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On the prime k-tuple conjecture Liu Fengsui Feb 12, 2002 Tel: 0086 10 65887888-7731# Fax: 0086 10 64951424, 64201375 Address: 6/F Full Link Plaza 18 Chao Yang Men Wai Avenue Chao Yang District, Beijing, China 100020 Email: [email protected]. Abstract: In this paper we defined some operations on the sets of natural numbers P (N ) , given a recursive formula T ' n of the set sequences, which approximates the set of prime k-tuples, and use the limit of the set sequences lim T ' n show the patterns of prime k-tuples k 41 infinitely often. Key words: second order arithmetic, model, prime formula, strict constructive proof, twin prime conjecture, triple prime conjecture, k-tuple prime conjecture, infinite, limit of set sequences, recursive formula, Dirichlet's Theorem, prime constellation. MBC2000: 11B37, 03C62. After all give some operations on the sets of natural numbers P (N ) Let A = < a1 , a 2 ,..., ai ,..., a n >, B = < b1 , b2 ,..., b j ,..., bm > be the finite sets of natural number, we define A + B = < a1 b1 , a 2 b1 ,..., ai b j ,..., a n 1 bm , a n bm >, A * B = < a1 * b1 , a 2 * b1 ,..., ai * b j ,..., a n 1 * bm , a n * bm >. A \ B be the set subtraction. Define the solution of the system of congruencies be X = < a1 , a 2 ,..., ai ,..., a n > mod a , X= < b1 , b2 ,..., b j ,..., bm > mod b X = D = < d11 , d 21 ,..., d ij ,..., d n 1m , d nm > mod a * b . 1 Where x = d ij mod a * b is the solution of the system of congruencies x = a i mod a , x = b j mod b . Now, we had founded a model of the second order arithmetic P(N ), ,, ,0,1, . Where N is the set of natural numbers and P (N ) is the power set of N . We identify the set < a >, which has only single element a , with the number a <a> = a. So that, we had isomorphic embedded the model N ,,, ,0,1 in the model P(N ), ,, ,0,1, . In the theory PA + ZF of the second order arithmetic P(N ), ,, ,0,1, , we shall show that the prime k-tuple conjecture with the limit of the set sequences lim T ' n . Perhaps, it is undecidable in the first order Peano theory PA . The prime k-tuple conjecture states that every admissible pattern for a prime constellation occurs infinitely often. To describe easily, we discuss the conjecture: There are infinity a such that all x2 x a are prime for x 0,1,2,..., k . [1] Example: k = 2, this is the Twin Prime Conjecture, k =3, this is the Triple Prime Conjecture, and so on. We use R(k , a ) denote that all x x a are prime for x 0,1,2,..., k , and say the 2 number a be a prime k-tuple. Let pn be n -th prime, p 0 = 2. For any prime p i > 2, we consider the set Bi = { a : p i | x 2 x a for some 0 xk} = { a : x x a = 0 mod p i for some 0 x k }. 2 2 ( pi 1 x) 2 ( pi 1 x) ( pi 1 x) * ( pi x) x 2 x mod p i , By it is easy to prove that, in the congruence x x a = 0 mod p i , when x runs through the 2 complete system mod p i , the number a runs through the classes of residues <0,-2,-6,..., ( pi 1) / 2 ( pi 1) / 4 > mod 2 pi . Thus the set Bi are the classes of residues: Bi = <0,-2,-6,..., ( pi 1) / 2 ( pi 1) 2 / 4 > mod p i , ( pi 1) / 2 k . Bi = <0,-2,-6,..., k k 2 > mod p i , ( pi 1) / 2 > k For example, let k 4 , the first few terms of the sets Bi are B1 = <0,-2> = <0,1> B2 =<0,-2,-6> =<0,3,4> mod 3, mod 5, B3 = <0,-2,-6,-12> = <0,5,1,2> mod 7, B4 = < 0,-2,-6,-12> = <0,9,5,10> mod 11. Let m n 1 = p0 * p1 * ... * pn , from the set of all odd numbers X = <1> mod 2 we cancel the classes B1 , B2 ,..., Bn successively, and obtain the class Tn 1 mod m n 1 such that Tn 1 = {a: x 2 x a =/= 0 mod p i for x 0,1,2,..., k and pi p1 , p2 ,..., pn }. Then the recursive formula of Tn 1 , which is the set of non negative representatives mod m n 1 is as follows: T1 = <1>, Tn 1 = ( Tn mn * 0,1,2,..., pn 1 ) \ Dn . Where X = D n mod m n 1 is the solution of the system of congruencies. X = Tn mod m n , X = Bn mod pn 3 For example, let k 4 , the first few terms of Tn 1 are T1 = <1>, T2 = (<1>+<0,2,4>) \ <1,3> = <5>, T3 = (<5> + <0,6,12,18,24>) \ <5,23,29> = <11,17>, T4 = (<11,17>+<0,30,60,...,180>) \ <47,71,77,107,131,,161,191,197> = <11,17,41,101,137,167>. Given any integer k 1 , take an integer s such that ( p s 1) / 2 s ( p s 1 1) / 2 , then the number of all elements of the set Tn 1 is Tn1 = ( p1 1) / 2 * ( p2 1) / 2 * ... * ( p s 1) / 2 * ( p s 1 k ) / 2 * ... * ( pn k ) / 2 . Now we had proved an elementary result: for every k and n there are number a such that x2 x a have not any prime factor p p n . This is the Ludovicus conjecture..[2] Obviously the criterion of prime k-tuple is R(k , a) iff a = pn mimTn k 1 Where mim Tn is the smallest number of the set Tn . This criterion R(k , a ) recursively enumerates all prime k-tuples. The recursive formula Tn expresses a process of sieve method. We can perform this sieve method infinite times on the entire set of natural numbers N in the second order arithmetic. This is difference from traditional sieve method. Let k 1 , then Bi =< 0> mod p i , we simplify the recursive formula Tn to be T1 = <1>, Tn 1 = ( Tn mn * 0,1,2,..., pn 1 )\ pn *Tn . 