关于丟番图方程x(x+1)(x+2)(x+3)=27y(y+1)(y+2)(y+3)  

On the Diophantine Equation x(x+1)(x+2)(x+3)=27y(y+1)(y+2)(y+3)

作　　者：张芊 ZHANG Qian(School of Mathematics and Statistics,Southwest University,Chongqing 400715,China)

机构地区：[1]西南大学数学与统计学院,重庆400715

出　　处：《数学的实践与认识》2023年第3期284-290,共7页Mathematics in Practice and Theory

摘　　要：先后运用了pell方程、勒让德符号,同余关系,递归序列、二次平方剩余,分类讨论的有关方法,并通过使用数学软件Mathematica进行计算,证明了以下结论:不定方程x(x+1)(x+2)(x+3)=27y(y+1)(y+2)(y+3)没有正整数解,并找出了该方程的全部16组整数解.In this paper,Pell equation,Legendre symbol,congruence relation,recursive sequence,quadratic squareresidue,and related methods of classified discussion have been usedsuccessively,and by using mathematical software Mathematica to calculate,thefollowing conclusions have been proved:the indefinite equation x(x+1)(x+2)(x+3)=27y(y+1)(y+2)(y+3) has no positive integer solution,and all 16 groups ofinteger solutions of the equation have been found.

关 键 词：不定方程 勒让德符号 整数解 平方剩余 分类讨论 递归序列 

分 类 号：O156.7[理学—数学]