Towards the Twin Prime Conjecture

Transcription

1 A talk given at the NCTS (Hsinchu, Taiwan, August 6, 2014) and Northwest Univ. (Xi an, October 26, 2014) and Center for Combinatorics, Nankai Univ. (Tianjin, Nov. 3, 2014) Towards the Twin Prime Conjecture Zhi-Wei Sun Nanjing University Nanjing , P. R. China zwsun Nov. 3, 2014

2 Abstract Prime numbers are the most basic objects in mathematics. They also are among the most mysterious, for after centuries of study, the structure of the set of prime numbers is still not well understood. Describing the distribution of primes is at the heart of much mathematics... Andrew Granville (1997) If p and p + 2 are both prime, then {p, p + 2} is called a twin prime pair. The famous Twin Prime Conjecture asserts that there are infinitely many twin prime pairs. In this talk we will give a survey of the developments towards the solution of the Twin Prime Conjecture. We will introduce Brun s theorem on twin primes, Chen s theorem on Chen primes, the recent breakthrough of Yitang Zhang and the Maynard-Tao theorem on m consecutive primes. We will also mention the Super Twin Prime Conjecture posed by the speaker and the recent work of Pan and Sun on consecutive primes and Legendre symbols. 2 / 48

3 Twin Primes Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,... Euclid (around 300 BC): There are infinitely many primes. For n = 1, 2, 3,... let p n denote the n-th prime. For example, p 1 = 2, p 2 = 3, p 3 = 5, p 4 = 7, p 5 = 11, p 6 = 13. If p and p + 2 are both prime, then we call {p, p + 2} (of the form {p n, p n+1 } with p n+1 p n = 2) a twin prime pair. For example, {3, 5}, {5, 7}, {11, 13}, {17, 19}, {29, 31}, {41, 43}, {59, 61}, {71, 73} are all the twin prime pairs below / 48

4 Twin Prime Conjecture and Cramér s Conjecture The Twin Prime Conjecture. There are infinitely many twin prime pairs. In other words, lim inf n (p n+1 p n ) = 2. de Polignac s Conjecture (1849). For each d = 2, 4, 6,..., there are infinitely many positive integers n with p n+1 p n = d. Cramér s Conjecture (1936). We have lim sup n p n+1 p n (log p n ) 2 = 1. Cramér showed that Riemann s Hypothesis implies that p n+1 p n = O( p n log p n ). 4 / 48

5 k-tuple Prime Conjecture and Admissible Sets Let h 1,..., h k be integers. If there are infinitely many integers n such that n + h 1,..., n + h k are all prime, then there is no prime p such that k p (n + h i ) for all n Z, i=1 i.e., k i=1 h i(mod p) Z for any prime p, where a(mod p) refers to a + pz. Definition. Let h 1,..., h k be distinct integers. If k i=1 h i(mod p) Z for any prime p, then we call H = {h 1,..., h k } an admissible set or an admissible k-tuple. k-tuple Prime Conjecture (Hardy and Littlewood, 1923): If H = {h 1,..., h k } is an admissible k-tuple, then there are infinitely many positive integers n such that are all prime. n + h 1, n + h 2,..., n + h k 5 / 48

6 Special Cases of the k-tuple Prime Conjecture If H = {h 1, h 2,..., h k } is an admissible k-tuple, then so is a + H = {a + h i : i = 1,..., k}, where a is an arbitrary integer. So we need only consider admissible set of the form H = {h 1 = 0 < h 2 <... < h k }. As {0, 2} is an admissible set, the k-tuple Prime Conjecture implies the Twin Prime Conjecture. Note that {0, 2, 4} is not admissible since 0(mod 3) 2(mod 3) 4(mod 3) = Z. But {0, 2, 6} and {0, 4, 6} are both admissible sets, so the k-tuple Conjecture implies the following conjecture. Prime Triplet Conjecture: There are infinitely many primes p with p + 2 and p + 6 both prime. Also, there are infinitely many primes p with p + 4 and p + 6 both prime. Examples. {11, 13, 17} and {7, 11, 13} are both prime triplets. 6 / 48

