Towards the Twin Prime Conjecture

Home , Prime triplet, Twin prime

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)

Zhi-Wei Sun

Nanjing University Nanjing 210093, P. R. China [email protected] http://math.nju.edu.cn/∼zwsun

Nov. 3, 2014 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 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 pn denote the n-th prime. For example,

p1 = 2, p2 = 3, p3 = 5, p4 = 7, p5 = 11, p6 = 13.

If p and p + 2 are both prime, then we call {p, p + 2} (of the form {pn, pn+1} with pn+1 − pn = 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 100.

3 / 48 Twin Prime Conjecture and Cram´er’sConjecture The Twin Prime Conjecture. There are inﬁnitely many twin prime pairs. In other words,

lim inf(pn+1 − pn) = 2. n→∞

de Polignac’s Conjecture (1849). For each d = 2, 4, 6,..., there are inﬁnitely many positive integers n with pn+1 − pn = d.

Cram´er’sConjecture (1936). We have

pn+1 − pn lim sup 2 = 1. n→∞ (log pn)

Cram´ershowed that Riemann’s Hypothesis implies that √ pn+1 − pn = O( pn log pn).

4 / 48 k-Tuple Prime Conjecture and Admissible Sets

Let h1,..., hk be integers. If there are inﬁnitely many integers n such that n + h1,..., n + hk are all prime, then there is no prime p such that k Y p (n + hi ) for all n ∈ Z,

i=1 Sk i.e., i=1 hi (mod p) 6= Z for any prime p, where a(mod p) refers to a + pZ.

Deﬁnition. Let h1,..., hk be distinct integers. If Sk i=1 hi (mod p) 6= Z for any prime p, then we call H = {h1,..., hk } an admissible set or an admissible k-tuple. k-Tuple Prime Conjecture (Hardy and Littlewood, 1923): If H = {h1,..., hk } is an admissible k-tuple, then there are inﬁnitely many positive integers n such that

n + h1, n + h2,..., n + hk are all prime. 5 / 48 Special Cases of the k-Tuple Prime Conjecture

If H = {h1, h2,..., hk } is an admissible k-tuple, then so is

a + H = {a + hi : i = 1,..., k}, where a is an arbitrary integer. So we need only consider admissible set of the form H = {h1 = 0 < h2 < . . . < hk }. 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 Dikson’s Conjecture

Dickson’s Conjecture: Let ai > 1 and bi be integers for all Qk i = 1,..., k. If there is no prime p dividing i=1(ai n + bi ) for all n ∈ Z, then there are inﬁnitely many integers n such that a1n + b1, a2n + b2,..., ak n + bk 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 Schinzel’s Hypothesis H and Bateman-Horn Conjecture

Schinzel’s Hypothesis H (1958). If f1(x),..., fk (x) are irreducible polynomials with integer coefficients and positive leading coefficients such that there is no prime dividing the product f1(q)f2(q)...fk (q) for all q ∈ Z, then there are infinitely + many n ∈ Z such that f1(n), f2(n),..., fk (n) are all primes. Remark. The hypothesis with k = 1 was a conjecture posed by Bunyakovsky in 1857.

Bateman-Horn Conjecture (1962). Let f1(x),..., fk (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 = f1 ··· fk . Then

|{1 6 n 6 x : f1(n),..., fk (n) are all prime}| Z x 1 Y 1 − Nf (p)/p dt ∼ , Qk (1 − 1/p)k (log t)k i=1 deg(fi ) p 2

where Nf (p) = |{0 6 x 6 p − 1 : f (x) ≡ 0 (mod p)}| < p. 8 / 48 Hardy-Littlewood Conjecture on π2(x) Prime Number Theorem. For x > 0 let π(x) denote the number of primes not exceeding x. Then x π(x) ∼ as x → ∞. log x

For x > 0, let

π2(x) := |{p 6 x : p + 2 is prime}|.

