Study on the Hilbert’s Eighth Problem Jinzhu Han1

Zaizhu Han2

1. Baotou North Hospital, Inner Mongolia, China 2. National Key Laboratory of Cognitive Neuroscience and Learning, Beijing Normal University, China

Abstract In this paper, we first prove Riemann Hypothesis and General Riemann Hypothesis by L’Hospital rule. Then, we improve the result of the prime number theorems for arithmetic progression. Finally, we provide a proof that Goldbach Conjecture and Twin Prime Conjecture are established. Thus, we have resolved the Hilbert’s Eighth Problem.

Keywords: Riemann Hypothesis Goldbach Conjecture Twin Prime Conjecture AMS: 11P32 Notation: ∞

ζ ( s ) = ∑ n− s

Riemann zeta Function

n =1

∞

L ( s, χ ) = ∑ χ ( n ) n − s

Dirichlet L -Function

∞ 1 ⎛ s⎞ = seγ s ∏ ⎜1 + ⎟e − s n n⎠ Γ (s) n =1 ⎝

Γ ( s ) as Γ -function

γ = 0.5772…

Euler constant

χ, χ 0, χ*

Character, principal character, primitive character

n =1

q

⎛n⎞

n =1

⎝ ⎠

τ ( χ ) = ∑ χ (n) e ⎜ ⎟ q f(

k)

(s) =

f (z) k! dz ∫ 2π i C ( z − s )k +1

One of Gauss sum

Cauchy derivative formula

⎛

φ ( n ) = n∏ ⎜1 − ⎝

pn

1⎞ ⎟ p⎠

Euler function

⎧log p, n = p m Λ (n) = ⎨ ⎩0, otherwise

Mangoldt function

ψ ( x) = ∑ Λ (n) n≤ x

θ ( x ) = ∑ log p p≤ x

π ( x ) = ∑1 p≤ x

∑

ψ ( x; q, l ) =

Λ (n)

n≤ x n ≡ l mod q

∑

ψ ( x; b, q, l ) =

Λ (n)

bn ≤ x bn ≡ l mod q

ψ ( x, χ ) = ∑ Λ ( n ) χ ( n ) n≤ x

θ ( x; q, l ) =

∑

log p

p ≡ l mod q 2< p ≤ x

∑

θ ( x; b, q, l ) =

log p

bp ≡ l mod q 2< p ≤ x

C(N) =

∏

2< p N

p −1 ⎛ 1 ⎞ 1 − ⎜ ⎟ ∏ p − 2 p > 2 ⎜⎝ ( p − 1)2 ⎟⎠

0. Introduction In 1900, Hilbert suggested 23 famous mathematic problems, whose Eighth Problem includes the following hypotheses: 1) Riemann Hypothesis (RH): all of non-trivial zero points of

ζ ( s ) have real

part equal to 1 2 ; 2) General Riemann Hypothesis (GRH): all of non-trivial zero points of L ( s, χ ) have real part equal to 1 2 ; 3) Goldbach Conjecture: any even integer >2 is sum of two primes; 4) Twin Prime Conjecture: the number of twin primes is infinite. Although a number of mathematicians have attempted to prove or disproved the above four

hypotheses [1-39], none of them has been completely resolved up to now. In this article, we will prove all of these hypotheses using some novel ways. To prove some relevant theorems, we need the following Lemmas. 1. Lemmas

Lemma 1: There is the well-known equation for

ζ (s) ,

ζ ( s ) = A ( s ) ζ (1 − s ) ,

(1.1.1)

where

A ( s ) = 2 ( 2π )

s −1

cos

π (1 − s ) 2

Γ (1 − s ) .

(1.1.2)

Let s = σ + it , then we have

σ = 12 ⇔ A ( s ) ≡ 1 ,

(1.1.3)

and

t t → ∞ ⇒ A( s) = 2π

⇒σ ≠

1

2

1

2 −σ

⎛ ⎛ 1 ⎞⎞ ⎜1 + O ⎜⎜ ⎟⎟ ⎟ ⎜ ⎟ ⎝ t ⎠⎠ ⎝

⇒ A( s) ≠ 1.