7 Dikson s Conjecture Dickson s Conjecture: Let a i > 1 and b i be integers for all i = 1,..., k. If there is no prime p dividing k i=1 (a in + b i ) for all n Z, then there are infinitely many integers n such that a 1 n + b 1, a 2 n + b 2,..., a k n + b k are all prime. Note that if a prime p divides an + b for all n Z, then both a and b are multiples of p. So Dirichlet s theorem on primes in arithmetic progressions is just Dickson s Conjecture in the case k = 1. Example. Any prime p does not divide n(2n + 1) with n = p 1, so Dickson s Conjecture implies that there are infinitely many primes p with 2p + 1 also prime. Such primes p are called Sophie Germain primes. 7 / 48

8 Schinzel s Hypothesis H and Bateman-Horn Conjecture Schinzel s Hypothesis H (1958). If f 1 (x),..., f k (x) are irreducible polynomials with integer coefficients and positive leading coefficients such that there is no prime dividing the product f 1 (q)f 2 (q)...f k (q) for all q Z, then there are infinitely many n Z + such that f 1 (n), f 2 (n),..., f k (n) are all primes. Remark. The hypothesis with k = 1 was a conjecture posed by Bunyakovsky in Bateman-Horn Conjecture (1962). Let f 1 (x),..., f k (x) be distinct irreducible polynomials with integer coefficients and positive leading coefficients such that there is no prime dividing f (q) for all q Z, where f = f 1 f k. Then {1 n x : f 1 (n),..., f k (n) are all prime} ( 1 ) 1 N f (p)/p x dt k i=1 deg(f i) (1 1/p) k 2 (log t) k, p where N f (p) = {0 x p 1 : f (x) 0 (mod p)} < p. 8 / 48

9 Hardy-Littlewood Conjecture on π 2 (x) Prime Number Theorem. For x > 0 let π(x) denote the number of primes not exceeding x. Then π(x) x as x. log x For x > 0, let π 2 (x) := {p x : p + 2 is prime}. The Bateman-Horn Conjecture with f 1 (x) = x and f 2 (x) = x + 2 yields the following conjecture on π 2 (x). Hardy-Littlewood Conjecture. We have x π 2 (x) 2C 2 log 2 as x, x where C 2 = ( ) 1 1 (p 1) p>2 9 / 48

10 The sieve method Let A = (a 1, a 2,...) be a finite integer sequence and let P be a set of some primes. The sift function S(A, P, z) := a A (a,pz )=1 1, where P z = p P p z Note that S(A, P, z) is the number of remaining terms of A after we sieve out those terms of A divisible by some primes p P with p z. Eratosthenes Sieve. Let z 2 and A = {n > z : n z 2 }, and let P be the set of all primes. Then S(A, P, z) = {z < n z 2 : n is divisible by no prime p z} =the number of primes in (z, z 2 ] = π(z 2 ) π(z). p. Remark. If S(A, P, z) > 0 for sufficiently large z, then we deduce that there are infinitely many primes. 10 / 48

11 Brun s sieve Inclusion-Exclusion Principle. Let S be a finite set, and let Then S k = {a S : a has the property P k } for k = 1,..., n. {a S : a has the property P k for no k = 1,..., n} n = S S k + S i S j... + ( 1) n S 1 S n. k=1 1 i<j n Brun s Observation. If m {1,..., n} is even, then n S \ n S k S S i + S i S j... k=1 + i=1 1 i 1 <...<i m n 1 i<j n ( 1) m S i1... S im. Remark. This is similar to the fact that π 4 = / 48

12 Brun s theorem Brun s Theorem (1920) There is a constant C > 0 such that π 2 (x) := {p x : p + 2 is prime} C x(log log x)2 (log x) 2 for all x 2. Idea of the Proof. Let 5 y < x, and let q 1, q 2,..., q r be all the distinct odd primes not exceeding y. If {n, n + 2} is a twin prime pair with y < n x, then n > q i and q i n(n + 2) for all i = 1,..., r. Thus where π 2 (x) y + {p (y, x] : p + 2 is prime} y + N(y, x), N(y, x) = {n x : n(n + 2) 0 (mod q i ) for all i = 1,..., r}. If we take y = x 1/(c log log x) for suitable c > 0, then we obtain via Brun s sieve x(log log x)2 π 2 (x) y + N(y, x) C (log x) 2 for some C > / 48

