检索结果-南通市图书馆

A New Proof for Congruent Number217;s Problem via Pythagorician Divisors

Advances in Pure Mathematics 2024年第4期14卷 283-302页

作者： Léopold Dèkpassi Keuméan François Emmanuel Tanoé UFR Mathematics and Computer Science F&#233lix Houphou&#235t Boigny University Abidjan Cote d&#8217Ivoire

Considering Pythagorician divisors theory which leads to a new parameterization, for Pythagorician triplets ( a,b,c )∈ ℕ 3∗ , we give a new proof of the well-known problem of these particular squareless numbers n∈ ℕ ∗ , called congruent numbers, characterized by the fact that there exists a right-angled triangle with rational sides: ( A α ) 2 + ( B β ) 2 = ( C γ ) 2 , such that its area Δ= 1 2 A α B β =n;or in an equivalent way, to that of the existence of numbers U 2 , V 2 , W 2 ∈ ℚ 2∗ that are in an arithmetic progression of reason n;Problem equivalent to the existence of: ( a,b,c )∈ ℕ 3∗ prime in pairs, and f∈ ℕ ∗ , such that: ( a−b 2f ) 2 , ( c 2f ) 2 , ( a+b 2f ) 2 are in an arithmetic progression of reason n;And this problem is also equivalent to that of the existence of a non-trivial primitive integer right-angled triangle: a 2 + b 2 = c 2 , such that its area Δ= 1 2 ab=n f 2 , where f∈ ℕ ∗ , and this last equation can be written as follows, when using Pythagorician divisors: (1) Δ= 1 2 ab= 2 S−1 d e ¯ ( d+ 2 S−1 e ¯ )( d+ 2 S e ¯ )=n f 2;Where ( d, e ¯ )∈ ( 2ℕ+1 ) 2 such that gcd( d, e ¯ )=1 and S∈ ℕ ∗ , where 2 S−1 , d, e ¯ , d+ 2 S−1 e ¯ , d+ 2 S e ¯ , are pairwise prime quantities (these parameters are coming from Pythagorician divisors). When n=1 , it is the case of the famous impossible problem of the integer right-angled triangle area to be a square, solved by Fermat at his time, by his famous method of infinite descent. We propose in this article a new direct proof for the numbers n=1 (resp. n=2 ) to be non-congruent numbers, based on an particular induction method of resolution of Equation (1) (note that this method is efficient too for general case of prime numbers n=p≡a ( ( mod8 ) , gcd( a,8 )=1 ). To prove it, we use a classical proof by induction on k , that shows the non-solvability property of any of the following systems ( t=0 , corresponding to case n=1 (resp. t=1 , corresponding to case n=2 )): ( Ξ t,k ){ X 2 + 2 t ( 2 k Y ) 2 = Z 2 X 2 + 2

关键词： prime numbers-diophantine equations of degree 2 & 4 Factorization Greater Common Divisor Pythagoras Equation Pythagorician Triplets Congruent numbers Inductive Demonstration Method Infinite Descent BSD Conjecture

来源：

维普期刊数据库评论

在线全文

维普期刊数据库

学校读者我要写书评

暂无评论

欢迎您,

建议与咨询留下您的常用邮箱和电话号码，以便我们向您反馈解决方案和替代方法

时间限定

文献类型

馆藏选择

核心期刊

语言

文献类型

帮助

文字说明：

检索规则说明：

检索范例：

分类表

所选分类

限定检索结果

文献类型

馆藏范围

日期分布

学科分类号

主题

机构

作者

语言

在线全文

请选择保存的检索档案：

请选择收藏分类：

通借通还

欢迎您,

建议与咨询 留下您的常用邮箱和电话号码，以便我们向您反馈解决方案和替代方法

时间限定

文献类型

馆藏选择

核心期刊

语言

文献类型

帮助

文字说明：

检索规则说明：

检索范例：

分类表

所选分类

限定检索结果

文献类型

馆藏范围

日期分布

学科分类号

主题

机构

作者

语言

在线全文

请选择保存的检索档案： 新增检索档案 确定 取消

请选择收藏分类： 新增自定义分类 确定 取消

通借通还

建议与咨询留下您的常用邮箱和电话号码，以便我们向您反馈解决方案和替代方法

请选择保存的检索档案：

请选择收藏分类：