(1.1.4)

Lemma 2: There is the equation for L ( s, χ ) ,

L ( s, χ ) = A ( s, χ ) L (1 − s, χ ) ,

(1.2.1)

where

A( s, χ ) = ( 2π ) q− sτ ( χ ) cos s −1

π (1 − s ) 2

{

}

Γ (1 − s ) e−iπ (1−s) 2 + χ ( −1) eiπ (1−s) 2 . (1.2.2)

Let s = σ + it , we obtain

σ = 1 2 ⇔ A ( s, χ ) ≡ 1 ,

(1.2.3)

and

qt t → ∞ ⇒ A ( s, χ ) = 2π

⇒σ ≠

1

2

1

2 −σ

⎛ ⎛ 1 ⎞⎞ ⎜1 + O ⎜⎜ ⎟⎟ ⎟ ⎜ ⎟ ⎝ t ⎠⎠ ⎝

⇒ A ( s, χ ) ≠ 1 .

(1.2.4)

Lemma 3: In complex region 0 ≤ Re ( s ) ≤ 1 , for k − th order derivative of ζ ( s ) , we have

ζ ( k ) ( s ) = ζ ( k ) (1 − s ) .

(1.3.1)

Proof: In complex region 0 ≤ Re ( s ) ≤ 1 , we select any of suitable simple closed contour

C = C1 + C2 , which is symmetry for real axis and line Re ( s ) =

1

2

, then

ζ ( z) ζ (1 − z ) k! k! dz = dz , k + 1 ∫ ∫ 2π i C1 ( z − s ) 2π i C2 ( z − s )k +1

(1.3.2)

and

ζ ( z) ζ (1 − z ) k! k! dz = dz . k + 1 ∫ ∫ 2π i C2 ( z − s ) 2π i C1 ( z − s )k +1

(1.3.3)

According to Cauchy derivative formula, we have

ζ (k ) ( s ) = =

ζ ( z) k! dz 2π i ∫C ( z − s )k +1 ζ ( z) ζ (z) k! k! dz + dz , (1.3.4) k +1 ∫ ∫ C C 2π i 1 ( z − s ) 2π i 2 ( z − s )k +1

and

ζ ( k ) (1 − s ) =

ζ (1 − z ) k! dz ∫ 2π i C ( z − s )k +1 =

ζ (1 − z ) ζ (1 − z ) k! k! dz + dz . (1.3.5) k + 1 ∫ ∫ 2π i C1 ( z − s ) 2π i C2 ( z − s )k +1

Through (1.3.2) – (1.3.5), we obtain

0 ≤ Re ( s ) ≤ 1 ⇒ ζ ( k ) ( s ) = ζ ( k ) (1 − s ) .

(1.3.6)

Lemma 3 is true. Lemma 4: In complex region 0 ≤ Re ( s ) ≤ 1 , for k − th order derivative of L ( s, χ ) , we have

L( k ) ( s, χ ) = L( k ) (1 − s, χ ) .

(1.4.1)

In a similar way, in complex region 0 ≤ Re ( s ) ≤ 1 , we select any suitable simple closed contour C ,

which is symmetry for real axis and line Re ( s ) =

L(

k)

( s, χ ) =

1

2

. By Cauchy derivative formula, we have

L ( z, χ ) k! k ! L (1 − z , χ ) = dz dz , 2π i ∫C ( z − s )k +1 2π i ∫C ( z − s )k +1

(1.4.2)

k ! L (1 − z , χ ) dz . 2π i ∫C ( z − s )k +1

(1.4.3)

and

L( k ) (1 − s, χ ) = So we obtain

0 ≤ Re ( s ) ≤ 1 ⇒ L( k ) ( s, χ ) = L( k ) (1 − s, χ ) .

(1.4.4)

Lemma 4 is established.

