关于丢番图方程x(x+1)(x+2)(x+3)=68y(y+1)(y+2)(y+3)  

On the Solution of Diophantine Equation x(x+1)(x+2)(x+3)=68y(y+1)(y+2)(y+3)

作　　者：林丽娟 LIN Li-juan(Department of General Studies Chongqing Chemical Industry Vocational College,Chongqing 401228,China)

机构地区：[1]重庆化工职业学院通识教育学院,重庆401228

出　　处：《唐山师范学院学报》2023年第6期7-9,共3页Journal of Tangshan Normal University

基　　金：重庆市教委科学技术研究项目(KJQN202204501);重庆化工职业学院校级科研课题(HZY2021-KJ03)。

摘　　要：利用Pell方程基本解性质、递推序列、同余思想以及二次剩余等初等方法得到并证明了在(M,N)=(1,68)时不定方程Mx(x+1)(x+2)(x+3)=Ny(y+1)(y+2)(y+3)仅有2组非平凡整数解(x,y)=(14,4),(14,-7)。Based on the elementary methods such as the basic solution of Pell Equation,Recursive Sequence and Congruence Theory,it was proved that the Diophantine Equation x(x+1)(x+2)(x+3)=68y(y+1)(y+2)(y+3)with(M,N)=(1,68)has two nontrivial integer solutions,(x,y)=(14,4),(14,-7).

关 键 词：不定方程 PELL方程 递推序列 整数解 

分 类 号：O156.1[理学—数学]