{"title":"On the quotient set of the distance set","authors":"A. Iosevich, D. Koh, Hans Parshall","doi":"10.2140/moscow.2019.8.103","DOIUrl":null,"url":null,"abstract":"Let ${\\Bbb F}_q$ be a finite field of order $q.$ We prove that if $d\\ge 2$ is even and $E \\subset {\\Bbb F}_q^d$ with $|E| \\ge 9q^{\\frac{d}{2}}$ then $$ {\\Bbb F}_q=\\frac{\\Delta(E)}{\\Delta(E)}=\\left\\{ \\frac{a}{b}: a \\in \\Delta(E), b \\in \\Delta(E) \\backslash \\{0\\} \\right\\},$$ where $$ \\Delta(E)=\\{||x-y||: x,y \\in E\\}, \\ ||x||=x_1^2+x_2^2+\\cdots+x_d^2.$$ If the dimension $d$ is odd and $E\\subset \\mathbb F_q^d$ with $|E|\\ge 6q^{\\frac{d}{2}},$ then $$ \\{0\\}\\cup\\mathbb F_q^+ \\subset \\frac{\\Delta(E)}{\\Delta(E)},$$ where $\\mathbb F_q^+$ denotes the set of nonzero quadratic residues in $\\mathbb F_q.$ Both results are, in general, best possible, including the conclusion about the nonzero quadratic residues in odd dimensions.","PeriodicalId":36590,"journal":{"name":"Moscow Journal of Combinatorics and Number Theory","volume":" ","pages":""},"PeriodicalIF":0.0000,"publicationDate":"2018-02-22","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"https://sci-hub-pdf.com/10.2140/moscow.2019.8.103","citationCount":"11","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Moscow Journal of Combinatorics and Number Theory","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.2140/moscow.2019.8.103","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q4","JCRName":"Mathematics","Score":null,"Total":0}
引用次数: 11
Abstract
Let ${\Bbb F}_q$ be a finite field of order $q.$ We prove that if $d\ge 2$ is even and $E \subset {\Bbb F}_q^d$ with $|E| \ge 9q^{\frac{d}{2}}$ then $$ {\Bbb F}_q=\frac{\Delta(E)}{\Delta(E)}=\left\{ \frac{a}{b}: a \in \Delta(E), b \in \Delta(E) \backslash \{0\} \right\},$$ where $$ \Delta(E)=\{||x-y||: x,y \in E\}, \ ||x||=x_1^2+x_2^2+\cdots+x_d^2.$$ If the dimension $d$ is odd and $E\subset \mathbb F_q^d$ with $|E|\ge 6q^{\frac{d}{2}},$ then $$ \{0\}\cup\mathbb F_q^+ \subset \frac{\Delta(E)}{\Delta(E)},$$ where $\mathbb F_q^+$ denotes the set of nonzero quadratic residues in $\mathbb F_q.$ Both results are, in general, best possible, including the conclusion about the nonzero quadratic residues in odd dimensions.