Lemma 5: There exists the inequality, for

χ * ≠ χ 0 , and ρ = β + iγ is non-trivial zero point of

L ( s, χ ) ,

ψ ( x, χ

*

)

⎛ x log 2 ( xqT ) ⎞ xβ + O⎜ ⎟. ∑ T γ ≤T 1 + γ ⎝ ⎠

(1.5.1)

Lemma 6: According to equivalent relation about θ ( x ) and ψ ( x ) , we have

θ ( x; q, l ) − x φ ( q )

∑ θ ( x; q, l ) − x φ ( q )

2≤ q ≤ D

ψ ( x; q, l ) − x φ ( q ) ,

(1.6.1)

∑ ψ ( x; q, l ) − x φ ( q ) ,

(1.6.2)

xb ∑ ψ ( x; b, q, l ) − φ ( q ) .

(1.6.3)

2≤ q ≤ D

and

xb ∑ θ ( x; b, q, l ) − φ ( q )

2≤ q ≤ D

2≤ q ≤ D

Lemma 7: We define a new weighted sieve function

S ( A, z ) =

∑

log p ,

p ≡ l mod a ( a , P( z ) ) =1

where P ( z ) =

∏

2≤ p ≤ z

p , and a ∈ A = {a : 2 ≤ a ≤ x} .

(1.7.1)

Let

ω (d ) =

d , then φ (d ) 0≤

ω ( p) p

=

1 < 1, φ ( p)

and

∑ ω ( p ) log p

p − log ( z w ) ≤ L , ( L is a positive constant).

w≤ p ≤ z

Therefore S ( A, z ) is a line sieve function. According to Rosser sieve method, let 2 ≤ z ≤ y

1

2

and X = θ ( x ) = x , we have

S ( A, z ) ≥ 2C ( N ) f ( u )

e −γ ⎛ ⎛ 1 ⎞ ⎞ ⎜1 + ⎜ ⎟ ⎟ − ∑ rd , log z ⎝ ⎝ log z ⎠ ⎠ d ≤ y

(1.7.2)

S ( A, z ) ≤ 2C ( N ) F ( u )

e−γ ⎛ ⎛ 1 ⎞ ⎞ ⎜1 + ⎜ ⎟ ⎟ − ∑ rd , log z ⎝ ⎝ log z ⎠ ⎠ d ≤ y

(1.7.3)

and

where

rd =

∑ Λ (a) − X

d a∈A

ω (d ) d

,

u = log y log z , f ( u ) = 2eγ u −1 log ( u − 1) , if 2 ≤ u ≤ 4 ,

(1.7.4)

(1.7.5) (1.7.6)

and

F ( u ) = 2eγ u −1 , if 2 ≤ u ≤ 3 .

(1.7.7)

2. Theorems Theorem 1: All of non-trivial zero points of ζ ( s ) have real part equal to 1 2 . Proof: It is well known that all of non-trivial zero points of ζ ( s ) are in complex region

0 ≤ Re ( s ) ≤ 1 . Let ρ = β + iγ is any of k − th order non-trivial zero points of ζ ( s ) , then the

functions ζ ( s ) and ζ (1 − s ) are both equal to zero at complex point

ρ = β + iγ . Namely,

ζ ( ρ ) = ζ (1 − ρ ) = 0 , A( ρ ) =

ζ (ρ) is the indeterminate form of ζ (1 − ρ )

0 0

(2.1.1)

.

According to L’Hospital Rule, we have

A ( ρ ) = Lim s→ρ

ζ (s)

ζ (1 − s )

= Lim s→ρ

ζ (k ) ( s )

ζ ( k ) (1 − s )

.

(2.1.2)

By Lemma 1 and lemma 3, we obtain

ζ

(k )

( s ) = ζ (1 − s ) ⇒ A ( ρ ) = Lim s→ρ (k )

ζ (k ) ( s )

ζ ( k ) (1 − s )

= 1,

(2.1.3)

and

ζ ( ρ ) =0 ⇒ A ( ρ ) = 1 ⇒ Re ( ρ ) = 1 2 .

