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A note on Jesmanowicz' conjecture concerning primitive Pythagorean triples

机译:关于Jesmanowicz关于原始毕达哥拉斯三元组猜想的注记

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摘要

Let (a, b, c) be a primitive Pythagorean triple such that a = -u(2) - v(2), b = 2uv, c = u(2) - v(2), where u, v are positive integers satisfying u > v, gcd(u, v) = 1 and 2 vertical bar uv. In 1956, L. Jesmanowicz conjectured that the equation (an)(x)+(bn)(y) = (cn)(z) has only the positive integer solutions (x, y, z, n) = (2, 2, 2, m), where m is an arbitrary positive integer. A positive integer solution (x, y, z, n) of the equation is called exceptional if (x, y, z) not equal (2, 2, 2) and n > 1. In this paper we prove the following results: (i) The equation has no positive integer solutions (x, y, z, n) which satisfy x = y, y > z and n > 1. (ii) If (x, y, z, n) is an exceptional solution of the question, then either y > z > x or x > z > y. (iii) If u = 2(r), v = 2(r) 1, where r is a positive.integer, then the equation has no exceptional solutions (x, y, z, n) with y > z > x. In particular, if 2(r) 1 is an odd prime, then the equation has no exceptional solutions. The last result means Jegmanowicz conjecture is true in this case. (C) 2015 Elsevier Inc. All rights reserved.
机译:令(a,b,c)为原始毕达哥拉斯三元组,使得a = -u(2)-v(2),b = 2uv,c = u(2)-v(2),其中u,v为正满足u> v的整数,gcd(u,v)= 1和2垂直线uv。 1956年,L.Jesmanowicz推测方程(an)(x)+(bn)(y)=(cn)(z)​​仅具有正整数解(x,y,z,n)=(2,2 ,2,m),其中m是任意正整数。如果(x,y,z)不等于(2,2,2)并且n> 1,则方程的正整数解(x,y,z,n)被称为例外。在本文中,我们证明以下结果: (i)该方程没有满足x = y,y> z和n> 1的正整数解(x,y,z,n)。(ii)如果(x,y,z,n)是一个例外解问题,则y> z> x或x> z> y。 (iii)如果u = 2(r),v = 2(r)1,其中r是一个正整数,则该方程没有y> z> x的例外解(x,y,z,n)。特别是,如果2(r)1是奇质数,则该方程式没有例外的解。最后的结果意味着在这种情况下Jegmanowicz猜想是正确的。 (C)2015 Elsevier Inc.保留所有权利。

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