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  • 8/7/2019 A Proof of Goldbach Conjecture

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    A Proof of Goldbach Conjecture

    Jinzhu Han1

    Zaizhu Han2

    1. Baotou North Hospital, Inner Mongolia, China

    2. State Key Laboratory of Cognitive Neuroscience and Learning,

    Beijing Normal University, China

    Abstract

    In this paper, the Goldbach Conjecture1, 1 is proved by the complex variable

    integration. To prove the conjecture, a new function is introduced into Dirichlet series.

    And then, by using the Perron Formula of Dirichlet Series and the Residue Theorem, we

    conclude that any larger even integer can be expressed as the sum of two primes.

    Notationp prime number

    N positive integer

    p N p exactly divides N

    ( ), 1p N = coprime between p and N

    1, 1 the even integer as the sum of two primes

    a, b the even integer as the sum of product of at most a primes and product of at

    most b primes

    ( )n the Euler function

    s it= + complex variable

    ( )s ( )1

    s

    n

    s n

    =

    = , the Riemann zeta function

    ( )s

    ( )

    ( )

    ( )

    ss

    s

    =

    ( )n ( )log ,

    0,

    mp if n p

    notherwise

    = =

    , the Mangoldt function

    ( )x ( ) 1p x

    x

    =

    ( )x ( ) logp x

    x p

    =

    ( )x ( ) ( )n x

    x n

    =

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    ( )C N 22

    2

    1 1( ) 1

    ( 1) 2p Npp

    pC N

    p p>>

    =

    c the convergence abscissa of Dirichlet series

    a the absolute convergent abscissa of Dirichlet series

    ( )resf s the residue of function ( )f s

    ( )( )N f s the number of zero point of function ( )f s

    ( ),B O A B A= there exists a calculable positive constant c , such that B cA

    ( )1o the constant tending to zero

    the Euler constant, 0.577 =

    the sufficiently small positive constant

    c calculable positive constant

    1. Introduction

    The Goldbach Conjecture 1, 1 is one of the most famous still unresolved problems in

    mathematics. The conjecture states that every even integer beyond 2 may be the sum of two prime

    numbers (e.g., 12 = 5 + 7, 20 = 3 + 17). In over past two hundreds years, several relevant proofs

    on this issue have been conducted. Using the sieve method, some mathematicians verified the

    results including9, 9(Brun, 1920) [1], 1, c(Renyi, 1947) [2], 1, 5(Pan, 1962) [3],

    1, 4(Wang, 1962) [4], 1, 3(Richert, 1969) [5], 1, 2(Chen, 1973) [6], etc. In 1975,

    Montgomery and Vaughan made the progress on the exceptional set in Goldbachs problem by the

    circle method [7]. However,1, 1 has not been proven up to now. In fact, one can enumerate

    that 1, 1 is established for small values, but there is not enough evidence to demonstrate that

    1, 1 is also established for larger values. In this case, once we can prove that any larger even

    integer can be written as the sum of two primes, 1, 1 is entirely verified. In current paper, by

    applying the complex variable integration, we are to prove that any larger even integer can

    decompose into the sum of two primes.

    First, we define a new function

    ( )0, mod

    1,

    p Nif n N p

    notherwise

    =

    , (1.1)

    where ( ), 1p N = , { }3,5,7,p = is the prime sequence. Hence, this function is

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    provided with properties of sieve function.

    Based on the series: ( )s , ( )x , ( )x and ( )x , we obtain the new following

    series:

    ( ) ( )1

    , s

    n

    s n n

    =

    = , (1.2)

    ( ) ( ),p x

    x p

    = , (1.3)

    ( ) ( ), logp x

    x p p

    = , (1.4)

    ( ) ( ) ( ),n x

    x n n

    = , (1.5)

    and

    ( ) ( ) ( )1

    , s

    n

    s n n n

    =

    = . (1.6)

    In this case, if x N= , such that ( ), 0N > , then 1, 1 is true.

