{"title":"基于几何原理的费马大定理证明","authors":"Yuriy Gevorkyan","doi":"10.21303/2461-4262.2023.003109","DOIUrl":null,"url":null,"abstract":"This paper provides another proof of Fermat's theorem. As in the previous work, a geometric approach is used, namely: instead of integers a, b, c, a triangle with side lengths a, b, c is considered. To preserve the completeness of the proof of the theorem in this work, the proof is repeated for the cases of right and obtuse triangles. In this case, the Fermat equation ap+bp=cp has no solutions for any natural number p>2 and arbitrary numbers a, b, c. When considering the case when the numbers a, b, c are sides of an acute triangle, it is proven that Fermat’s equation has no solutions for any natural number p>2 and non-zero integer numbers a, b, c. Numbers a=k, b=k+m, c=k+n, where k, m, n are natural numbers that satisfy the inequalities n>m, n<k+m, exhaust all possible variants of natural numbers a, b, c, which are the sides of the triangle. In an acute triangle, the following condition is additionally satisfied:
 To study the Fermat equation, an auxiliary function f(k,p)=kp+(k+m)p–(k+n)p, is introduced, which is a polynomial of natural degree p in the variable k. The equation f(k,p)=0 has a single positive root for any natural
 A recurrent formula connecting the functions f(k,p+1) and f(k,p) has been proven: f(k,p+1)=kf(k,p)-[n(k+n)p-m(k+m)p]. The proof of the main proposition 2 is based on considering all possible relationships between the assumed integer solution of the equation f(k,p+1)=0 and the number corresponding to this solution
 The proof was carried out using the mathematical apparatus of number theory, elements of higher algebra and the foundations of mathematical analysis. These studies are a continuation of the author’s works, in which some special cases of Fermat’s theorem were proved","PeriodicalId":11804,"journal":{"name":"EUREKA: Physics and Engineering","volume":null,"pages":null},"PeriodicalIF":0.0000,"publicationDate":"2023-09-29","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"The proof of Fermat’s last theorem based on the geometric principle\",\"authors\":\"Yuriy Gevorkyan\",\"doi\":\"10.21303/2461-4262.2023.003109\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"This paper provides another proof of Fermat's theorem. As in the previous work, a geometric approach is used, namely: instead of integers a, b, c, a triangle with side lengths a, b, c is considered. To preserve the completeness of the proof of the theorem in this work, the proof is repeated for the cases of right and obtuse triangles. In this case, the Fermat equation ap+bp=cp has no solutions for any natural number p>2 and arbitrary numbers a, b, c. When considering the case when the numbers a, b, c are sides of an acute triangle, it is proven that Fermat’s equation has no solutions for any natural number p>2 and non-zero integer numbers a, b, c. Numbers a=k, b=k+m, c=k+n, where k, m, n are natural numbers that satisfy the inequalities n>m, n<k+m, exhaust all possible variants of natural numbers a, b, c, which are the sides of the triangle. In an acute triangle, the following condition is additionally satisfied:
 To study the Fermat equation, an auxiliary function f(k,p)=kp+(k+m)p–(k+n)p, is introduced, which is a polynomial of natural degree p in the variable k. The equation f(k,p)=0 has a single positive root for any natural
 A recurrent formula connecting the functions f(k,p+1) and f(k,p) has been proven: f(k,p+1)=kf(k,p)-[n(k+n)p-m(k+m)p]. The proof of the main proposition 2 is based on considering all possible relationships between the assumed integer solution of the equation f(k,p+1)=0 and the number corresponding to this solution
 The proof was carried out using the mathematical apparatus of number theory, elements of higher algebra and the foundations of mathematical analysis. These studies are a continuation of the author’s works, in which some special cases of Fermat’s theorem were proved\",\"PeriodicalId\":11804,\"journal\":{\"name\":\"EUREKA: Physics and Engineering\",\"volume\":null,\"pages\":null},\"PeriodicalIF\":0.0000,\"publicationDate\":\"2023-09-29\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"EUREKA: Physics and Engineering\",\"FirstCategoryId\":\"1085\",\"ListUrlMain\":\"https://doi.org/10.21303/2461-4262.2023.003109\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"Q3\",\"JCRName\":\"Engineering\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"EUREKA: Physics and Engineering","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.21303/2461-4262.2023.003109","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q3","JCRName":"Engineering","Score":null,"Total":0}
引用次数: 0
摘要
本文给出了费马定理的另一个证明。与之前的工作一样,使用了几何方法,即:考虑边长为a, b, c的三角形,而不是整数a, b, c。为了保持本工作中定理证明的完整性,对直角三角形和钝角三角形重复证明。在这种情况下,费马方程ap + bp = cp没有任何自然数p>解决方案2和任意数字a, b, c。当考虑当数字,b, c是一个锐角三角形,这是证明费马方程没有任何自然数p>解决方案2和非零的整数数字a, b, c。数字= k b = k + m, c = k + n, k, m, n是自然数,满足不等式n> m, n< k + m,耗尽所有可能的变体的自然数,b, c,它们是三角形的两条边。在锐角三角形中,还满足下列条件:
为了研究费马方程,引入一个辅助函数f(k,p)=kp+(k+m)p - (k+n)p,它是变量k的自然次p的多项式,方程f(k,p)=0对任意自然
都有一个正根;证明了一个连接函数f(k,p+1)和f(k,p)的循环公式:f(k,p+1)=kf(k,p)-[n(k+n)p-m(k+m)p]。主要命题2的证明是基于考虑方程f(k,p+1)=0的假设整数解与该解对应的数
之间的所有可能关系;这个证明是用数论的数学工具、高等代数的基本原理和数学分析的基础来完成的。这些研究是作者对费马定理的一些特殊情况的证明工作的延续
The proof of Fermat’s last theorem based on the geometric principle
This paper provides another proof of Fermat's theorem. As in the previous work, a geometric approach is used, namely: instead of integers a, b, c, a triangle with side lengths a, b, c is considered. To preserve the completeness of the proof of the theorem in this work, the proof is repeated for the cases of right and obtuse triangles. In this case, the Fermat equation ap+bp=cp has no solutions for any natural number p>2 and arbitrary numbers a, b, c. When considering the case when the numbers a, b, c are sides of an acute triangle, it is proven that Fermat’s equation has no solutions for any natural number p>2 and non-zero integer numbers a, b, c. Numbers a=k, b=k+m, c=k+n, where k, m, n are natural numbers that satisfy the inequalities n>m, n