13 Brun s constant Choose a constant C > 0 such that π 2 (x) C x(log log x)2 (log x) 2 C x (log x) 1.5 for all x 2. Let {t n, t n + 2} be the n-th twin prime pair. Then t n n = π 2 (t n ) C (log t n ) 1.5 and hence 1 t n It follows that n=1 1 t n C n=1 1 <. n(log n) 1.5 Brun s constant: ( ) t n t n + 2 n=1 C (log n) / 48

14 Chen primes Via a very sophisticated weighted linear sieve, the Chinese mathematician Jing-run Chen established the following famous result. Chen s Theorem (Jing-run Chen, 1973) (i) Large even numbers can be written as p + q, where p is a prime, and q is either a prime or a product of two primes. (ii) There are infinitely many primes p such that p + 2 is either a prime or a product of two primes. Remark. Part (i) is the best record on Goldbach s conjecture, while part (ii) is close to the Twin Prime Conjecture. Chen prime: A prime p is called a Chen prime if p + 2 is a product of at most two primes. Example. 13 is a Chen prime since = / 48

15 Bombieri-Vinorgradov Theorem Prime Number Theorem for Arithmetic Progressions. Let 1 a q with (a, q) = 1, then π(x; a, q) := 1 π(x) ϕ(q) Li(x) ϕ(q), p x p a (mod q) where x dt Li(x) = 2 log t x log x. Bombieri-Vinorgradov Theorem (1965). For any A > 0, there is a constant B > 0 depending on A such that ( E(q, x) = O q x/(log x) B where E(q, x) := max 1 a q (a,q)=1 x (log x) A ), Li(x) π(x; a, q) ϕ(q). 15 / 48

16 Elliott-Hlberstam Conjecture The Bombieri-Vinorgradov Theorem plays a very important role in analytic number theory. It led Bombieri and Vinorgradov to obtain on Goldbach s conjecture. Jing-run Chen also needed the Bombieri-Vinorgradov Theorem in his proof of Chen s theorem ( ). Elliott-Halberstam Conjecture (1973). For any 0 < θ < 1, we have the following property (called EH(θ)): For any A > 0, there is a constant C A > 0 such that x E(x, q) C A (log x) A. q x θ The Bombieri-Vinorgradov Theorem indicates that EH(θ) holds for any 0 < θ < 1 2. Up to now, nobody succeeds to prove EH(θ) for some θ > / 48

17 Selberg s upper bound sieve Selberg s Sieve. Let A = (a 1, a 2,...) be a finite sequence and let A be its length. Let P be a set of some primes, and let P(z) = p z p P p for z 2. For a squarefree positive integer d, let A d denote the subsequence of A consisting of terms divisible by d. Let g be a multiplicative arithmetic function with 0 < g(p) < 1 for all p P. And let g 1 be a completely multiplicative function with g 1 (p) = g(p) for all p P. Then S(A, P, z) A G(z) + 3 ω(d) r(d), d z 2 where ω(d) is the number of distinct prime divisors of d, G(z) = m z p m p P d Pz g 1 (m) and r(d) = A d g(d) A. 17 / 48

18 Starting point of the proof Let z 2, and let λ be a real arithmetic function with λ(1) = 1 and λ(z) = 0 for z > D. Observe that S(A, P, z) = a A (a,pz )=1 a A = 1 ( d (a,p z ) d 1,d 2 P z λ(d 1 )λ(d 2 ) = ) 2 λ(d) = a A [d 1,d 2 ] a a A d 1 (a,p z ) 1 = w(n) = ( d n λ(d) ) 2 is called a weight of n. λ(d 1 ) d 2 (a,p z ) λ(d 2 ) d 1,d 2 P z λ(d 1 )λ(d 2 ) A [d1,d 2 ] To get an ideal upper bound for S(A, P, z), we should manage to optimize the choice of the auxiliary function λ(d). 18 / 48