(2.1.4)

This deduction is suitable for all of non-trivial zero points of ζ ( s ) . Consequently, Theorem 1 is established and RH is true. Theorem 2: All of non-trivial zero points of L ( s, χ ) have real part equal to 1 2 . Proof: All of non-trivial zero points of L ( s, χ ) are in complex region 0 ≤ Re ( s ) ≤ 1 . Let

ρ = β + iγ is any of non-trivial zero points of L ( s, χ ) , then L ( ρ , χ ) = L (1 − ρ , χ ) = 0 ,

(2.2.1)

L (1 − ρ , χ ) = L (1 − ρ , χ ) = 0 .

(2.2.2)

L ( ρ , χ ) = A ( ρ , χ ) L (1 − ρ , χ ) = 0 .

(2.2.3)

and

So we have

According to L’Hospital Rule, Lemma 2 and lemma 4, we have

L ( s, χ ) L( ) ( s, χ ) A ( ρ , χ ) = Lim = Lim ( k ) = 1, s → ρ L (1 − s, χ ) s→ρ L (1 − s, χ ) k

and

(2.2.4)

L ( ρ , χ ) = 0 ⇒ A ( ρ , χ ) = 1 ⇒ Re ( ρ ) =

1

2

.

(2.2.5)

This deduction is suitable for all of non-trivial zero points of L ( s, χ ) . Theorem 2 is proved, and

GRH is true. Theorem 3: There exists a calculable constant for O and

, such that

(

)

ψ ( x; q, l ) = x φ ( q ) + O x log 2 x , 1

2

(2.3.1)

or

E ( x; q, l ) = ψ ( x; q, l ) − x φ ( q ) Proof: Let χ ≠

1

x 2 log 2 x .

(2.3.2)

χ 0 , ( q, l ) = 1 , then

ψ ( x; q, l ) = xφ −1 ( q ) + φ −1 ( q )

∑ χ ( l )ψ ( x, χ ) ,

(2.3.3)

χ mod q

and

E ( x; q, l ) = φ −1 ( q )

φ −1 ( q ) Since

φ −1 ( q )

∑ 1 < 1,

χ mod q

∑ χ ( l )ψ ( x, χ )

χ mod q

∑ χ ( l ) ψ ( x, χ ) .

χ ( n ) = 1 , we have E ( x; q, l )

maxψ ( x, χ )

maxψ ( x, χ * ) + O ( log x log q ) . Let

(2.3.4)

χ mod q

(2.3.5)

ρ = 1 2 + iγ is non-trivial zero point of L ( s, χ ) , we obtain E ( x; q, l )

1 ⎛ x log 2 ( xqT ) ⎞ x2 + O ⎜ ⎟ + O ( log x log q ) . ∑ T γ ≤T 1 + γ ⎝ ⎠

(2.3.6)

Let 2 ≤ q ≤ x , T = x 2 , then 1

E ( x; q, l )

1

x 2 log 2 x .

(2.3.7)

Theorem 3 is true. Theorem 4: There exist positive numbers B ≥ A + 2 , such that

∑

2 ≤ q ≤ x log − B x

E ( x; q, l )

x log − A x .

(2.4.1)

Proof: Let ρ ( q ) =

1

2

+ iγ ( q ) is non-trivial zero point of L ( s, χ mod q ) , then

⎛ xD log 2 ( xDT ) ⎞ +O⎜ ⎟ + O ( D log x log D ) (2.4.2) ∑ ∑ T 2≤ q ≤ D γ ( q ) ≤T 1 + γ ( q ) ⎝ ⎠

x

∑ E ( x; q, l )

2≤ q ≤ D

Since

1

2

χ * ( q1 ) ≠ χ * ( q2 ) ⇒ L ( s, χ * mod q1 ) ≠ L ( s, χ * mod q2 ) , all of non-trivial zero point of

L ( s, χ mod qn ) are different, then

∑ ∑

1

x

2 ≤ q ≤ D γ ( q ) ≤T

x

∑

2

1+ γ (q)

1

1

x2 ， ∑ γ ≤T 1 + γ

2

1 + γ ( qn )

γ ( qn ) ≤T

(2.4.3)

where γ ≤ T includes all of the different non-trivial zero points of L ( s, χ mod qn ) . Furthermore we obtain 1

x2 ∑ γ ≤T 1 + γ

x

1

2

1 dt −T 1 + t

∫

T

1

x 2 log T .

