vert {\mathbf{x}}^1+{\mathbf{x}}^2+\cdots +{\mathbf{x}}^k\Vert \in {\mathbb{F}}_q:{\mathbf{x}}^j\in E_j,j=1,2,\dots ,k\right\},$$ where $\Vert \mathbf{y}\Vert ={\mathbf{y}}_1^2+\cdots +{\mathbf{y}}_d^2$ for $\mathbf{y}=({\mathbf{y}}_1,\dots ,{\mathbf{y}}_d)\in {\mathbb{F}}_q^d.$ We prove that if $\prod _{j=1}^3|E_j|\ge C{q}^{3\left(\frac{d+1}{2}-\frac{1}{6d+2}\right)}$ for $d=4,6$ with a sufficiently large constant $C>0$, then $|{\Delta }_3(E_1,E_2,E_3)|\ge cq$ for some constant $0<c\le 1,$ and if $\prod _{j=1}^4|E_j|\ge C{q}^{4\left(\frac{d+1}{2}-\frac{1}{6d+2}\right)}$ for even $d\ge 8,$ then $|{\Delta }_4(E_1,E_2,E_3,E_4)|\ge cq.$ This generalizes the previous result in \cite{CKP16}. We also show that if $\prod _{j=1}^3|E_j|\ge C{q}^{3\left(\frac{d+1}{2}-\frac{1}{9d-18}\right)}$ for even $d\ge 8,$ then $|{\Delta }_3(E_1,E_2,E_3)|\ge cq.$ This result improves the previous work in \cite{CKP16} by removing $\epsilon >0$ from the exponent.