19 Applying the Cauchy-Schwarz inequality Cauchy-Schwarz Inequality. Let a i, b i R for i = 1,..., n. Then ( n ) 2 ( n )( n ) a i b i. i=1 If a j 0 for some j {1,..., n}, then the equality holds if and only for some t R we have b i = ta i for all i = 1,..., n. Sketch of the Proof. 0 (a i b j a j b i ) 2 = 1 i<j n i=1 n i=1 a 2 i a 2 i n j=1 i=1 b 2 i ( n ) 2 bj 2 a i b i. The Cauchy-Schwarz Inequality implies the following lemma needed for Selberg s sieve. Lemma. Let a 1,..., a n be positive reals and let b 1,..., b n R. Under the restriction b 1 y b n y n = 1, Q(y 1,..., y n ) = a 1 y a n yn 2 has the minimum m = ( n i=1 b2 i /a i) 1. And the minimum is attained if and only if y i = mb i /a i for all i = 1,..., n. 19 / 48 i=1

20 The work of Goldston-Pintz-Yildirim Let P be the set of all primes, and let χ P (n) take 1 or 0 according as n is prime or not. D.A. Goldston, J. Pintz, C.Y. Yildirim (posted to arxiv in 2005): Let H = {h 1,..., h k } be an admissible k-tuple. Choose λ(d) = µ(d)p(log D d ) with d D for suitable polynomial P with P(1) = 1, and set W (n) = ( d k j=1 (n+h j ) λ(d))2. If EH(θ) holds for some θ > 1/2, then for large N we have χ P (n + h j )W (n) > 1 W (n) for all j = 1,..., k, k N n<2n hence N n<2n j=1 N n<2n k χ p (n + h j )W (n) > N n<2n W (n) and thus there are 1 i < j k such that p = n + h i and q = n + h j are both prime. Note that p q does not exceed d(h) = max H min H (the diameter of H). 20 / 48

21 Main Results of Goldston-Pintz-Yildirim D.A. Goldston, J. Pintz, C.Y. Yildirim [Annals of Math. 170(2009)]: We have lim inf n p n+1 p n log p n = 0. Under the Elliott-Halberstam conjecture, lim inf n (p n+1 p n ) 16. D.A. Goldston, J. Pintz, C.Y. Yildirim [Acta Math. 204(2010)]: lim inf n p n+1 p n log pn (log log p n ) 2 <. 21 / 48

22 Discrepancy Let f be an arithmetic function with supp(f ) = {n : f (n) 0} finite. For a primitive residue class a(q) = a + qz (with (a, q) = 1), define the discrepancy (f ; a(q)) := n a (mod q) f (n) 1 ϕ(q) (n,q)=1 f (n). Below we let 1 I (n) take 1 or 0 according as n I or not. Fouvry and Iwaniec [Mathematica 27(1980)]: Let A > 0 and let x be large. Let f (n) = 1 if no prime divisor of n is smaller than x 1/883, and f (n) = 0 otherwise. Then q x 1/2+1/42 max (f 1 1 a q [1,x], a(q)) = O (a,q)=1 ( x (log x) A Friedlander and Iwaniec [Annals of Math. 34(1985)]: Let A > 0, and let q x 1/2+1/230 and (a, q) = 1. Then ( ) x (τ 3 1 [1,x], a(q)) = O q(log x) A, where τ 3 (n) = Σ abc=n 1. ). 22 / 48

23 The work of Motohashi and Pintz In 2008, Y. Motohashi and J. Pintz published a paper with the title A smoothed GPY sieve in Bull. Lond. Math. Soc. 40(2008), This paper was posted to arxiv in Motohashi-Pintz (2006): Let f (n) = log n if n is a prime, and let f (n) = 0 otherwise. If there is a θ > 1/2 and an admissible H = {h 1,..., h k } such that for any A > 0 and large x we have q x θ q p x θ/2 1/4 p 1 a q, (a,q)=1 q k j=1 (a+h j ) ( (f 1 [x,2x], a(q)) = O x (log x) A then there are infinitely many n such that {n + h 1,..., n + h k } contains at least two primes, and hence lim inf n (p n+1 p n ) d(h) <. ), 23 / 48