The Bateman-Horn Conjecture with f1(x) = x and f2(x) = x + 2 yields the following conjecture on π2(x). Hardy-Littlewood Conjecture. We have x π2(x) ∼ 2C2 as x → ∞, log2 x where Y 1 C = 1 − ≈ 0.66. 2 (p − 1)2 p>2 9 / 48 The sieve method

Let A = (a1, a2,...) be a ﬁnite integer sequence and let P be a set of some primes. The sift function X Y S(A, P, z) := 1, where Pz = p. a∈A p∈P (a,Pz )=1 p6z 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 6 z. 2 Eratosthenes’ Sieve. Let z > 2 and A = {n > z : n 6 z }, and let P be the set of all primes. Then

2 S(A, P, z) =|{z < n 6 z : n is divisible by no prime p 6 z}| =the number of primes in (z, z2] = π(z2) − π(z).

Remark. If S(A, P, z) > 0 for sufficiently large z, then we deduce that there are infinitely many primes. 10 / 48 Brun’s sieve Inclusion-Exclusion Principle. Let S be a finite set, and let

Sk = {a ∈ S : a has the property Pk } for k = 1,..., n. Then

|{a ∈ S : a has the property Pk for no k = 1,..., n}| n X X n =|S| − |Sk | + |Si ∩ Sj | − ... + (−1) |S1 ∩ · · · ∩ Sn|. k=1 16i 0 such that x(log log x)2 π (x) := |{p x : p + 2 is prime}| C 2 6 6 (log x)2 for all x > 2. Idea of the Proof. Let 5 6 y < x, and let q1, q2,..., qr be all the distinct odd primes not exceeding y. If {n, n + 2} is a twin prime pair with y < n 6 x, then n > qi and qi - n(n + 2) for all i = 1,..., r. Thus

π2(x) 6 y + |{p ∈ (y, x]: p + 2 is prime}| 6 y + N(y, x), where

N(y, x) = |{n 6 x : n(n + 2) 6≡ 0 (mod qi ) for all i = 1,..., r}|. If we take y = x1/(c log log x) for suitable c > 0, then we obtain via Brun’s sieve x(log log x)2 π (x) y + N(y, x) C for some C > 0. 2 6 6 (log x)2 12 / 48 Brun’s constant Choose a constant C 0 > 0 such that x(log log x)2 x π (x) C C 0 for all x 2. 2 6 (log x)2 6 (log x)1.5 >

Let {tn, tn + 2} be the n-th twin prime pair. Then 0 0 tn 1 C n = π2(tn) 6 C 1.5 and hence 6 1.5 . (log tn) tn (log n) It follows that ∞ ∞ X 1 X 1 C 0 < ∞. t 6 n(log n)1.5 n=1 n n=1 Brun’s constant: ∞ X 1 1 + ≈ 1.9021604. t t + 2 n=1 n n

13 / 48 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 13 + 2 = 3 × 5.

14 / 48 Bombieri-Vinorgradov Theorem Prime Number Theorem for Arithmetic Progressions. Let 1 6 a 6 q with (a, q) = 1, then X π(x) Li(x) π(x; a, q) := 1 ∼ ∼ , ϕ(q) ϕ(q) p6x p≡a (mod q) where Z x dt x Li(x) = ∼ . 2 log t log x Bombieri-Vinorgradov Theorem (1965). For any A > 0, there is a constant B > 0 depending on A such that X x E(q, x) = O A , √ (log x) q6 x/(log x)B where Li(x) E(q, x) := max π(x; a, q) − . 16a6q ϕ(q) (a,q)=1

15 / 48 Elliott-Hlberstam Conjecture

The Bombieri-Vinorgradov Theorem plays a very important role in analytic number theory. It led Bombieri and Vinorgradov to obtain “1 + 3” on Goldbach’s conjecture. Jing-run Chen also needed the Bombieri-Vinorgradov Theorem in his proof of Chen’s theorem (“1 + 2”). 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 CA > 0 such that X x E(x, q) C . 6 A (log x)A q6xθ

