...
首页> 外文期刊>International Journal of Discrete Mathematics >On the Set of Primitive Triples of Natural Numbers Satisfying the Diophantine Equation of Pythagor
【24h】

On the Set of Primitive Triples of Natural Numbers Satisfying the Diophantine Equation of Pythagor

机译:关于满足Pythagoras衍生线方程的自然数的原始三元组

获取原文
           

摘要

The main task considered in the article is to find the condition primitive integer solutions of the Diophantine Pithagorean equation x~2+y~2=z~2 It is known that for this purpose it is enough to find primive solution of x, y such that x is even and y is odd. In this paper, in particular, we proved that the z of a primitive solution is a Prime number of the form 4k+1. It is prove in this paper that any right triangle with integer side lengths has a hypotenuse equal to a Prime of the form 4k+1and we show with the help of the descent axiom how to find primitive solutions of x and y in this case. We divide the search for primitive solutions (x, y, z) of right triangles into two cases: 1) the hypotenuse of such triangles is a Prime number of the form 4k+1 and 2) the hypotenuse of such triangles is a composite number. In section 3 we use formulas known to the ancient Hindus to find primitive solutions of Pithagorean equations in cases where m and nm are mutually Prime numbers of different parity, and this happens in cases where, for example, n=4, and m is a Prime number ending in 3 or 7. In such cases, the hypotenuse z=m~2+n~2 is an compaund number ending in 5. To find primes ending in 3 and 7, we refer the reader to our paper, which presents algorithms for constructing all primes and twin primes. The proposed paper also presents a generalization of Euclid's fundamental result on the infinity of the set of Primes, namely, it is shown that all twin primes are in residue classes (1, 3), (2, 4), (4, 1), and there are infinity many such twins.
机译:在文章中考虑的主要任务是找到衍生的原始整数解决方案的二色子型Pithagorean等式x〜2 + y〜2 = z〜2,已知为此目的,它足以找到x,y的原始解决方案x是偶数,y是奇数。在本文中,特别是,我们证明了原始溶液的Z是形式4K + 1的素数。本文证明,具有整数侧长度的任何正确的三角形具有等于形式4K + 1的素数的斜边,我们在下降公理的帮助下显示了在这种情况下如何找到X和Y的原始解。我们将搜索右三角形的原始解决方案(x,y,z)分为两种情况:1)这种三角形的斜边是形式4k + 1和2)的素数,这种三角形的斜边是复合数字。在第3节中,在古代印度教徒所知的公式中,在M和N

著录项

相似文献

  • 外文文献
  • 中文文献
  • 专利
获取原文

客服邮箱:kefu@zhangqiaokeyan.com

京公网安备:11010802029741号 ICP备案号:京ICP备15016152号-6 六维联合信息科技 (北京) 有限公司©版权所有
  • 客服微信

  • 服务号