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 210093, P. R. China
zwsun@nju.edu.cn
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 Cramér’s Conjecture
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.
Cramér’s Conjecture (1936). We have
lim sup
n→∞
pn+1 − pn
= 1.
(log pn )2
Cramér showed 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
(n + hi ) for all n ∈ Z,
p 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
h
(mod
p) 6= Z for any prime p, then we call
i=1 i
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
for all
Qintegers
k
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
a1 n + b1 , a2 n + 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}|
Y
Z x
1 − Nf (p)/p
dt
1
,
∼ Qk
k
k
(1
−
1/p)
(log
t)
deg(f
)
2
i
p
i=1
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 2
as x → ∞,
log x
where
Y
1
≈ 0.66.
C2 =
1−
(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
(a,Pz )=1
p∈P
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.
Eratosthenes’ Sieve. Let z > 2 and A = {n > z : n 6 z 2 }, and
let P be the set of all primes. Then
S(A, P, z) =|{z < n 6 z 2 : n is divisible by no prime p 6 z}|
=the number of primes in (z, z 2 ] = π(z 2 ) − 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
=|S| −
|Sk | +
|Si ∩ Sj | − . . . + (−1)n |S1 ∩ · · · ∩ Sn |.
k=1
16i<j6n
Brun’s Observation. If m ∈ {1, . . . , n} is even, then
n
n
[
X
X
S \
S
6|S|
−
|Si | +
|Si ∩ Sj | − . . .
k
i=1
k=1
+
X
16i<j6n
(−1)m |Si1 ∩ . . . ∩ Sim |.
16i1 <...<im 6n
Remark. This is similar to the fact that
1 1 1
1 1
π
= 1 − + − + ··· 6 1 − + .
4
3 5 7
3 5
11 / 48
Brun’s theorem
Brun’s Theorem (1920) There is a constant C > 0 such that
π2 (x) := |{p 6 x : p + 2 is prime}| 6 C
x(log log x)2
(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
If we take y =
Brun’s sieve
x 1/(c log log x)
(mod qi ) for all i = 1, . . . , r }|.
for suitable c > 0, then we obtain via
π2 (x) 6 y + N(y , x) 6 C
x(log log x)2
for some C > 0.
(log x)2
12 / 48
Brun’s constant
Choose a constant C 0 > 0 such that
π2 (x) 6 C
x(log log x)2
x
6 C0
(log x)2
(log x)1.5
for all x > 2.
Let {tn , tn + 2} be the n-th twin prime pair. Then
n = π2 (tn ) 6 C 0
1
C0
tn
and
hence
6
.
(log tn )1.5
tn
(log n)1.5
It follows that
∞
∞
X
X
1
1
6 C0
< ∞.
tn
n(log n)1.5
n=1
n=1
Brun’s constant:
∞ X
1
n=1
1
+
tn tn + 2
≈ 1.9021604.
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
Li(x)
π(x)
∼
,
π(x; a, q) :=
1∼
ϕ(q)
ϕ(q)
p6x
p≡a
where
(mod q)
Z
x
dt
x
∼
.
log x
2 log t
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
,
(log x)A
√
B
Li(x) =
q6 x/(log x)
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
E (x, q) 6 CA
q6x θ
x
.
(log x)A
The Bombieri-Vinorgradov Theorem indicates that EH(θ) holds for
any 0 < θ < 21 . Up to now, nobody succeeds to prove EH(θ) for
some θ > 12 .
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
X
|A|
+
3ω(d) |r (d)|,
S(A, P, z) 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
Let z > 2, and let λ be a real arithmetic function with λ(1) = 1
and λ(z) = 0 for z > D. Observe that
X
S(A, P, z) =
1
a∈A
(a,Pz )=1
6
X X
a∈A
=
2
λ(d)
=
d1 ,d2 |Pz
λ(d1 )λ(d2 )
λ(d1 )
a∈A d1 |(a,Pz )
d|(a,Pz )
X
X X
X
a∈A
[d1 ,d2 ]|a
1=
X
X
λ(d2 )
d2 |(a,Pz )
λ(d1 )λ(d2 )|A[d1 ,d2 ] |
d1 ,d2 |Pz
=···
w (n) =
2
λ(d)
is called a weight of n.
d|n
P
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
X
2 X
X
n
n
n
2
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.
X
2
n
n
n
X
X
X
2
2
2
(ai bj − aj bi ) =
ai
bj −
ai bi .
06
16i<j6n
i=1
j=1
i=1
The Cauchy-Schwarz Inequality implies the following lemma
needed for Selberg’s sieve.
Lemma. Let a1 , . . . , an be positive reals and let b1 , . . . , bn ∈ R.