The Bombieri-Vinorgradov Theorem indicates that EH(θ) holds for 1 any 0 < θ < 2 . Up to now, nobody succeeds to prove EH(θ) for 1 some θ > 2 . 16 / 48 Selberg’s upper bound sieve

Selberg’s Sieve. Let A = (a1, a2,...) be a ﬁnite sequence and let |A| be its length. Let P be a set of some primes, and let Y P(z) = p for z > 2. p6z p∈P

For a squarefree positive integer d, let Ad 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 g1 be a completely multiplicative function with g1(p) = g(p) for all p ∈ P. Then |A| X S(A, P, z) + 3ω(d)|r(d)|, 6 G(z) 2 d6z d|Pz where ω(d) is the number of distinct prime divisors of d, X G(z) = g1(m) and r(d) = |Ad | − g(d)|A|. m6z p|m⇒p∈P 17 / 48 Starting point of the proof

P 2 w(n) = d|n λ(d) is called a weight of n. 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 Applying the Cauchy-Schwarz inequality Cauchy-Schwarz Inequality. Let ai , bi ∈ R for i = 1,..., n. Then n 2 n n X X 2 X 2 ai bi 6 ai bi . i=1 i=1 i=1

If aj 6= 0 for some j ∈ {1,..., n}, then the equality holds if and only for some t ∈ R we have bi = tai for all i = 1,..., n. Sketch of the Proof. n n n 2 X 2 X 2 X 2 X 0 6 (ai bj − aj bi ) = ai bj − ai bi . 16i

Let P be the set of all primes, and let χP (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 = {h1,..., hk } be an admissible k-tuple. Choose D λ(d) = µ(d)P(log d ) with d 6 D for suitable polynomial P with P 2 P(1) = 1, and set W (n) = ( Qk λ(d)) . If EH(θ) holds d| j=1(n+hj ) for some θ > 1/2, then for large N we have X 1 X χ (n + h )W (n) > W (n) for all j = 1,..., k, P j k N6n<2N N6n<2N hence k X X X χp(n + hj )W (n) > W (n) N6n<2N j=1 N6n<2N and thus there are 1 6 i < j 6 k such that p = n + hi and q = n + hj are both prime. Note that |p − q| does not exceed d(H) = max H − min H (the diameter of H). 20 / 48 Main Results of Goldston-Pintz-Yildirim

D.A. Goldston, J. Pintz, C.Y. Yildirim [Annals of Math. 170(2009)]: We have p − p lim inf n+1 n = 0. n→∞ log pn Under the Elliott-Halberstam conjecture,

lim inf(pn+1 − pn) 16. n→∞ 6

D.A. Goldston, J. Pintz, C.Y. Yildirim [Acta Math. 204(2010)]: p − p lim inf √ n+1 n < ∞. 2 n→∞ log pn(log log pn)

21 / 48 Discrepancy Let f be an arithmetic function with supp(f ) = {n : f (n) 6= 0} ﬁnite. For a primitive residue class a(q) = a + qZ (with (a, q) = 1), deﬁne the discrepancy X 1 X ∆(f ; a(q)) := f (n) − f (n). ϕ(q) n≡a (mod q) (n,q)=1

Below we let 1I (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 x1/883, and f (n) = 0 otherwise. Then X x max |∆(f 1[1,x], a(q))| = O A . 16a6q (log x) q6x1/2+1/42 (a,q)=1 Friedlander and Iwaniec [Annals of Math. 34(1985)]: Let A > 0, 1/2+1/230 and let q 6 x and (a, q) = 1. Then x ∆(τ31[1,x], a(q)) = O A , where τ3(n) = Σabc=n1. q(log x) 22 / 48 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), 298-310. This paper was posted to arXiv in 2006. 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 = {h1,..., hk } such that for any A > 0 and large x we have