24 Yitang Zhang s breakthrough In 2013, Yitang Zhang rediscovered the approach of Motohashi and Pintz. Moreover, he proved that the condition holds for θ = and k = To deduce this, he needed to bound incomplete exponential sums in the form e 2πi c1 n+c 2 n+l q, N n 2N where n denotes the inverse of n modulo q. In this step, Zhang employed some deep results like Deligne s theorem which extends the Weil bound on Kloosterman sums. Zhang noted that H = {p π(k)+j : j = 1,..., k} is an admissible k-tuple. In fact, for any prime p k, we have p π(k)+j 0 (mod p) for all j = 1,..., k; for any prime p > k obviously k j=1 p π(k)+j(mod p) Z. For k = , p π(k)+k < Zhang s Theorem (2013). lim inf n (p n+1 p n ) < / 48

25 Some comments on Zhang s work The main results are of the first rank. The author had proved a landmark theorem in the distribution of prime numbers. One of the referees Basically no one knows him. Now suddenly he has proved one of the great result in the history of number theory. Andrew Granville 25 / 48

26 Maynard s approach In Oct. 2013, the young number theorist James Maynard announced a new approach to bounded gaps between primes. On Nov. 19, 2013 he posted a preprint Small gaps between primes on arxiv. Maynard did not follow Zhang s approach. Instead, he modified Goldston-Yildirim s original unsuccessful approach. Instead of using weights of the form ( ) 2 W (n) = λ(d) d k i=1 (n+h i ) (where H = {h 1,..., h k } is an admissible k-tuple), Maynard employed the weights of the new form ( ) 2 w(n) = λ d1,...,d k. d i n+h i (i=1,...,k) 26 / 48

27 Maynard s approach Let N be large and set W = p log log log N p. As H = {h 1,..., h k } is admissible, for any prime p W, there is an integer r p k i=1 h i( (mod p)). By the Chinese Remainder Theorem, there is an integer ν such that ν r p (mod p) for all p W and hence W is coprime to k i=1 (ν + h i). Maynard restricted his attention only to those n ν (mod W ). Let S 1 = N n<2n n ν (mod W ) w(n), S 2 = N n<2n n ν (mod W ) ( k i=1 ) χ P (n + h i ) w(n). If S 2 > ms 1, then at least m + 1 of the numbers n + h 1,..., n + h k are primes. 27 / 48

28 Maynard s approach Let F be a piecewise differentiable function with F (x 1,..., x k ) 0 for all x 1,..., x k 0 with x x k 1. Let θ > 0 and R = N θ/2 δ for some small fixed δ > 0. For integers d 1,..., d k > 0, if ( k i=1 d i, W ) = 1 then put ( k ) λ d1,...,d k := µ(d i )d i and let λ d1,...,d k J (s) i=1 I k (F ) := 1 k (F ) := 0 r i 0 (mod d i ) (0<i k) ( k i=1 r i,w )=1 = 0 otherwise. Set µ( k i=1 r i) 2 ( log k i=1 ϕ(r i) F r1 log R,..., log r ) k, log R w(n) := ( Σ di n+h i (i=1,...,k)λ d1,...,d k ) 2, ( F (t 1,..., t k ) 2 dt 1 dt 2 dt k, F (t 1,..., t k )dt s ) 2 dt 1 dt s 1 dt s+1 dt k. 28 / 48

29 Maynard s approach Suppose that EH(θ) holds. Provided I k (F ) k s=1 K (s) k ϕ(w )k S 1 W k+1 N(log ϕ(w R)k )k I k (F ), S 2 W k+1 N log N and thus Define ( S 2 θ S 1 2 δ ) k s=1 J(s) k (F ) I k (F ) k s=1 J(s) as N. (F ) 0, (log R)k+1 M k = k (F ). I k (F ) F Then there are infinitely many n ν (mod W ) such that {n + h i : i = 1,..., k} contains at least m = θm k /2 primes, in particular lim inf (p n+m 1 p n ) d(h) <. n k s=1 J (s) k (F ), 29 / 48

30 Maynard-Tao Theorem Theorem (Maynard, 2013, arxiv: ). (i) M 5 > 2, thus the EH conjecture implies that lim inf n (p n+1 p n ) 12 since {0, 2, 6, 8, 12} is admissible. (ii) M 105 > 4 and thus lim inf n (p n+1 p n ) 600 (since EH(θ) holds fore all 0 < θ < 1 2 and there is an admissible 105-tuple with diameter 600). (iii) For large values of k, we have M k > log k 2 log log k 2. Maynard-Tao Theorem (2013). There is an absolute constant C > 0 such that lim inf n (p n+m p n ) Cm 3 e 4m for all m = 1, 2, 3,.... Polymath. lim inf n (p n+1 p n ) 246 (P. Nielsen), and M k > log k + O(1). 30 / 48