(2.4.4)

Therefore we have

∑

E ( x; q, l )

2≤ q ≤ D

Let T = x , D = x log

−B

⎛ xD log 2 ( xDT ) ⎞ 1 x 2 log T + O ⎜ ⎟ + O ( D log x log D ) . T ⎝ ⎠

(2.4.5)

x , B ≥ A + 2 , then

∑ E ( x; q, l )

x log − A x .

(2.4.6)

2≤ q ≤ D

Theorem 4 is established. Theorem 5: there exist positive numbers B ≥ A + 2 , such that

∑

2 ≤ q ≤ x log

−B

∑

x 1≤b ≤ x

1

E ( x; b, q, l )

x log − A x ,

(2.5.1)

2

where

E ( x; b, q, l ) = ψ ( x; b, q, l ) − ( x b ) φ ( q ) . Proof: Let χ ≠

χ 0 , then E ( x; b, q, l ) = φ −1 ( q )

∑ χ (l ) χ (b ) ∑ Λ ( n) χ ( n )

χ mod q

max ∑ Λ ( n ) χ ( n ) bn ≤ x

bn ≤ x

(2.5.2)

maxψ ( x b, χ )

maxψ ( x b, χ * ) + O ( log ( x b ) log q ) .

(2.5.3)

Furthermore, we have

( x b) ∑ γ ( q ) ≤T 1 + γ ( q ) 1

E ( x; b, q, l )

⎛ x b log 2 ( xqT ) ⎞ +O⎜ ⎟ + O ( log ( x b ) log q ) , T ⎝ ⎠

2

(2.5.4)

and

∑ ∑

E ( x; b, q, l )

2 ≤ q ≤ D 1≤ b ≤ x 1 2

⎛ xD log x log 2 ( xDT ) ⎞ 1 2 x 2 log x log T + O ⎜ ⎟ + O ( D log x log D ) . T ⎝ ⎠ Let T = x log x , D = x log

−B

(2.5.5)

x , B ≥ A + 2 , then

∑ ∑

2 ≤ q ≤ D 1≤b ≤ x

1

E ( x; b, q, l )

x log − A x .

(2.5.6)

2

Theorem 5 is proved. Theorem 6: Any large even integer is sum of two primes, and we have

∑

log p > 4 ( 2 log 2 − log 3) C ( N )

p = N − p1

⎛ N ⎞ N + O⎜ ⎟, A log N ⎝ log N ⎠

(2.6.1)

where 2 < p < N . Proof: Let N is any large even integer, A = {a : a = N − p, 2 < p ≤ N } , then

∑

(

log p ≥ S A, x

p = N − p1

1

2

) ≥ S ( A, x ) − ∑ 1

1

ω (d ) =

and

θ ( N; d, N ) −

1

1

3

).

(2.6.2)

2

d , then φ (d )

( )=

W N

rd

1

N 3 ≤ p≤ N p2 p = N − p1

Let X = N , X ( p ) = N p , and

(

S A( p), x

3

3

6C ( N ) , log N

N φ (d )

E ( N; d, N ) ,

(2.6.3)

(2.6.4)

rd ( p )

θ ( N , p, d , N ) −

N p φ (d )

ψ ( N , p, d , N ) −

N p . φ (d )

(2.6.5)

By Theorems 5 and 6, we have

∑

d ≤ N log − B N

∑

rd

E ( N; d, N )

d ≤ N log − B N

N log − A N ,

(2.6.6)

and

∑

∑

∑

rd ( p )

d ≤ N log − B N N 13 ≤ p ≤ N 1 2

∑

d ≤ N log − B N N 13 ≤ p ≤ N 1 2

E ( N ; p, d , N )

N log − A N . (2.6.7)