X X x |∆(f 1 , a(q))| = O , [x,2x] (log x)A θ 1 a q, (a,q)=1 q6x 6 6 Q Qk q| θ/2−1/4 p q| (a+hj ) p6x j=1

then there are inﬁnitely many n such that {n + h1,..., n + hk } contains at least two primes, and hence

lim inf(pn+1 − pn) d(H) < ∞. n→∞ 6

23 / 48 Yitang Zhang’s breakthrough In 2013, Yitang Zhang rediscovered the approach of Motohashi and Pintz. Moreover, he proved that the condition holds for 1 1 θ = + and k = 3.5 × 106. 2 584 To deduce this, he needed to bound incomplete exponential sums in the form X 2πi c1n¯+c2n+l e q , N6n62N wheren ¯ 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 = {pπ(k)+j : j = 1,..., k} is an admissible k-tuple. In fact, for any prime p 6 k, we have pπ(k)+j 6≡ 0 (mod p) for all j = 1,..., k; for any prime p > k obviously Sk 6 7 j=1 pπ(k)+j (mod p) 6= Z. For k = 3.5 × 10 , pπ(k)+k < 7 × 10 . 7 Zhang’s Theorem (2013). lim infn→∞(pn+1 − pn) < 7 × 10 . 24 / 48 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 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 X W (n) = λ(d) Qk d| i=1(n+hi )

(where H = {h1,..., hk } is an admissible k-tuple), Maynard employed the weights of the new form

2 X w(n) = λd1,...,dk .

di |n+hi (i=1,...,k)

26 / 48 Maynard’s approach

Let N be large and set W = Q p. As p6log log log N H = {h1,..., hk } is admissible, for any prime p | W , there is an Sk integer rp 6∈ i=1 hi ( (mod p)). By the Chinese Remainder Theorem, there is an integer ν such that ν ≡ −rp (mod p) for all Qk p | W and hence W is coprime to i=1(ν + hi ). Maynard restricted his attention only to those n ≡ ν (mod W ). Let

k X X X S1 = w(n), S2 = χP (n + hi ) w(n). N6n<2N N6n<2N i=1 n≡ν (mod W ) n≡ν (mod W )

If S2 > mS1, then at least m + 1 of the numbers n + h1,..., n + hk are primes.

27 / 48 Maynard’s approach

Let F be a piecewise diﬀerentiable function with F (x1,..., xk ) 6= 0 for all x1,..., xk > 0 with x1 + ··· + xk 6 1. Let θ > 0 and R = Nθ/2−δ for some small ﬁxed δ > 0. For integers Qk d1,..., dk > 0, if ( i=1 di , W ) = 1 then put k Y λd1,...,dk := µ(di )di i=1 Qk 2 X µ( ri ) log r1 log rk × i=1 F ,..., , Qk ϕ(r ) log R log R ri ≡0 (mod di ) (0

and let λd1,...,dk = 0 otherwise. Set 2 w(n) := Σdi |n+hi (i=1,...,k)λd1,...,dk , Z 1 Z 1 2 Ik (F ) := ··· F (t1,..., tk ) dt1dt2 ··· dtk , 0 0 Z 1 Z 1 Z 1 2 (s) Jk (F ) := ··· F (t1,..., tk )dts dt1 ··· dts−1dts+1 ··· dtk . 0 0 0 28 / 48 Maynard’s approach Qk (s) Suppose that EH(θ) holds. Provided Ik (F ) s=1 Kk (F ) 6= 0,

k k k ϕ(W ) ϕ(W ) N X (s) S ∼ N(log R)k I (F ), S ∼ · (log R)k+1 J (F ), 1 W k+1 k 2 W k+1 log N k s=1 and thus S θ Pk J(s)(F ) 2 → − δ s=1 k as N → ∞. S1 2 Ik (F ) Deﬁne Pk (s) X s=1 Jk (F ) Mk = . Ik (F ) F Then there are inﬁnitely many n ≡ ν (mod W ) such that {n + hi : i = 1,..., k} contains at least m = dθMk /2e primes, in particular lim inf(pn+m−1 − pn) d(H) < ∞. n→∞ 6

29 / 48 Maynard-Tao Theorem

Theorem (Maynard, 2013, arXiv:1311.4600). (i) M5 > 2, thus the EH conjecture implies that lim infn→∞(pn+1 − pn) 6 12 since {0, 2, 6, 8, 12} is admissible. (ii) M105 > 4 and thus lim infn→∞(pn+1 − pn) 6 600 (since EH(θ) 1 holds fore all 0 < θ < 2 and there is an admissible 105-tuple with diameter 600).

