{"title":"Green’s problem on additive complements of the squares","authors":"Yuchen Ding","doi":"10.5802/crmath.107","DOIUrl":null,"url":null,"abstract":"Let A and B be two subsets of the nonnegative integers. We call A and B additive complements if all sufficiently large integers n can be written as a +b, where a ∈ A and b ∈ B . Let S = {12,22,32, · · ·} be the set of all square numbers. Ben Green was interested in the additive complement of S. He asked whether there is an additive complement B = {bn }n=1 ⊆Nwhich satisfies bn = π 2 16 n 2+o(n2). Recently, Chen and Fang proved that if B is such an additive complement, then limsup n→∞ π2 16 n 2 −bn n1/2 logn ≥ √ 2 π 1 log4 . They further conjectured that limsup n→∞ π2 16 n 2 −bn n1/2 logn =+∞. In this paper, we confirm this conjecture by giving a much more stronger result, i.e., limsup n→∞ π2 16 n 2 −bn n ≥ π 4 . 2020 Mathematics Subject Classification. 11B13, 11B75. Manuscript received 3rd August 2020, revised 19th August 2020, accepted 20th August 2020.","PeriodicalId":0,"journal":{"name":"","volume":null,"pages":null},"PeriodicalIF":0.0,"publicationDate":"2020-12-03","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"2","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"","FirstCategoryId":"100","ListUrlMain":"https://doi.org/10.5802/crmath.107","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 2
Abstract
Let A and B be two subsets of the nonnegative integers. We call A and B additive complements if all sufficiently large integers n can be written as a +b, where a ∈ A and b ∈ B . Let S = {12,22,32, · · ·} be the set of all square numbers. Ben Green was interested in the additive complement of S. He asked whether there is an additive complement B = {bn }n=1 ⊆Nwhich satisfies bn = π 2 16 n 2+o(n2). Recently, Chen and Fang proved that if B is such an additive complement, then limsup n→∞ π2 16 n 2 −bn n1/2 logn ≥ √ 2 π 1 log4 . They further conjectured that limsup n→∞ π2 16 n 2 −bn n1/2 logn =+∞. In this paper, we confirm this conjecture by giving a much more stronger result, i.e., limsup n→∞ π2 16 n 2 −bn n ≥ π 4 . 2020 Mathematics Subject Classification. 11B13, 11B75. Manuscript received 3rd August 2020, revised 19th August 2020, accepted 20th August 2020.