关于不定方程5x(x+1)(x+2)(x+3)=18y(y+1)(y+2)(y+3)  

On the Diophantine Equation 5x(x+1)(x+2)(x+3)=18y(y+1)(y+2)(y+3)

作　　者：杨晓柳[1] 牟全武[1] YANG Xiao-liu;MU Quan-wu(School of Science, Xi'an Polytechnic University, Xi'an 710048, China)

机构地区：[1]西安工程大学理学院,西安710048

出　　处：《西南大学学报（自然科学版）》2019年第4期92-96,共5页Journal of Southwest University(Natural Science Edition)

基　　金：国家自然科学基金项目(11271283);西安工程大学基础课程质量提升项目(104020184)

摘　　要：运用Pell方程、递推序列、同余式及平方剩余等初等数论知识,证明了不定方程5x(x+1)(x+2)(x+3)=18y(y+1)(y+2)(y+3)仅有4组非平凡整数解(x,y)=(6, 4),(-9, 4),(6,-7),(-9,-7),同时给出该不定方程的全部整数解,分别为(x,y)=(0, 0),(0,-1),(0,-2),(0,-3),(-1, 0),(-1,-1),(-1,-2),(-1,-3),(-2, 0),(-2,-1),(-2,-2),(-2,-3),(-3, 0),(-3,-1),(-3,-2),(-3,-3),(6, 4),(-9, 4),(6,-7),(-9,-7).Using some techniques of elementary number theory involving Pell’s equation, recurrent sequence, congruence and quadratic residues, this paper proves that the diophantine equation 5x(x+1)(x+2)(x+3)=18y(y+1)(y+2)(y+3) has only four non-trivial solutions, i.e.(x, y)=(6, 4),(-9, 4),(6,-7),(-9,-7), and gives all of its integer solutions, i.e.(x, y)=(0, 0),(0,-1),(0,-2),(0,-3),(-1, 0),(-1,-1),(-1,-2),(-1,-3),(-2, 0),(-2,-1),(-2,-2),(-2,-3),(-3, 0),(-3,-1),(-3,-2),(-3,-3),(6, 4),(-9, 4),(6,-7),(-9,-7).

关 键 词：不定方程 整数解 递推序列 平方剩余 

分 类 号：O156.2[理学—数学]