(iii) For large values of k, we have Mk > log k − 2 log log k − 2. Maynard-Tao Theorem (2013). There is an absolute constant C > 0 such that

3 4m lim inf(pn+m − pn) Cm e n→∞ 6 for all m = 1, 2, 3,....

Polymath. lim infn→∞(pn+1 − pn) 6 246 (P. Nielsen), and Mk > log k + O(1). 30 / 48 Consecutive primes and Legendre symbols

Theorem (H. Pan & Z.-W. Sun, arXiv:1405.0290) Let m be any positive integer and let δ1, δ2 ∈ {1, −1}. Then, for some constanst Cm > 0 there are inﬁnitely many integers n > 1 with pn+m − pn 6 Cm such that pn+i pn+j = δ1 and = δ2 pn+j pn+i

· for all 0 6 i < j 6 m, where pk denotes the k-th prime, and ( p ) denotes the Legendre symbol for any odd prime p. + Conjecture. Let m ∈ Z , δ ∈ {1, −1}, and δij ∈ {±1} for all 0 6 i < j 6 m. Then, there are inﬁnitely many integers n > 1 such that pn+i pn+j = δij = δ for all 0 6 i < j 6 m. pn+j pn+i

31 / 48 Examples Example 1. The smallest integer n > 1 with p n+i = 1 for all i, j = 0,..., 6 with i 6= j pn+j

is 176833. The 7 consecutive primes p176833, p176834,..., p178639 have concrete values: 2434589, 2434609, 2434613, 2434657, 2434669, 2434673, 2434681. Example 2. The smallest integer n > 1 with p n+i = −1 for all i, j = 0,..., 5 with i 6= j pn+j is 2066981, and the 6 consecutive primes p2066981, p2066982,..., p2066986 have concrete values: 33611561, 33611573, 33611603, 33611621, 33611629, 33611653.

32 / 48 Examples (continued) Example 3. The smallest integer n > 1 with pn+i pn+j − = 1 = for all 0 6 i < j 6 6 pn+j pn+i is 7455790, and the 7 consecutive primes p7455790, p7455791,..., p7455796 have concrete values: 131449631, 131449639, 131449679, 131449691, 131449727, 131449739, 131449751. Example 4. The smallest integer n > 1 with pn+i pn+j = 1 = − for all 0 6 i < j 6 5 pn+j pn+i is 59753753, and the 6 consecutive primes p59753753, p59753754,..., p59753758 have concrete values: 1185350899, 1185350939, 1185350983, 1185351031, 1185351059, 1185351091.

33 / 48 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 = {hi : i = 1,..., k} is an admissible set of cardinality k and W = q Q p (with q ∈ +) is relatively prime to Qk h with 0 p6w 0 Z i=1 i 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 + h1, n + h2,..., n + hk . Lemma 2 (Pan-Sun) Let k > 1 be an integer. Then there is an admissible set H = {h1,..., hk } with h1 = 0 < h2 < . . . < hk which has the following properties: Q (i) All those h1, h2,..., hk are multiples of K = 4 p<2k p.

(ii) Each hi − hj with 1 6 i < j 6 k has a prime divisor p > 2k 2 with hi 6≡ hj (mod p ). (iii) If 1 6 i < j 6 k, 1 6 s < t 6 k and {i, j}= 6 {s, t}, then no prime p > 2k divides both hi − hj and hs − ht .

34 / 48 Proof of the Theorem

By Lemma 1, there is an integer k = km > m depending on m such that for any admissible set H = {h1,..., hk } of cardinality k Qk if x is suﬃciently large and i=1 hi is relatively prime to W = 4 Q p then for some integer n ∈ [x/W , 2x/W ] there are p6w more than m primes among Wn + h1, Wn + h2,..., Wn + hk , where w = log log log x.

