素数无穷大的简单证明

IF 0.1 Q4 MATHEMATICS
F. Lemmermeyer
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引用次数: 0

摘要

下面是对Barnes[1]给出的用连分式证明质数无穷的简化。假设只有有限个素数,即2,p1 = 3,…pn。设q = p1···pn为所有奇素数之积;那么q + 1不能被任何奇素数整除,因此它一定是2的幂。因为q +1≡2 mod 4,我们必须有q +1 = 2,因此q = 1:矛盾。因为没有奇数素数p≡3 mod 4能除q + 1,所以这个证明实际上表明有无穷多个素数p≡1 mod 4。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
A simple proof of the infinitude of primes
The following is a simplification of the proof of the infinitude of primes using continued fractions given by Barnes [1]. Assume that there are only finitely many prime numbers, namely 2, p1 = 3, . . . , pn. Let q = p1 · · · pn be the product of all odd primes; then q + 1 is not divisible by any odd prime, hence must be a power of 2. Since q +1 ≡ 2 mod 4, we must have q +1 = 2 and therefore q = 1: contradiction. Since no odd prime p ≡ 3 mod 4 can divide q + 1, the proof actually shows that there are infinitely many primes p ≡ 1 mod 4.
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