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Goldbach conjecture

  1. Sep 21, 2003 #1

    Goldbach conjecture: ”Each even number greater than six, could be written as sum of two primes”
    This conjecture is almost experimentally proven on computer , but we still miss theoretical proof. I’ll try to reach it by the method showed below.
    Let define set S: the S is set of primes a1,a2,a3,a4,a5…an ,such that a1<a2<a3<a4<a5<…<an. Now we need interval of numbers L (0,a1*a2*a3*a4*a5*…*an). We have to prove that in interval L exists one more prime number a_(n+1):
    If we take number m=a1*a2*a3*a4*a5…*an, and increase it for 1,we got the number
    n=m+1=a1*a2*a3*a4*a5…an+1.Number n is odd and it’s obvious that it cannot be combination of the elements of the set S.
    a2* ((a2)^(p-1)(a3)^q(a4)^r(a5)^s(a6)^t-a1a3a4a5a6a7)=1
    but it’s obvious that difference between is >=1 so we have the proof that in interval L has to be one more prime at least to form the number n,so we can write that:
    a1*a2*a3*a4*a5*…*a_(n-1)>a_n ........................................(i)
    That means
    so the inequality i could be wriiten as
    if we approximate 6 with 2^3
    if we approximate 3 with 2^2
    we go that
    according to
    2^k>k^2 for k>5
    the largest value an can have to satisfy inequality (D) is
    because (D) is satisfied for all k
    so that means
    and again repeat (U)
    and it is easy to prove that
    Proof is the same as to prove (A):
    t=a3*a4*a5*a6*a7+1 isn't divisible by the elements of the set B=
    (a3,a4,a5,a6,a7) because if it is divisible then t could be written as
    a3* ((a3)^(q-1)(a4)^r(a5)^s(a6)^t-a4a5a6a7)=1
    it's obvious that expression in the brackets is higher or equal 1.
    end of proof.
    But that doesn't mean that t isn't divisible by 6.
    And it's obvious that the smallest number which is divisible by set B
    is a3*a4*a5*a6*a7+a3 because
    a3* ((a3)^(q-1)(a4)^r(a5)^s(a6)^t-a4a5a6a7)=a3
    which is possible
    odd number isn't divisible by 2
    so the number
    t+1(or +3) isn't divisible by 2
    because number t-1 is odd.
    So if t+1 is divisible by 3 then it could be written as 3^k and t+3
    is 3^k+2 and it cannot be divisible by 3 and as it is odd it cannot
    been divisible by 2 so a3*a4*a5*a6>a7
    or for all n
    End of proof.
    according to (G)
    to simplify inequality we it is allowed to say:
    We have to determine number of numbers that can be sum of k pairs
    of different primes. For example 24=13+11=17+7=23+1.the number of
    that combinations is n*(n+1)/2
    when we sub it from n^2 we get that of n primes could be at least
    formed n(n-1)/2
    now let sub of an, n (because there are n prime numbers in
    interval a_n) and then divide it by 2 to eliminate odd numbers and
    using (Z)
    where (a_n-n)/2 is actually number od even numbers in interval
    (0,an] so the GoldBach conjecture is proven.

    Attached Files:

  2. jcsd
  3. Sep 25, 2003 #2


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    I haven't read through your proof (I will) but I notice that you haven't even stated Goldbach's conjecture correctly: there is no need to say "greater than six": 4= 2+2 and 6= 3+3.

    The conjecture is: "Every even number larger than 2 can be written as the sum of two primes"
  4. Sep 25, 2003 #3
    I have stated original statement of GoldBach in 1700.I know that it stands for even number greater than two,nowdays.
    Thanks and best wishes
  5. Sep 25, 2003 #4


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    You mean Goldbach didn't KNOW that 3+ 3= 6??
  6. Sep 25, 2003 #5


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    When you say "2^k>k^2 for k>5" and then:"2^(2n)>an" so
    "(2n)^2>an, (4) 2^2>an, n^2>an/4 " You have the inequality going the wrong way: it doesn't follow.
  7. Sep 25, 2003 #6
    Last edited: Sep 25, 2003
  8. Sep 26, 2003 #7


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    If you are posting this with the address of Zhejiang Ocean University, have you discussed it with mathematics faculty there?
  9. Sep 26, 2003 #8

    It is an arXiv paper.

  10. Sep 26, 2003 #9
    The Goldbuch's conjecture right as follow don't worse the time so as it is hard so

    Axiom 1: every natural number factorization is sole.
    No same factors theorem: between every both odd number from 2n to 2n- square root(2n), it is not same factorization.
    The certification summary as follow:
    To consider both odd numbers A1, A2. Its all primary factors are same.
    As this reason: A1+p1=2n; A2+p2=2n; p1 The one surplus number of p1 and p2 corresponding the A1, A2 common factor, should be large than the A1, A2 common factor.
    It is imposible.
    As the next condition axiom1,
    the no same factors theorem is right.

