关于丢番图方程x^3+1=143y^2  

On the Diophantine Equation x^3+1=143y^2

作　　者：吴小英 周科[2] 

机构地区：[1]四川师范大学,四川成都610068 [2]广西师范学院,广西南宁530023

出　　处：《广西师范学院学报（自然科学版）》2017年第4期46-49,共4页Journal of Guangxi Teachers Education University(Natural Science Edition)

基　　金：国家自然科学基金项目(11671283)

摘　　要：该文运用简单同余法、分解因子法、Pell方程法等初等方法求解丢番图方程x^3+1=143y^2.首先运用因式分解法把丢番图方程x^3+1=11×13y^2分解为与之等价的8个方程组,然后运用同余、转化、勒让德符号等初等数论的基础知识、方法,证明前7个方程组无解,最后运用递归数列以及Pell方程的解的性质证明最后一个方程组仅有唯一解,由此得到丢番图方程x^3+1=143y^2有且仅有整数解(x,y)=(-1,0).In order to obtain integer solutions of Diophantine equation x3＋1 =143 ; y 2 ,w e use a simple congruence factor decomposition, Pell equation and so on. During the solving process firstly, we turn the Diophantine equation x3 ＋1 = 1 43; y 2 into eight equations which are equaled bdecomposition method. Secondly,we use the basic knowledge of elementary number theory which in-cludes congruence, conversion,Legendre symbol and so on to prove that the previous seven equations has no solution. Finally,we use Recursion sequence and the nature ofthat the last equation has the only solution Therefore,we can come to a conclusion that the Diophan- tine equation x 3 ＋1 = 1 43; y2 has the only integer solut ion,which is x 3 ＋1 = 1 43; y 2.

关 键 词：丢番图方程 整数解 平方剩余 同余式 

分 类 号：O156.7[理学—数学]