    By the Prime Number Theorem [8], we have

    ( ) ( ) ( ) 12

    , , log ,x

    x x x u u du = , (1.7)

    ( ) ( )( ) ( ) ( )11 2

    2, , log , log

    x

    x x x x u u du

    = + , (1.8)and

    ( ) ( ) ( )12, ,x x O x = + . (1.9)

    So we can estimate the value of function ( ),x by function ( ),x .

    In present paper, by using both the Perron Formula of Dirichlet Series and the Residue

    Theorem, we will verify the following theorems.

    Theorem 1.If N is any larger even integer, then

    ( )( ) ( )log

    ( , ) 4 1 1 ( ) elog

    c NNN e o C N O N N

    = + + . (1.10)

    Theorem 2.If N is any larger even integer, then

    ( ) ( )( ) ( ) ( )log2, 4 1 1 logc NN

    N e o C N O NeN

    = + + . (1.11)

    Remark: The propositions (1.10) and (1.11) are equivalent.

    By Theorem 2 we have ( ), 0N > , the Goldbach Conjecture (i.e., {1, 1}) is established.

    To prove Theorem 1 and Theorem 2, we need the following lemmas.

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    ( ) ( ), ,s s , (2.8)

    and

    ( ) ( ) ( ), ,s s s + = . (2.9)

    By the Rouch Theorem, we deduce

    ( ) ( )( , ) ( )N s N s = . (2.10)

    Since ( )it + has not zero point in the range ( )111 log 2c t + , then ( ),it +

    has not zero point in the same range ( )111 log 2c t + too.

    This proves Lemma 2.

    Lemma 3. For Dirichlet series ( ) ( ) ( )1

    ,s

    ns n n n

    = =

    , we have

    1c a = = . (2.11)

    Proof. By the Prime Number Theorem for Arithmetic Progressions, when x and

    ( ), 1q N = , we have

    ( )( )mod

    limx

    n N q

    n x

    xn

    q

    = (2.12)

    ( ) ( )( )

    ( ), 1

    1lim 1x

    n x p N

    p N

    n n xp

    =

    =

    ( ), 1

    11

    1p Np N

    xp

    =

    =

    ( )1

    4 1 log log

    xe C N O N N

    = + , (2.13)

    Hence

    ( ) ( )11

    lim log log 1x

    cx

    n

    x n n

    =

    = = . (2.14)

    Since ( ) 0n , 1c a = = .

    This proves Lemma 3.

    Lemma 4.At 1s = , ( ) ( ) ( )1

    , s

    n

    s n n n

    =

    = exists a pole, and the residue of function

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    ( ),s is that

    ( ) ( )1

    1 1, 4 1

    log logsres s e C N O

    N N

    =

    = +

    . (2.15)

    Proof. Because

    ( ) ( ) 11

    1

    lim 1 1s

    n

    s n n

    =

    = , (2.16)

    and

    ( )( )

    ( )1 11

    mod

    1limx

    n x n

    n N q

    n n n nq

    =

    = , (2.17)

    consequently

    ( ) ( )( )

    ( )

    ( )1 11

    , 1

    1lim 1x

    n x np N

    p N

    n n n n np

    =

    =

    =

    ( )

    ( ) 11

    , 1

    11

    1 np Np N

    n np

    =

    =

    =

    ( ) ( ) 11

    1 14 1

    log log ne C N O n n

    N N

    =

    = +

    . (2.18)

    Therefore

    ( ) ( ) ( ) ( ) 111

    1

    , lim 1ss

    n

    res s s n n n

    ==

    =

    ( ) ( ) ( ) 11

    1

    1 14 1 lim 1

    log log s ne C N O s n n

    N N

    =

    = +

    ( )1 1

    4 1log log

    e C N ON N

    = +

    . (2.19)

    Lemma 4 is proved.

    3. Proof of Theorem 1

    For Dirichlet series (1.6)

    ( ) ( ) ( )1

    , s

    n

    s n n n

    =

    = ,

    let ( )111 log 2a c T= + , 11 logb x= + , ( )

    11log logT x += ( )0 1< < , we have

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    ( ) logH u u , (3.1)

    and

    ( ) 2 logB u c x , (3.2)

    where2c is a positive constant.

