丢番图方程(75n)x+ (308n)y= (317n)z
On the Diophantine Equation (75n)x+ (308n)y= (317n)z
DOI: 10.12677/PM.2023.1311348, PDF,   
作者: 黄日娣, 邓乃娟:湛江幼儿师范专科学校数学系,广东 湛江
关键词: Jesmanowicz猜想丢番图方程正整数解Jesmanowicz’s Conjecture Diophantine Equation Positive Integer Solution
摘要: 设a,b,c是两两互素的正整数且a2+b2=c2。Jesmanowicz猜想:对于任意给定的正整数n,方程(an)x+(bn)y=(cn)z只有正整数解(x,y,z)=(2,2,2)。本文利用数论中的一些方法证明了:对任意的正整数n,方程(75n)x+ (308n)y= (317n)z只有正整数解(x,y,z)=(2,2,2),即当(a,b,c)=(75,308,317)时,Jesmanowicz猜想成立。
Abstract: Let a,b,c be a primitive Pythagogrean triples such that a2+b2=c2. Jesmanowicz conjectured that, for any positive integer n, the Diophantine equation (an)x+(bn)y=(cn)z has only positive integer solution (x,y,z)=(2,2,2). In this paper, by using some methods of number theory,we prove that, for any positive integer n, the Diophantine equation (75n)x+ (308n)y= (317n)z has only positive integer solution (x,y,z)=(2,2,2), that is the Jesmanowicz conjecture is true, when (a,b,c)=(75,308,317).
文章引用:黄日娣, 邓乃娟. 丢番图方程(75n)x+ (308n)y= (317n)z[J]. 理论数学, 2023, 13(11): 3358-3364. https://doi.org/10.12677/PM.2023.1311348

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