Let H = {h1,..., hk } with h1 = 0 < h2 < . . . < hk be an admissible set satisfying the conditions (i)-(iii) in Lemma 2. Clearly K = 4 Q p ≡ 0 (mod 8). Let x be suﬃciently large with the p62k interval (hk , w] containing more than hk − k primes. Note that 8 | W since w > 2.

Let δ := δ1δ2. For any integer b ≡ δ (mod K) and each prime p < 2k, clearly b + hi ≡ δ + 0 (mod p) and hence gcd(b + hi , p) = 1 for all i = 1,..., k.

35 / 48 Proof of the Theorem (continued) For any 1 6 i < j 6 k, the number hi − hj has a prime divisor 2 pij > 2k with hi 6≡ hj (mod pij ). Suppose that p > 2k is a prime dividing Q (h − h ), then there is a unique pair {i, j} with 16i

So, for any integer b ≡ rp − hi (mod p), we have b + hs 6≡ 0 (mod p) for all s = 1,..., k.

Assume that S = {h1, h1 + 1,..., hk }\H is a set {ai : i = 1,..., t} of cardinality t > 0. Clearly t 6 hk − k + 1 and hence we may choose t distinct primes q1,..., qt ∈ (hk , w]. If b ≡ −ai (mod qi ), then b + hs ≡ hs − ai 6≡ 0 (mod qi ) for all s = 1,..., k since 0 < |hs − ai | < hk < qi . 36 / 48 Proof of the Theorem (continued) Let

Y Q = p ∈ (2k, w]: p - (hi − hj ) \{qi : i = 1,..., t}. 16i

For any prime q ∈ Q, there is an integer rq 6≡ −hi (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 = δ1δ2 (mod K). (2) b ≡ rp − hi ≡ rp − hj (mod p) if p > 2k is a prime dividing hi − hj with 1 6 i < j 6 k. (3) b ≡ −ai (mod qi ) for all i = 1,..., t. (4) b ≡ rq (mod q) for all q ∈ Q. Qk By the above analysis, s=1(b + hs ) is relatively prime to W .

37 / 48 Proof of the Theorem (continued) 0 As H = {b + hs : 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 + hs (s = 1,..., k). For ai ∈ S, we have

Wn + b + ai ≡ 0 − ai + ai = 0 (mod qi )

and hence Wn + b + ai is composite since W > qi . Therefore, there are consecutive primes pN , pN+1,..., pN+m with pN+i = Wn + b + hs(i) for all i = 0,..., m, where 1 6 s(0) < s(1) < . . . < s(m) 6 k. Note that

pN+m−pN = (Wn+b+hs(m))−(Wn+b+hs(0)) = hs(m)−hs(0) 6 hk .

For each s = 1,..., k, clearly Wn + b + hs ≡ 0 + δ + 0 = δ (mod 8) and hence −1 2 = δ and = 1. Wn + b + hs Wn + b + hs 38 / 48 Proof of the Theorem (continued)

As pN+i = Wn + b + hs(i) ≡ δ (mod 8) for all i = 0,..., m, by the Quadratic Reciprocal Law we have pn+j pn+i = δ for all 0 6 i < j 6 m. pN+i pN+j

Let 0 6 i < j 6 m. Then p h − h h N+i = s(i) s(j) = δ ij , pN+j Wn + b + hs(j) Wn + b + hs(j)

where hij is the odd part of hs(j) − hs(i). For any prime divisor p of hij , clearly p 6 hk 6 w and p Wn + b + h b + h = δ(p−1)/2 s(j) = δ(p−1)/2 s(j) . Wn + b + hs(j) p p

If p < 2k, then p | K, hence b + hs(j) ≡ δ + 0 (mod p) and thus p b + h δ = δ(p−1)/2 s(j) = δ(p−1)/2 = 1. Wn + b + hs(j) p p 39 / 48 Proof of the Theorem (continued) If p > 2k, then by the choice of b we have

p b + h r =δ(p−1)/2 s(j) = δ(p−1)/2 p Wn + b + hs(j) p p ( r δ δ if p = p , = p = 2 s(i),s(j) p 1 otherwise.

