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arXiv 1402.6440math.NT
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毕达哥拉斯三元组的Frobenius数

Frobenius numbers of Pythagorean triples

Byung Keon Gil, Ji-woo Han, Tae Hyun Kim, Ryun Han Koo, Bon Woo Lee, Jaehoon Lee, Kyeong Sik Nam, Hyeon Woo Park, Poo-Sung Park

摘要:

给定互素整数$a_1, \dotsc, a_n$,Frobenius数$g(a_1, \dotsc, a_n)$定义为不能表示为$x_1 a_1 + \dotsb + x_n a_n$且$x_i$为非负整数的最大整数.本文给出了本原毕达哥拉斯三元组的Frobenius数。即,\[ g(m^2-n^2, 2mn, m^2+n^2) = (m-1)(m^2-n^2) + (m-1)(2mn) - (m^2 + n^2). \]

英文摘要:

Given relatively prime integers $a_1, \dotsc, a_n$, the Frobenius number $g(a_1, \dotsc, a_n)$ is defined as the largest integer which cannot be expressed as $x_1 a_1 + \dotsb + x_n a_n$ with $x_i$ nonnegative integers. In this article, we give the Frobenius number of primitive Pythagorean triples. That is, \[ g(m^2-n^2, 2mn, m^2+n^2) = (m-1)(m^2-n^2) + (m-1)(2mn) - (m^2 + n^2). \]

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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