31 Consecutive primes and Legendre symbols Theorem (H. Pan & Z.-W. Sun, arxiv: ) Let m be any positive integer and let δ 1, δ 2 {1, 1}. Then, for some constanst C m > 0 there are infinitely many integers n > 1 with p n+m p n C m such that ( ) ( ) pn+i pn+j = δ 1 and = δ 2 p n+j p n+i for all 0 i < j m, where p k denotes the k-th prime, and ( p ) denotes the Legendre symbol for any odd prime p. Conjecture. Let m Z +, δ {1, 1}, and δ ij {±1} for all 0 i < j m. Then, there are infinitely many integers n > 1 such that ( ) ( ) pn+i pn+j = δ ij = δ for all 0 i < j m. p n+j p n+i 31 / 48

32 Examples Example 1. The smallest integer n > 1 with ( ) pn+i = 1 for all i, j = 0,..., 6 with i j p n+j is The 7 consecutive primes p , p ,..., p have concrete values: , , , , , , Example 2. The smallest integer n > 1 with ( ) pn+i = 1 for all i, j = 0,..., 5 with i j p n+j is , and the 6 consecutive primes p , p ,..., p have concrete values: , , , , , / 48

33 Examples (continued) Example 3. The smallest integer n > 1 with ( ) ( ) pn+i pn+j = 1 = for all 0 i < j 6 p n+j p n+i is , and the 7 consecutive primes p , p ,..., p have concrete values: , , , , , , Example 4. The smallest integer n > 1 with ( ) ( ) pn+i pn+j = 1 = for all 0 i < j 5 p n+j p n+i is , and the 6 consecutive primes p , p ,..., p have concrete values: , , , , , / 48

34 Two Lemmas Lemma 1 (Maynard-Tao) Let m be any positive integer. Then there is an integer k > m depending only on m such that if H = {h i : i = 1,..., k} is an admissible set of cardinality k and W = q 0 p w p (with q 0 Z + ) is relatively prime to k i=1 h i with w = log log log x large enough, then for some integer n [x, 2x] with W n there are more than m primes among n + h 1, n + h 2,..., n + h k. Lemma 2 (Pan-Sun) Let k > 1 be an integer. Then there is an admissible set H = {h 1,..., h k } with h 1 = 0 < h 2 <... < h k which has the following properties: (i) All those h 1, h 2,..., h k are multiples of K = 4 p<2k p. (ii) Each h i h j with 1 i < j k has a prime divisor p > 2k with h i h j (mod p 2 ). (iii) If 1 i < j k, 1 s < t k and {i, j} {s, t}, then no prime p > 2k divides both h i h j and h s h t. 34 / 48

35 Proof of the Theorem By Lemma 1, there is an integer k = k m > m depending on m such that for any admissible set H = {h 1,..., h k } of cardinality k if x is sufficiently large and k i=1 h i is relatively prime to W = 4 p w p then for some integer n [x/w, 2x/W ] there are more than m primes among Wn + h 1, Wn + h 2,..., Wn + h k, where w = log log log x. Let H = {h 1,..., h k } with h 1 = 0 < h 2 <... < h k be an admissible set satisfying the conditions (i)-(iii) in Lemma 2. Clearly K = 4 p 2k p 0 (mod 8). Let x be sufficiently large with the interval (h k, w] containing more than h k k primes. Note that 8 W since w 2. Let δ := δ 1 δ 2. For any integer b δ (mod K) and each prime p < 2k, clearly b + h i δ + 0 (mod p) and hence gcd(b + h i, p) = 1 for all i = 1,..., k. 35 / 48