Recall that ps(i),s(j)khij . Therefore, pN+i hij = δ = δδ2 = δ1 pN+j Wn + b + hs(j)

and pN+j pN+i = δ = δ2. pN+i pN+j This concludes the proof.

40 / 48 Artin’s Primitive Root Conjecture

Artin’s Primitive Root Conjecture (1927). Let g 6= −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:1404.4007). Let g 6= −1 be an integer which is not a square. Let q1 < q2 < . . . denote the sequence of primes having g as a primitive root. For any positive integer m, there is a constant Cm > 0 not depending on g such that

lim inf(qn+m − qn) Cm. n→+∞ 6

41 / 48 Consecutive primes and primitive roots

Conjecture (Z.-W. Sun, 2014). For any positive integer m, there are inﬁnitely many positive integers n such that pn+i is a primitive root modulo pn+j for any distinct i and j among 0, 1,..., m. + Example. The least n ∈ Z with pn+i a primitive root modulo pn+j for any distinct i and j among 0, 1, 2, 3 is 8560. Note that

p8560 = 88259, p8561 = 88261 and p8562 = 88289.

Theorem (H. Pan & Z.-W. Sun, arXiv:1405.0290). The conjecture holds under the Extended Riemann Hypothesis.

42 / 48 A Firoozbakht-type conjecture for twin primes √ n Firoozbakht’s Conjecture (1982). The sequence ( pn)n>1 is strictly decreasing.

Conjecture (Z.-W. Sun, 2012) (i) If {t1, t1 + 2},..., {tn, tn + 2} are the first n pairs of twin primes, then the first prime t in the √ n+1√ 1+1/n n n+1 next pair of twin primes is smaller than tn , i.e., tn > tn+1. n+1p pn (ii) The sequence ( T (n + 1)/ T (n))n>9 is strictly increasing Pn with limit 1, where T (n) = k=1 tk . √ √ n n+1 Remark. Via Mathematica I verified that tn > tn+1 for all n = 1,..., 500000, and

n+1pT (n + 1)/pn T (n) < n+2pT (n + 2)/ n+1pT (n + 1)

for all n = 9, 10,..., 500000. Note that t500000 = 115438667. After I made the conjecture public, Marek Wolf veriﬁed the √ √ n n+1 inequality tn > tn+1 for all the 44849427 pairs of twin primes below 234 ≈ 1.718 × 1010. 43 / 48 Uniﬁcation of Goldbach’s conjecture and the twin prime conjecture

Uniﬁcation of Goldbach’s Conjecture and the Twin Prime Conjecture (Sun, 2014-01-29). For any integer n > 2, there is a prime q with 2n − q and pq+2 + 2 both prime. We have veriﬁed the conjecture for n up to 2 × 108. 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 pq+2 + 2 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 = 3 + 17 with 3, 17 and p3+2 + 2 = 11 + 2 = 13 all prime.

44 / 48 Graph for a(n) = |{q < 2n : q, 2n − q, pq+2 + 2 are all prime}|

45 / 48 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, 2014-02-05). Any integer n > 2 can be written as k + m with k and m positive integers such

that pk + 2 and ppm + 2 are both prime.

Example. 22 = 20 + 2 with p20 + 2 = 71 + 2 = 73 and

pp2 + 2 = p3 + 2 = 5 + 2 = 7 both prime.

Remark. If all those positive integer m with ppm + 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 pk(j) + 2 and ppm(j) + 2 are both prime, and hence k(j) ∈ ((j − 1)N, jN) since m(j) < N; thus ∞ ∞ X 1 X 1 > , p pjN j=1 k(j) j=1 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 Graph for a(n) = |{0 < k < n : pk + 2 and ppn−k + 2 are both prime}|

47 / 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,..., 109. In my opinion, the solution of the Super Twin Prime Conjecture might be beyond the intelligence of human beings! Thank you!

48 / 48