Under the restriction b1 y1 + · · · + bn yn = 1,
Q(y1 , . . . , yn ) = a1 y12 + . . . + an yn2
P
has the minimum m = ( ni=1 bi2 /ai )−1 . And the minimum is
attained if and only if yi = mbi /ai for all i = 1, . . . , n.
19 / 48
The work of Goldston-Pintz-Yildirim
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
6 D for suitable polynomial P with
λ(d) = µ(d)P(log Dd ) with dP
P(1) = 1, and set W (n) = ( d|Qk (n+hj ) λ(d))2 . If EH(θ) holds
j=1
for some θ > 1/2, then for large N we have
X
1 X
χP (n + hj )W (n) >
W (n)
k
N6n<2N
for all j = 1, . . . , k,
N6n<2N
hence
X
k
X
N6n<2N j=1
χp (n + hj )W (n) >
X
W (n)
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
lim inf
n→∞
pn+1 − pn
= 0.
log pn
Under the Elliott-Halberstam conjecture,
lim inf (pn+1 − pn ) 6 16.
n→∞
D.A. Goldston, J. Pintz, C.Y. Yildirim [Acta Math. 204(2010)]:
lim inf √
n→∞
pn+1 − pn
< ∞.
log pn (log log pn )2
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
X
1
f (n).
∆(f ; a(q)) :=
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
x 1/883 , and f (n) = 0 otherwise. Then
X
x
max |∆(f 1[1,x] , a(q))| = O
.
16a6q
(log x)A
1/2+1/42 (a,q)=1
q6x
Friedlander and Iwaniec [Annals of Math. 34(1985)]: Let A > 0,
and let q 6 x 1/2+1/230 and (a, q) = 1. Then
x
∆(τ3 1[1,x] , a(q)) = O
, where τ3 (n) = Σabc=n 1.
q(log x)A
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[x,2x] , a(q))| = O
,
(log x)A
θ
16a6q, (a,q)=1
Q q6x
p
q|
p6x θ/2−1/4
Q
q| k (a+hj )
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 ) 6 d(H) < ∞.
n→∞
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
c n̄+c n+l
2πi 1 q 2
,
e
N6n62N
where n̄ 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
Sk p) for all j = 1, . . . , k; for any prime p 6> k obviously
7
j=1 pπ(k)+j (mod p) 6= Z. For k = 3.5 × 10 , pπ(k)+k < 7 × 10 .
Zhang’s Theorem (2013). lim inf n→∞ (pn+1 − pn ) < 7 × 107 .
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)
d|
Qk
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
Q
Let N be large and set W = p6log log log N p. As
H = {h1 , . . .S
, hk } is admissible, for any prime p | W , there is an
integer rp 6∈ ki=1 hi ( (mod p)). By the Chinese Remainder
Theorem, there is an integer ν such that ν ≡ −rp (mod p) for all
Q
p | W and hence W is coprime to ki=1 (ν + hi ). Maynard
restricted his attention only to those n ≡ ν (mod W ). Let
S1 =
X
N6n<2N
n≡ν (mod W )
w (n), S2 =
X
N6n<2N
n≡ν (mod W )
X
k
χP (n + hi ) w (n).
i=1
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 = N θ/2−δ for some
Q small fixed δ > 0. For integers
d1 , . . . , dk > 0, if ( ki=1 di , W ) = 1 then put
Y
k
λd1 ,...,dk :=
µ(di )di
i=1
X
×
ri ≡0 (mod di ) (0<i6k)
Q
( k ri ,W )=1
i=1
Q
µ( ki=1 ri )2
log r1
log rk
F
,...,
,
Qk
log R
log R
i=1 ϕ(ri )
and let λd1 ,...,dk = 0 otherwise. Set
2
w (n) := Σdi |n+hi (i=1,...,k) λd1 ,...,dk ,
Z 1
Z 1
Ik (F ) :=
···
F (t1 , . . . , tk )2 dt1 dt2 · · · dtk ,
(s)
Jk (F )
Z
1
Z
0
0
1Z 1
···
:=
0
F (t1 , . . . , tk )dts
0
0
2
dt1 · · · dts−1 dts+1 · · · dtk .
28 / 48
Maynard’s approach
Suppose that EH(θ) holds. Provided Ik (F )
(s)
s=1 Kk (F )
Qk
6= 0,
k
X (s)
ϕ(W )k N
ϕ(W )k
k
k+1
N(log
R)
I
(F
),
S
∼
·
(log
R)
Jk (F ),
S1 ∼
2
k
W k+1
W k+1 log N
s=1
and thus
S2
→
S1
θ
−δ
2
(s)
s=1 Jk (F )
Pk
Ik (F )
Define
Mk =
X
F
as N → ∞.
