《人类第一次打开素数大门(iii》由会员分享，可在线阅读，更多相关《人类第一次打开素数大门(iii（8页珍藏版）》请在金锄头文库上搜索。
1、人类第一次打开素数大门（人类第一次打开素数大门（III） One Has First Entered The Door Of The Prime Numbers(III) 蒋春暄 在素数大门有两个猜想：孪生素数猜想和哥德巴赫猜想。到今天仍没有解决，那么素数分布研究成果几乎等于零。陶哲轩和格林证明了存在任意长的素数等差数列。陶哲轩获得 2006 年国际数学家大会菲尔茨奖。加拿大著名数论专家Andrew Granville 对陶哲轩和格林工作评价： There are certain cases that they are not able to attack as yet: that there
2、 are infinitely many pairs of primes (the twin prime conjecture); for any large even integer there are pairs of primes (the Goldbach conjecture), and that there infinitely many pairs of primes (Sophie Germain twins). 他们不能证明孪生素数猜想和哥德巴赫猜想， 他们只能在素数大门外瞎猜。2006 年 9 月 25 日王元对陶哲轩工作评价： “我不敢想象天下会有这样伟大的成就。 ”我们
3、看到陶哲轩和格林 65 页的论文，他们只得出一个下限，谁是变量也没有交待。2006 年 10 月 5 日我们花了半天证明本文定理 1。 证明了存在任意长的素数等差数列， 只用 8 个公式就证明了定理 1。 公式 （5）证明了存在无限多素数和使得都是素数。公式（8）得出计算和个数的公式。1982 年 Grosswald 对定理 1 作了全面的研究。比陶哲轩和格林的结果更好。他们没有提到 Grosswald 的工作。定理 2 是另一类存在任意长的素数等差数列。这个问题到今天无人研究。数学爱好者你们先把 6 个例子仔细推导一下，而后再研究定理 1 和定理 2。你们一定会证明定理 1 和定理 2。这是
4、目前国际最重要问题，受到国内外高度重视，美国普林斯顿高级研究院把定理 1 作为 2007年研究项目，美国 Clay 数学研究所把它作为支助项目。实际上，这是两个非常简单的数学问题。本文是英文，主要目的在国外发表和传播，一寄到国外反映非常, p p+2PN, p Np,21pp +1P2P3,kP L1P2P1好，国内不允许发表我的论文，通过网络宣传和普及我们先进数论思想。 The Simplest Proofs of Both Arbitrarily Long Prime Arithmetic Progressions Chun-Xuan Jiang P. O. Box 3924, Beiji
5、ng 100854 P. R. China J Abstract We define arbitrarily long prime arithmetic progressions: . Using Jiang function 11,iPPdi+=+1,2,1,1,ikd=LNL3( )J we prove that there exist infinitely many primes and such that are all primes and find the best asymptotic formula (8). We define another arbitrarily long
6、 prime arithmetic progressions: 1P2P3,kP L PP1112,1,2,2,gigggP PPPi iP+ =+=L. Using Jiang function 2( )J we prove that there exist infinitely many primes such that are all primes and find the best asymptotic formula (25). 1P12, gPPP +L12In prime numbers theory there are both well-known conjectures t
7、hat there exist arbitrarily long prime arithmetic progressions. In this paper using Jiang functions 3( )J and 2( )J we obtain the simplest proofs of both arbitrarily long prime arithmetic progressions. Theorem 1. We define prime arithmetic progressions: 121311,2 ,(1) ,1,kP PPd PPdPPkd dN,=+=+=+LL. (
8、1) We rewrite (1) . （2） 21(1)(2),3jPjPjPj= kPWe have Jiang function 1 2 33( )(1)( ) PJPX = . （3） ( )X P denotes the number of solutions for the following congruence 213(1)(2)0(mod),kjjqjqP = （4） where 121,2,1;1,2,1.qPqP=LLFrom (4) we have 33( )(1)(1)(1) P kk PJPPPk for large . kGrosswald and Zagier
9、obtain heuristically even asymptotic formulae 2. Green and Tao obtain only lower bound 3. Example 1. Let 3k =. From (2) we have . （9） 322PP=1PFrom (5) we have 33( )(1)(2) PJPP = as . （10） We prove that there exist infinitely many primes and such that are primes. From (8) we have the best asymptotic
10、formula 1P2P3P2222331(,3)2 1(1(1)1.32032(1(1)(1)loglogNNNoPNN=+=+o. （11） Example 2. Let 4k =. From (2) we have , 322PP=1P1.4232PPP= （12） From (5) we have 35( )2(1)(3) PJPP = as . （13） 4We prove that there exist infinitely many primes and such that and are all primes. From (8) we have the best asympt
11、otic formula 1P2P3P4P2233459(3)(,3)(1(1)2(1)logPPPNNoPN =+. （14） Example 3. Let 5k =. From (2) we have 322PP1P=, , 4232PP=1P1P5243PP=. （15） From (5) we have 35( )2(1)(4) PJPP = as . （16） We prove that there exist infinitely many primes and such that , and are all primes. From (8) we have the best as
12、ymptotic formula 1P2P3P4P5P32445527(4)(,3)(1(1)2(1)logPPPNNoPN =+. （17） Theorem 2. We define another prime arithmetic progressions1, 4: 11,1,2,igPPi ik1+=+=L （18） where 2ggP P = is called a common difference, gP g th prime. We have Jiang function 1, 4 23( )(1( ) PJPX P = , （19） ( )X P denotes the nu
13、mber of solutions for the following congruence （20） 11()0(modkgiqiP=+),where . 1,2,1qP=L5If gP, then ( )0X P =; ( )1X Pk= otherwise. From (20) we have 123( )(1)()ggP PPPJPPk+ =. （21） If 1gkP+= then , 21()gJP+= 02( )0J=, there exist finite primes such that are all primes. If then 1P2,kP L P11gkP+=2(
14、)0J there exist infinitely many primes such that are all primes. 1P2,kP L PLet . From (18) we have 11gkP+=111,1,2,2iggPPi iP+=+=L. （22） From (21) we have 1, 4 123( )(1)(1)gggP PPPJPPP1+ =+ , as . （23） We prove that there exist infinitely many primes such that are all primes for all 1P 121, gPPP +L1gP+. We have the best asymptotic formula 1, 4 111111112 2 11(,2)(prime,12,( )(1(1).( ( )(log)ggggPggPPPNPiiPPJNoN +=+= =+N. （24） Substituting (7) and (23) into (24) we have 111111122 1 112(,2)(1)(1(1).1(1)(log)gggggggPPP g PPP PPPNPPPPNoPPN + + =+ =