36 Proof of the Theorem (continued) For any 1 i < j k, the number h i h j has a prime divisor p ij > 2k with h i h j (mod pij 2 ). Suppose that p > 2k is a prime dividing 1 i<j k (h i h j ), then there is a unique pair {i, j} with 1 i < j k such that h i h j (mod p). Note that p h k. All the k 2 < (p 3)/2 numbers h i h s with 1 s k and s i, j are relatively prime to p, so there is an integer r p h i h s (mod p) for all s = 1,..., k such that ( ) { rp δ δ 2 if p = p ij, = p 1 otherwise. So, for any integer b r p h i (mod p), we have b + h s 0 (mod p) for all s = 1,..., k. Assume that S = {h 1, h 1 + 1,..., h k } \ H is a set {a i : i = 1,..., t} of cardinality t > 0. Clearly t h k k + 1 and hence we may choose t distinct primes q 1,..., q t (h k, w]. If b a i (mod q i ), then b + h s h s a i 0 (mod q i ) for all s = 1,..., k since 0 < h s a i < h k < q i. 36 / 48

37 Proof of the Theorem (continued) Let Q = { p (2k, w] : p 1 i<j k } (h i h j ) \ {q i : i = 1,..., t}. For any prime q Q, there is an integer r q h i (mod q) for all i = 1,..., k since H is admissible. By the Chinese Remainder Theorem, there is an integer b satisfying the following (1)-(4). (1) b δ = δ 1 δ 2 (mod K). (2) b r p h i r p h j (mod p) if p > 2k is a prime dividing h i h j with 1 i < j k. (3) b a i (mod q i ) for all i = 1,..., t. (4) b r q (mod q) for all q Q. By the above analysis, k s=1 (b + h s) is relatively prime to W. 37 / 48

38 Proof of the Theorem (continued) As H = {b + h s : s = 1,..., k} is also an admissible set of cardinality k, for large x there is an integer n [x/w, 2x/W ] such that there are more than m primes among Wn + b + h s (s = 1,..., k). For a i S, we have Wn + b + a i 0 a i + a i = 0 (mod q i ) and hence Wn + b + a i is composite since W > q i. Therefore, there are consecutive primes p N, p N+1,..., p N+m with p N+i = Wn + b + h s(i) for all i = 0,..., m, where 1 s(0) < s(1) <... < s(m) k. Note that p N+m p N = (Wn+b+h s(m) ) (Wn+b+h s(0) ) = h s(m) h s(0) h k. For each s = 1,..., k, clearly Wn + b + h s 0 + δ + 0 = δ (mod 8) and hence ( ) ( ) 1 2 = δ and = 1. Wn + b + h s Wn + b + h s 38 / 48

39 Proof of the Theorem (continued) As p N+i = Wn + b + h s(i) δ (mod 8) for all i = 0,..., m, by the Quadratic Reciprocal Law we have ( ) ( ) pn+j pn+i = δ for all 0 i < j m. p N+i p N+j Let 0 i < j m. Then ( ) ( ) pn+i hs(i) h s(j) = p N+j Wn + b + h s(j) ( = δ h ij Wn + b + h s(j) where h ij is the odd part of h s(j) h s(i). For any prime divisor p of h ij, clearly p h k w and ( ) ( ) ( ) p Wn + b + = δ (p 1)/2 hs(j) b + = δ (p 1)/2 hs(j) Wn + b + h s(j) p p If p < 2k, then p K, hence b + h s(j) δ + 0 (mod p) and thus ( ) ( ) ( ) p b + = δ (p 1)/2 hs(j) δ = δ (p 1)/2 = 1. Wn + b + h s(j) p p ), 39 / 48

40 Proof of the Theorem (continued) If p > 2k, then by the choice of b we have ( ) ( ) ( ) p b + =δ (p 1)/2 hs(j) = δ (p 1)/2 rp Wn + b + h s(j) p p ( ) { rp δ δ 2 if p = p = = s(i),s(j), p 1 otherwise. Recall that p s(i),s(j) h ij. Therefore, and ( pn+i p N+j ) ( ) h ij = δ = δδ 2 = δ 1 Wn + b + h s(j) ( pn+j p N+i This concludes the proof. ) ( ) pn+i = δ = δ 2. p N+j 40 / 48