(s)
s=1 Jk (F )
Pk
Ik (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 ) 6 d(H) < ∞.
n→∞
29 / 48
Maynard-Tao Theorem
Theorem (Maynard, 2013, arXiv:1311.4600).
(i) M5 > 2, thus the EH conjecture implies that
lim inf n→∞ (pn+1 − pn ) 6 12 since {0, 2, 6, 8, 12} is admissible.
(ii) M105 > 4 and thus lim inf n→∞ (pn+1 − pn ) 6 600 (since EH(θ)
holds fore all 0 < θ < 12 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
lim inf (pn+m − pn ) 6 Cm3 e 4m
n→∞
for all m = 1, 2, 3, . . . .
Polymath. lim inf n→∞ (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+j
pn+i
= 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+j
pn+i
= δij = δ
for all 0 6 i < j 6 m.
pn+j
pn+i
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Examples
Example 1. The smallest integer n > 1 with
pn+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
pn+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.
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Examples (continued)
Example 3. The smallest integer n > 1 with
pn+j
pn+i
−
=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+j
pn+i
=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.
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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 Q
: i = 1, . . . , k} is an admissible set of cardinality
Q k and
W = q0 p6w p (with q0 ∈ Z+ ) is relatively prime to ki=1 hi 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 + 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
with hi 6≡ hj (mod p 2 ).
(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 admissibleQ
set H = {h1 , . . . , hk } of cardinality k
k
if x is sufficiently
large
and
i=1 hi is relatively prime to
Q
W = 4 p6w p then for some integer n ∈ [x/W , 2x/W ] there are
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
Q set satisfying the conditions (i)-(iii) in Lemma 2. Clearly
K = 4 p62k p ≡ 0 (mod 8). Let x be sufficiently large with the
interval (hk , w ] containing more than hk − k primes. Note that
8 | W since w > 2.
Let δ := δ1 δ2 . 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
pij > 2k Q
with hi 6≡ hj (mod pij2 ). Suppose that p > 2k is a prime
dividing 16i<j6k (hi − hj ), then there is a unique pair {i, j} with
1 6 i < j 6 k such that hi ≡ hj (mod p). Note that p 6 hk . All
the k − 2 < (p − 3)/2 numbers hi − hs with 1 6 s 6 k and s 6= i, j
are relatively prime to p, so there is an integer rp 6≡ hi − hs
(mod p) for all s = 1, . . . , k such that
(
δ2 if p = pij ,
rp δ
=
p
1 otherwise.
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
Q=
p ∈ (2k, w ] : p -
Y
(hi − hj ) \ {qi : i = 1, . . . , t}.
16i<j6k
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 = δ1 δ2 (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.
Q
By the above analysis, ks=1 (b + hs ) is relatively prime to W .
37 / 48
Proof of the Theorem (continued)
As H0 = {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
2
−1
= 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
hs(i) − hs(j)
hij
pN+i
=
=δ
,
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
(p−1)/2 Wn + b + hs(j)
(p−1)/2 b + hs(j)
=δ
=δ
Wn + b + hs(j)
p
p
If p < 2k, then p | K , hence b + hs(j) ≡ δ + 0 (mod p) and thus
p
(p−1)/2 δ
(p−1)/2 b + hs(j)
=δ
=δ
= 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
(p−1)/2 b + hs(j)
(p−1)/2 rp
=δ
=δ
Wn + b + hs(j)
p
p
(
δ2 if p = ps(i),s(j) ,
rp δ
=
=
p
1 otherwise.
Recall that ps(i),s(j) khij . Therefore,
and
pN+i
pN+j
=δ
pN+j
pN+i
hij
Wn + b + hs(j)
=δ
pN+i
pN+j
= δδ2 = δ1
= δ2 .
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 ) 6 Cm .
n→+∞
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.
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A Firoozbakht-type conjecture for twin primes
√
Firoozbakht’s Conjecture (1982). The sequence ( n 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 tn+1 in the
√
√
1+1/n
next pair of twin primes
is smaller p
than tn
, i.e., n tn > n+1 tn+1 .
p
(ii) The sequence ( n+1 T (n P
+ 1)/ n T (n))n>9 is strictly increasing
with limit 1, where T (n) = nk=1 tk .
√
√
Remark. Via Mathematica I verified that n tn > n+1 tn+1 for all
n = 1, . . . , 500000, and
p
p
p
p
n+1
T (n + 1)/ n T (n) < n+2 T (n + 2)/ n+1 T (n + 1)
for all n = 9, 10, . . . , 500000. Note that t500000 = 115438667.
After I made the conjecture public, Marek Wolf verified the
√
√
inequality n tn > n+1 tn+1 for all the 44849427 pairs of twin primes
below 234 ≈ 1.718 × 1010 .
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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
X
1
1
>
,
pk(j)
pjN
j=1
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