关于单位分数的Lazar问题  

On a problem of Lazar on unit fractions

作　　者：卢健 李懋[2] 邱敏 LU Jian;LI Mao;QIU Min(School of Mathematics, Sichuan University, Chengdu 610064, China;School of Mathematics and Statistics, Southwest University, Chongqing 400715, China;School of Science, Xihua University, Chengdu 610039, China)

机构地区：[1]四川大学数学学院,成都610064 [2]西南大学数学与统计学院,重庆400715 [3]西华大学理学院,成都610039

出　　处：《四川大学学报（自然科学版）》2020年第6期1067-1072,共6页Journal of Sichuan University(Natural Science Edition)

基　　金：国家自然科学基金(11771304)。

摘　　要：设n为任意正整数.Erdös-Straus猜想是指当n≥2时,Diophantine方程4n=1x+1y+1z总有正整数解(x,y,z).设p≥5为任意素数.最近,Lazar证明Diophantine方程4p=1x+1y+1z在区域xy<z/2内没有x与y互素的正整数解(x,y,z).同时,Lazar提出问题:在上述方程中以5/p替换4/p,是否有类似结果?这也是Sierpinski提出的一个猜想.本文证明Diophantine方程ap=1x+1y+1z没有满足x,y互素且xy<z/2的正整数解(x,y,z),其中a为满足a<7≤p的正整数.这回答了上述Lazar问题,推广了Lazar的结果.证明方法和工具主要是利用有理数ap的连分数表示.Let n be a positive integer.The well-known Erdös-Straus conjecture asserts that the positive integral solution of the Diophantine equation 4n=1x+1y+1z always exists when n≥2.Recently,Lazar investigated some properties of the solutions to the above Diophantine equation in the special case that n is a prime number.Let p≥5 be a prime number.Lazar showed that there are no triple of positive integers(x,y,z)which is solution of the Diophantine equation 4p=1x+1y+1z in the range xy<z/2 and x,y=1.Meanwhile,Lazar pointed out that it would be interesting to find an analog of this result for 5/p instead of 4/p,which is also a conjecture due to Sierpinski.In this paper,we answer Lazar's question affirmatively and also extended Lazar's result by showing that the Diophantine equation ap=1x+1y+1z does not have any integer solution(x,y,z)such that x and y are coprime and xy<z/2,where a is a positive integer such that a<7≤p.Our proof mainly uses the continued fraction expansion of ap.

关 键 词：DIOPHANTINE方程 连分数 渐近分数 Erdös-Straus猜想 

分 类 号：O156.1[理学—数学] O156.7[理学—基础数学]