{"title":"素数无穷大的简单证明","authors":"F. Lemmermeyer","doi":"10.4171/em/407","DOIUrl":null,"url":null,"abstract":"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.","PeriodicalId":41994,"journal":{"name":"Elemente der Mathematik","volume":"75 1","pages":"80-80"},"PeriodicalIF":0.1000,"publicationDate":"2020-04-08","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"https://sci-hub-pdf.com/10.4171/em/407","citationCount":"0","resultStr":"{\"title\":\"A simple proof of the infinitude of primes\",\"authors\":\"F. Lemmermeyer\",\"doi\":\"10.4171/em/407\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"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.\",\"PeriodicalId\":41994,\"journal\":{\"name\":\"Elemente der Mathematik\",\"volume\":\"75 1\",\"pages\":\"80-80\"},\"PeriodicalIF\":0.1000,\"publicationDate\":\"2020-04-08\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"https://sci-hub-pdf.com/10.4171/em/407\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"Elemente der Mathematik\",\"FirstCategoryId\":\"1085\",\"ListUrlMain\":\"https://doi.org/10.4171/em/407\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"Q4\",\"JCRName\":\"MATHEMATICS\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"Elemente der Mathematik","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.4171/em/407","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q4","JCRName":"MATHEMATICS","Score":null,"Total":0}
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。
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.