摘要
主要运用Pell方程、递归数列、同余式及(非)平方剩余等一些初等的证明方法,证明了不定方程x(x+1)(x+2)·(x+3)=13y(y+1)(y+2)(y+3)无正整数解。在证明该结论的过程中,对不定方程进行变形和整理,将其化为Pell方程形式。根据得到的Pell方程整数解的情况,从而得到6类整数解。根据原不定方程的情况舍去了两类,剩余4类整数解。本文逐一对每一类整数解用同余式及平方剩余的证明方法进行讨论和证明,最后得到原不定方程无正整数解的结论。根据本文的结论也能得到这个不定方程的全部整数解,它们都为其平凡解,由于比较简单,故文中没有再给出。同时本文证明了不定方程(x2+3x+1)2-13y2=-12仅有整数解(x,±y)=(0,1),(-3,1),(-2,1),(-1,1),(-14,43),(11,43)。本文进一步完善了此类不定方程的正整数解的研究。
In this paper, with the elementary method of recurrence sequence, congruent form and quadratic residue, the author has shown that the Diophantine equationx (x+1) (x+2) (x+3) =13y (y+1) (y+2) (y+3) has no positive integer solution. In the process of proving this conclusion, the authors change the Diophantine equation to a form of Pell-equation. According to the in- teger solution of the Pell-equation, we get six classes of integer solution. Because of the Diophantine equation, we get four classes of integer solution. Then, elementary method of congruent form and quadratic residue are used to research them. After that, we argue that the Diophantine equation x (x+1) (x+2) (x+-3) =13y (y+l) (y+2) (y+3)has no positive integer solution. Besides, we can easily find all integer solution of this Diophantine equation, which are trivial solutions. Elementary theory of numbers is used through all the process of proving. Also, the total integer solutions of Diophantine equation (x2+3x+1)2 -13y2= -12 are (x, ± y) = (0, 1), (-3, 1), (--2, 1) , (--1, 1), (--14, 43), (11, 43)hasbeenproved. In a word, the positiveintegersolu tion of such Diophantine equation research becomes more perfective through our work.
出处
《重庆师范大学学报(自然科学版)》
CAS
CSCD
北大核心
2013年第5期101-105,共5页
Journal of Chongqing Normal University:Natural Science
关键词
不定方程
整数解
递归数列
diophantine equation
integer solution
recurrence sequence