4 From this formula we obtain a prime formula p1 3 pn S (Tn ) . Where the function S (Tn ) takes the second element of the set Tn . S (Tn ) = S t1 , t 2 ,...... t 2 . This formula provides a direct strict constructive proof, that the number of primes is infinite. [3][4]. If we canceled all classes B1 , B2 ,..., Bi ,...... , from entire set of natural numbers N , will obtain what result? In other word, we consider the limit of the sequences of sets T1 , T2 ,…, Tn ,.....… In the classes of residues there is include relation N T1 T2 ... Ti ...... Thus the sequences of sets T1 , T2 ,..., Tn ,...... have lim Tn .[3] It is easy to prove lim Tn is empty, since that when we canceled classes B1 , B2 ,..., Bi ,...... to sift the prime k-tuple a , we removed the prime k-tuple a also by pn = a and pn | a . We modified the set B j to be B' j B' j = { a : p j | x 2 x a for some 0 x k but p j =/= x 2 x a } = { a : x x a = 0 mod p j for some 0 x k but p j =/= x x a }. 2 2 And we modified the set Tn 1 to be T ' n1 T ' n1 = { a : x 2 x a =/= 0 mod p j but p j = x 2 x a for x 0,1,2,..., k and p j p1 , p 2 ,..., p n ,}. Except reserving the prime k-tuples a , Tn and T ' n are same, so that | Tn || T ' n | . Now the recursive formula T ' n will approximates the set of all prime k-tuples lim T ' n if n go to infinity.If we have known that the set of prime k-tuples is not empty, example k 41 , we 5 try prove the pattern R(k , a ) occurs infinitely often with the lim T ' n . Proof: Suppose that the number of patterns R(k , a ) is finite and not 0, then there is a maximum number a 0 , such that for every a > a 0 , R(k , a ) is false. From the set of all odd numbers X 1 mod 2 , We cancel the classes B1 , B2 ,..., Br 1 successively, and obtain the class Tr such that a 0 = mim Tr = p r , Then for every j r , in the class T j there is not any a such that R(k , a ) , and for every a a0 there is not any a such that R(k , a) . From the class Tr we continue cancel the set Br , B' r 1 ,..., B' j ,...... . Now as n goes to infinity, the recursive formula T ' n approximated all prime k-tuple a a0 . Because lim | Tn | is infinite, and | Tn || T ' n | , so that lim | T ' n | is infinite and lim T ' n is not empty. Take a number e belongs lim T ' n , e lim T ' n , then x2 x e does not contain any prime p i or p j as factor except itself for x 0,1,2,..., k Thus R(k , a ) hold by definition and e > a0 . This is a contradiction, so we have proved the prime k-tuple conjecture: The pattern R(k , a ) , k 41 , occurs infinitely often. # Let k 1 , as a practical example of this method, we provided an abstract proof that the number of primes is infinite with lim T ' n . Thus we can prove that the prime and prime k-tuple both are infinite with same method. Change the sets Bi , we can prove the infinity of various patterns with above method. 6 Example: the primes in arithmetic sequence ax b , ( a, b) 1 is infinite. Give any prime p i , except p i | a , let Bi = { x : ax b 0 mod p i }, then | Tn | = ( p1 1) * ( p2 1) * ... * ( pn1 1) , except p i | a . Repeat above proof, we obtained the Dirichlet's Theorem again. If the set of prime k-tuples is empty for some patterns, it is no meaning to discuss if they are infinite, to extend above method to empty set, we describe the empty set with the sentence. There is no prime k-tuple a a0 or there is no prime k-tuple a a0 . Assume the component of disjunction "there is no prime k-tuple a a0 ", run above proof, if we do not obtain contradiction, we can not prove anything, if we obtain a contradiction, then we have proved the component of disjunction " there is no prime k-tuple a a0 " is true, namely we have proved the set of prime k-tuples is empty again. There is an Euler's formula x 2 x 41 all are primes for x 0,1,2,..., k . So that, for k 41 , the set of prime k-tuples is not empty, by above proof, the pattern R(k , a ) , k 41 , occurs infinitely often. For k 41 , we do not know if there is a number a such that x2 x a all are primes for x 0,1,2,..., k , we do not know if they are infinite. If we found one a , such that R(k , a ) , then they would infinite. Reference : [1] 华罗庚,堆垒素数论(中国科学院 1953) 204-205。 [2] http://www.primepuzzles.net/conjectures/conj_017.htm [3] http://www.primepuzzles.net/problems/prob_037.htm [4] P.Ribenboim, The new book of prime number records, 3rd edition, Springer- Verlag, New Tork. NY,(1995)213. [5] K. Kuratowsky and A. Mostowsky , Set Theory, With an introduction to descriptive set theory, North-Holland Publishing com. (1976) 118-120. 7 8