By Lemma 7,

(

1

S A, N

(

S A( p), N

1

3

) ≥ 6e (1 + o (1) ) f ( 3) C ( N ) logNN + O ⎛⎜⎝ logN N ⎞⎟⎠ , −γ

A

3

⎛

N p + O⎜ ∑ ) ≤ 6e (1 + o (1) ) F ( 3) C ( N ) log ⎜ N −γ

⎝ d ≤ N log− B N

(2.6.8)

⎞ rd ( p ) ⎟ , (2.6.9) ⎟ ⎠

and

N

1

∑

(

S A( p), N

1 3 ≤ p≤ N 2

1

3

) ≤ 6e (1 + o (1) ) F ( 3) C ( N ) logNN ∑ −γ

N

1

1 3 ≤ p≤ N 2

⎛ N ⎞ 1 + O⎜ ⎟. A p ⎝ log N ⎠ (2.6.10)

Since

( ) ( ) 1

N

1

∑

1 3 ≤ p≤ N 2

log N 2 ⎛ 1 ⎞ 1 = log + O⎜ ⎟ 1 3 p log N ⎝ log N ⎠ ⎛ 1 ⎞ = log 3 − log 2 + O ⎜ ⎟, ⎝ log N ⎠

(2.6.11)

therefore

N

1

∑

1 3 ≤ p≤ N 2

(

S A( p), N

1

3

) ≤ 6e

−γ

( log 3 − log 2 ) F ( 3) C ( N )

⎛ N ⎞ N + O⎜ ⎟. A log N ⎝ log N ⎠

(2.6.12)

Since f ( 3) = 23 e log 2 , and F ( 3) = 23 e , we obtain γ

γ

∑

p = N − p1

(

log p ≥ S A, x

1

3

)− ∑ N

1

(

S A( p), x 1

1

3

)

3 ≤ p≤ N 2

≥ 4 ( 2 log 2 − log 3) C ( N )

⎛ N ⎞ N + O⎜ ⎟. A log N ⎝ log N ⎠

(2.6.13)

Since

∑

∑

log p > 0 ⇒

p = N − p1

1> 0,

(2.6.14)

N = p + p1

Theorem 6 is proved, and Goldbach Conjecture is true. Theorem 7: Let 2 < p < N , we have

∑

log p > 4 ( 2 log 2 − log 3) C ( N )

p = p1 + 2

⎛ N ⎞ N + O⎜ ⎟. A log N ⎝ log N ⎠

(2.7.1)

Proof: Let A = {a : a = p + 2, 2 < p ≤ N } , then

∑

(

log p ≥ S A, N

p = p1 + 2

1

2

) ≥ S ( A,N ) − ∑ 1

1

N 3 ≤ p≤ N p2 p = p1 + 2

Let X = N , X ( p ) = N p , and

(

S A( p), N

3

1

1

3

).

(2.7.2)

2

d , then φ (d )

ω (d ) =

rd = ψ ( N ; d , 2 ) −

N = E ( N ; d , 2) , φ (d )

(2.7.3)

and

rd ( p )

N

ψ ( N ; p, d , 2 ) −

pφ ( d )

= E ( N ; p, d , 2 ) .

(2.7.4)

By Theorems 5 and 6, we have

∑

d ≤ N log

−B

∑

rd N

d ≤ N log

−B

E ( N ; d , 2)

N log − A N ,

(2.7.5)

N

and

∑

d ≤ N log

−B

N N

1

∑

∑

rd ( p ) 1

3 ≤ p≤ N 2

d ≤ N log

−B

N N

1

∑

E ( N ; p, d , 2 )

N log − A N .

(2.7.6)

1

3 ≤ p≤ N 2

By Lemmas 7, we obtain

∑

log p ≥ 4 ( 2 log 2 − log 3) C ( N )

p = p1 + 2

⎛ N ⎞ N + O⎜ ⎟. A log N ⎝ log N ⎠

(2.7.7)

Theorem 7 is proved. In addition, we have

N →∞⇒

∑

p = p1 + 2

log p → ∞ ⇒

∑

1→ ∞ .

p = p1 + 2

That is to say, the number of twin primes is infinite, and Twin Prime Conjecture is true.

(2.7.8)

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