    By (2.5) in Lemma 1, we have

    ( ) ( ) ( ),n x

    x n n

    =

    ( ) 11

    ,2

    sb iT

    b iT

    xs ds R

    i s

    +

    = + , (3.3)

    where

    22

    1

    loglog ,0 1

    x xR x x

    T

    < , ( ) ( )2

    1

    logs

    n

    n n it t

    =

    = + (see [8]).

    By Lemma 2 and Lemma 4, ( ),s has not zero point in closed contour . So function

    ( ),

    sx

    ss

    only exists a pole at 1s = , we obtain

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    ( ) ( )1

    1, ,

    2

    s s

    s

    x xres s s ds

    s i s

    = =

    14 ( ) 1

    log log

    xe C N O

    N N

    = +

    . (3.7)

    By (3.5) and (3.7), we have

    1 2

    1( , ) 4 ( ) 1

    log log

    xx e C N O R R

    N N

    = + + +

    . (3.8)

    Thus, combining (3.8) with (3.4) and (3.6) we have

    ( )log1( , ) 4 ( ) 1log log

    c xxx e C N O O xe

    N N

    = + +

    . (3.9)

    Let x N= , and N is any larger number, we then obtain (1.10), namely,

    ( )( ) ( )log( , ) 4 1 1 ( ) elog

    c NNN e o C N O N

    N

    = + + .

    This completely proves Theorem 1.

    4. Proof of Theorem 2

    Combining (3.9) with (1.7), (1.8) and (1.9) we have

    ( ) ( ) ( )( )

    3, 4 1 1log log

    xx e C N o R

    N x

    = + + (4.1)

    where

    ( ) ( )log log3 2 22 2

    ,

    log log log

    c xx x

    c xu uxeR du du xe

    u u x u u

    + + . (4.2)

    By the Prime Number Theorem

    ( ) ( )logc xx x O xe = + , (4.3)we have

    ( ) log log2 2 22 2 2log log log

    c ux x x

    c xu u uedu du du xe

    u u u u u u

    + . (4.4)

    By (4.1), (4.2) and (4.4), we have

    ( ) ( ) ( )( ) ( )log, 4 1 1log log

    c xxx e C N o O xe

    N x

    = + + . (4.5)

    Let x N= , and N is any larger number, we then obtain (1.11), i.e.,

    ( ) ( )( ) ( )( )

    log

    2, 4 1 1

    log

    c NNN e o C N O Ne

    N

    = + + .

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    Theorem 2 is proved. Therefore, we have ( ), 0N > . Thus, the Goldbach Conjecture

    1, 1 is established.

    References

    [1] V. Brun, De cribe d'Eratosth6ne et le th6orhme de Goldbach, Videnskabs-selskabt: Kristiania

    Skrifter I, Mate-Naturvidenskapelig Klasse. 3 (1920) 136.

    [2] A. Renyi. On the representation of an even number as the sum of a single prime and a single

    almost-prime number. Dokl. Akad. Nauk SSSR. 56 (1947), 455458.

    [3] C.T. Pan, On the representation of an even number as the sum of a prime and of an almost

    prime, Acta Mathematica Sinica. 12 (1962) 95106.

    [4] Y. Wang, On the representation of a large integer as the sum of a prime and of an almost prime,

    Sci. Sin. 11 (1962) 10331054.

    [5] H.E. Richert, Selbergs sieve with weights, Mathemetika. 16 (1969) 122.

    [6] J.R. Chen, On the representation of a large even integer as the sum of a prime and the product

    of at most two primes, Sci. Sin. 17 (1973) 157176.

    [7] H.L. Montgomery, R.C. Vaughan, The exceptional set in Goldbachs problem, Acta. Arith. 27

    (1975) 353370.

    [8] C.D. Pan, C.B. Pan, Basic Analytic Number Theory, Science Press, Beijing, 1999.