    The reasoning result 1: every even number, it is equal to both primary number add, one primary of the both less than the even number square root.
    The right key as follow:
    2n, the even number,
    From 2n to 2n - square root 2n, every both odd number factor group is different.
    As 2 is beside the add, according to the primary number serious from 1 to 2n square root, there is a odd number according to the factor 1 also, that is right.
    There is a both add primary numbers to equal to the 2n, it is also right.
    The Goldbuch's conjecture ir right.

    [The ¸çµÂ°ÍºÕ suspect is right.

    ×öΪÎÞÏàͬÒò×Ó¶¨ÀíµÄÒ»¸öÍÆÂÛ£¬Ò»¸ö´óżÊý 2n ¿ÉÖÁÉÙ·Ö½âΪ int£¨(2nµÄƽ·½¸ù£©/2£©¶ÔËØÊýÖ®ºÍ¡£ ]

    Hey . Like
    Last edited: Jan 6, 2004
  11. Sep 27, 2003 #10
    Sorry,butthe last line cannot be read!Could you attach it in doc or pdf,please.
    Best wishes SVRadic
  12. Sep 27, 2003 #11
    It's obvious that 2^(2^n)>a1*a2*a3*a4...*a_(n-1) (1),and if we use INEQUALITY
    2^k>k^2 which is correct for k>5,aand inequality (1) is correct for all n,so we may write 2^(2n)>=a1*a2*a3*a4...*a_(n-1),but 2^(2n)>an,so
  13. Sep 28, 2003 #12


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    To Organic: Yes, I could see your paper was on the "ArXiv" website. My question was not about the website. Most universities are very careful about what papers bear their address since that implies (perhaps incorrectly) that they are representative of the university. My question was: If you are posting this with the address of Zhejiang Ocean University, have you discussed it with mathematics faculty there?

    More to the point, do you in fact have permission of the University to use their address on a paper?
  14. Sep 28, 2003 #13


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    To Radic: I've looked over your paper and have a few comments.

    You start out by giving a complicated proof, involving factorization into primes to show that the product of primes,plus one, cannot have any of those primes as a factor. I don't know why you did that, it is sufficient to argue (as Euclid did 2000 years ago) that, if you divide a1a2...an+ 1 by any of those primes, you get a remainder of 1. You conclusion, that there must be another prime, less than a1a2...an+1 that is a factor is not quite correct: it might be a prime itself.

    You then go on to argue that the product of primes a1, a2,..., an must be larger than an+1. I'm not completely convince by your argument, but, in fact, there is no reason to think that is not true.

    (By the way, you define "a1,a2,...,an as "Set of primes such that a12< a2<...< an. From what you do later, it's clear that you intend that a1= 2, a2= 3,... In other words that the set is the set of the first n primes. If that is what you mean, you should say so. The way you define the set it could be {3, 17, 37}. That is a set of primes such that 3< 17< 37.)

    You arrive, finally, at the conclusion that n2> an. Okay, that seems reasonable: 22= 4> 3, 32= 9> 5, 42= 16> 7 and the difference is getting larger. I'm willing to accept it.

    You then assert "We have todetermine number of numbers that can be sum of k pairs of different primes. For example 24= 13+ 11= 17+ 7= 23+1. The number of that combinations is n(n+1)/2." This where I got lost.

    First, a minor point: your example is incorrect. 23+1 is not a sum of primes because 1 is not a prime. However, you did not include 19+ 5 so your you could replace 23+1 with that: 24 can be written as a sum of two primes in 3 different ways.
    However, my first reaction on reading that was "What is n? What happened to k?" You talked about the "number of numbers that can be sum of k pairs of different primes" and from your example I thought you meant "numbers that can be written as a sum of two primes in exactly k different ways". Even if we assume n was accidently typed for k, that formula doesn't work. In fact, I don't see any reason not to think that the "number of integers that can be written as the sum of two primes in exactly 3 different ways" isn't infinite.

    In any case, I THINK what you actually meant was number of diffent possible combinations of the first n prime numbers, taken 2 at a time, and not considering permutations (so that we do not count both 17+ 7 and 7+ 17), is n(n-1)/2. That's correct. (But does not count such possiblities as 17+17 or 7+ 7.)

    Then you go on to say: "when we sub it from n2 we get that of n primes could be at least n(n-1)/2 different even numbers.