41 Artin s Primitive Root Conjecture Artin s Primitive Root Conjecture (1927). Let g 1 be an integer which is not a square. Then there are infinitely many primes p for which g is a primitive root modulo p. C. Hooley (1967): Artin s conjecture holds under the Extended Riemann Hypothesis for Dedekind zeta functions. By combining Hooley s work with the Manard-Tao method, P. Pollack obtained the following result. P. Pollack (arxiv: ). Let g 1 be an integer which is not a square. Let q 1 < q 2 <... denote the sequence of primes having g as a primitive root. For any positive integer m, there is a constant C m > 0 not depending on g such that lim inf (q n+m q n ) C m. n + 41 / 48

42 Consecutive primes and primitive roots Conjecture (Z.-W. Sun, 2014). For any positive integer m, there are infinitely many positive integers n such that p n+i is a primitive root modulo p n+j for any distinct i and j among 0, 1,..., m. Example. The least n Z + with p n+i a primitive root modulo p n+j for any distinct i and j among 0, 1, 2, 3 is Note that p 8560 = 88259, p 8561 = and p 8562 = Theorem (H. Pan & Z.-W. Sun, arxiv: ). The conjecture holds under the Extended Riemann Hypothesis. 42 / 48

43 A Firoozbakht-type conjecture for twin primes Firoozbakht s Conjecture (1982). The sequence ( n p n ) n 1 is strictly decreasing. Conjecture (Z.-W. Sun, 2012) (i) If {t 1, t 1 + 2},..., {t n, t n + 2} are the first n pairs of twin primes, then the first prime t n+1 in the next pair of twin primes is smaller than tn 1+1/n, i.e., n t n > n+1 t n+1. (ii) The sequence ( n+1 T (n + 1)/ n T (n)) n 9 is strictly increasing with limit 1, where T (n) = n k=1 t k. Remark. Via Mathematica I verified that n t n > n+1 t n+1 for all n = 1,..., , and n+1 T (n + 1)/ n T (n) < n+2 T (n + 2)/ n+1 T (n + 1) for all n = 9, 10,..., Note that t = After I made the conjecture public, Marek Wolf verified the inequality n t n > n+1 t n+1 for all the pairs of twin primes below / 48

44 Unification of Goldbach s conjecture and the twin prime conjecture Unification of Goldbach s Conjecture and the Twin Prime Conjecture (Sun, ). For any integer n > 2, there is a prime q with 2n q and p q both prime. We have verified the conjecture for n up to Clearly, it is stronger than Goldbach s conjecture. Now we explain why it implies the twin prime conjecture. In fact, if all primes q with p q prime are smaller than an even number N > 2, then for any such a prime q the number N! q is composite since N! q 0 (mod q) and N! q q(q + 1) q > q. Example. 20 = with 3, 17 and p = = 13 all prime. 44 / 48

45 Graph for a(n) = {q < 2n : q, 2n q, p q are all prime} 45 / 48

46 Super Twin Prime Conjecture If p, p + 2 and π(p) are all prime, then we call {p, p + 2} a super twin prime pair. Super Twin Prime Conjecture (Sun, ). Any integer n > 2 can be written as k + m with k and m positive integers such that p k + 2 and p pm + 2 are both prime. Example. 22 = with p = = 73 and p p2 + 2 = p = = 7 both prime. Remark. If all those positive integer m with p pm + 2 prime are smaller than an integer N > 2, then by the conjecture, for each j = 1, 2, 3,..., there are positive integers k(j) and m(j) with k(j) + m(j) = jn such that p k(j) + 2 and p pm(j) + 2 are both prime, and hence k(j) ((j 1)N, jn) since m(j) < N; thus j=1 1 p k(j) j=1 1 p jn, which is impossible since the series on the right-hand side diverges while the series on the left-hand side converges by Brun s theorem. 46 / 48

47 Graph for a(n) = {0 < k < n : p k + 2 and p pn k + 2 are both prime} 47 / 48

48 Concluding remarks The current methods of Yitang Zhang or Mynard-Tao could not be modified to prove the Twin Prime Conjecture, To solve the Twin Prime Conjecture, number theorists must invent new tools and build a new powerful theory! There is a long way to go! I have verified the Super Twin Prime Conjecture for all n = 3,..., In my opinion, the solution of the Super Twin Prime Conjecture might be beyond the intelligence of human beings! Thank you! 48 / 48