数论中的应用

R. Specifically, R. S T G, Sha Thb
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引用次数: 1

摘要

如果a和b是正整数,且gcd(a, b) = d,则存在使d = s×a + t×b的整数s和t。我们首先通过反转欧几里得算法的计算来说明找到这些乘数s和t的方法。稍后我们将展示一种使用扩展欧几里得算法求s和t的直接方法。扩展欧几里得算法可以用来求两个数的gcd,并将其表示为两个数的线性组合。它使用辅助数字1和0以及两个起始条件来生成一个产生所需结果的不变表达式G = S×A + T×B。我们可以显示gcd(356,252) = 4, 4 =(17)356 +(-24)252在下面的表格中,前两行将A和B表示为它们自己的线性组合。计算从第三行开始,其中Q n = floor (R (n - 2) / R (n -1))和A = R(-1)和B = R(0)。其他每一列都使用Q n来查找后续条目,并且每行重复此过程。A 356 1 0 356 = 1ha + 0hb b 252 0 1 252 = 0ha + 1hb
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Applications in Number Theory
Greatest common divisor as a linear combination Theorem If a and b are positive integers and gcd(a, b) = d then there are integers s and t such that d = s×a + t×b. We illustrate first a method of finding these multipliers s and t by reversing the calculations of the Euclidean Algorithm. Later we show a direct way of finding s and t using the Extended Euclidean Algorithm. The Extended Euclidean Algorithm can be used to find the gcd of two numbers and express it as a linear combination of those numbers. It uses auxiliary numbers 1 and 0 and two starting conditions to produce an invariant expression G = S×A + T×B that yields the desired result. Example We can show that gcd(356, 252) = 4 and that 4 = (17)356 + (-24)252 In the following tableau, the first two lines express A and B as linear combinations of themselves. The calculation begins in the third line where Q n = floor (R (n–2) / R (n–1)) and A = R (-1) and B = R (0). Each of the other columns uses Q n to find the subsequent entry, and the process is repeated for each line. A 356 1 0 356 = 1HA + 0HB B 252 0 1 252 = 0HA + 1HB
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