    Do you mean "subtract it from ... ". Where did you get the n2. How did subtracting n(n-1)/2 from n2 give you information about n(n-1)/2? (n2- n(n-1)/2= n(n+1)/2.)
    In any case, it is clear that "from n primes" we CANNOT form
    n(n-1)/2 DIFFERENT even numbers. That is the figure for all possible sums of two distinct prime numbers of the first n. As your example showed they are not all different. Surely the correct figure is much smaller than n(n-1)/2.

    "now let sub of an,n (because there are in interval an and divide it by 2 to eliminate odd numbers"

    We may have a language problem here. I have absolutely no idea what you mean by "let sub of an,n". Is "sub" here "subtract"? Are you subtracting n from an or exactly what?

    I think, from what you say following, that you are subtracting n (the number of primes less than or equal to an) from an to determine how many "non-primes" there are less than an. You then divide by 2 "to eliminate odd numbers". Excuse me? Are you assuming that of the numbers left exactly half are even and half are odd? How can that be when you have just removed a fair number of ODD numbers from the set?

    For example, if n= 7, a7= 17. There are, of course, 17 numbers less than that and 7 of them are prime. 17- 7= 10 of them are composite. They are: 1,4, 6, 8, 9, 10, 12, 14, 15, 16.
    Exactly 7 of those are even (4, 6, 8, 10, 12, 14, 16) and only 3 (1, 9, 15) are odd.

    You end by declaring that the total number of "different even numbers" that can be formed by adding two primes less than or equal to an is n(n-1)/2 while the total number of even numbers less than an is (an-n)/2 and
    n(n-1)/2> (an-n)/2. Finally you assert that since
    n(n-1)/2 > (an-n)/2, Goldbach's conjecture follows.

    Both of those numbers are incorrect. Since n(n-1)/2 counts different pairs of primes that give the same sum, it is too large.
    Since (an-n)/2 assumes that the number of even and odd numbers in the composites less than an are the same (and they are not), it is too small.

    By the way, if your assertions WERE true, those two numbers should have been exactly the same. Why did you think that
    n(n-1)/2 should be greater than (an-n)/2 in order for Goldbach's conjecture to be true?
  15. Sep 28, 2003 #14
    Hi, I am intrigued at how this is almost proven on a computer? Since you have to test all the even integers to infinity, I am assuming you guys are almost at infinity?

    As for the last line of hey. like's post I have a screen shot...
  16. Sep 29, 2003 #15
    To HallsofIvy

    let use interval [0,24],and number 24.
    We ca represent each even number as y=2*x where x is some positive integer,y is even positive integer and y>x.So as
    y=2x=x+x=x+m+x-m,where m is such that x+m and x-m are primes.So the number of m ,called w where max(w)=n/2 (n is number of primes in interval [0,24]) defines number of two integer combinations that give the same number (in this case y).I call numbers x+m and x-m symmetric primes of x.So if in interval exists number such that w=n/2,other numbers could have maximaly (n-1)/2,(n-2)/2....1,or if we sum it then maximal number of combination that gives same numbers are n(n+1)/4.
    Now about where comes n^2.There are n(n-1)/2 combinations of two different prime numbers,but there is 2+2,3+3,5+5 etc.. so u have more n combinations,and when you add it to n(n-1)/2 you'll get
    (n^2+n)/2 which is actually number of combinations of two primes that gives even number.Now to get combinations in which pairs of primes gives the different even number,we have to do:
    (4n^2+4n-n^2-n)/4=3(n^2+n)/4.....(A) combinations.
    By the way in interval [0,a_n] there is a_n/2 even numbers and as
    n^2>a_n then
    and as
    3(n^2+n)/4 - n^2/2=(n^2+3n)/4>0 or
    that means that number of combination of two primes which pairs give different even number(I meant on sum of two primes),equation (A) is
    greater then number of even numbers in interval [0,a_n],or in other words for (ALL y=2k where k ELEMENT N)(EXISTS two primes) SUCH THAT
  17. Sep 30, 2003 #16
  18. Oct 1, 2003 #17
    To Organic:

    I think that this address already has been sent onto the topic.
  19. Oct 2, 2003 #18
    Hi Radic,

    Do you have any remarks on Shi Kaida's paper ?

    Last edited: Oct 2, 2003
  20. Oct 26, 2003 #19
    This topic isn't still close,or I conviced you about the proof?
  21. Dec 2, 2003 #20
    I'm interested in the Goldbach conjecture but your notation is difficult to follow. I can't tell, for example, whether 4n^2+4n-n^2-n
    means 4 times n to the power 2 plus 4 times n minus n to the power 2, minus n or n to the power (2-n). I'd also appreciate a more explanator treatment of your proof so I can follow it better.
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