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Thread: The Goldbach Conjecture

  1. #1 The Goldbach Conjecture 
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    The Goldbach Conjecture states that every even integer greater than 4 is the sum of two odd prime numbers.

    In other versions, I have seen the word odd excluded, and sometimes the number 4 itself has undergone changes.

    What exactly is the correct version of the Goldbach Conjecture? Having checked Wikipedia, I have become confused as to what the modern version of the conjecture is.


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    "In mathematics, a prime number (or a prime) is a natural number which has exactly two distinct natural number divisors: 1 and itself. An infinitude of prime numbers exists, as demonstrated by Euclid around 300 BC. The first twenty-five prime numbers are:
    2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97" - Wikipedia

    1 and 4 are the first two non-prime numbers, but only 4 is even and non-prime. All prime numbers are odd except for 2 (AFAIK), so that might be where the problem comes in. 4 is the sum of 2 + 2, which are even, so it would be wrong to include the odd bit in the definition then.

    I don't know what the definition is exactly, but as far as I can see it can only be written in one of these ways:

    The Goldbach Conjecture states that every even integer greater than 4 is the sum of two odd prime numbers.

    The Goldbach Conjecture states that every even integer greater than and including 4 is the sum of two prime numbers.

    Of course, I know virtually no maths at all, so I could be talking bollocks. :wink:


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  4. #3  
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    The Goldbach Conjecture states that every even integer greater than 4 is the sum of two odd prime numbers.

    The Goldbach Conjecture states that every even integer greater than and including 4 is the sum of two prime numbers.
    Are these two conjectures equivalent, however? If we were to prove any one, would it be correct to say that the other would be proved as well?

    Looking at the two statements you've given me, I would say no. But I'd like your opinion.

    Also, if we could prove that the sum of any two odd primes is equivalent to the sum of any two even numbers and then show that all even numbers are the sum of two even numbers, can we then say that all even numbers are the sum of two odd primes (including one, of course)?

    Of course, I know virtually no maths at all, so I could be talking bollocks.
    Hey, at the very least, thanks for trying.
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  5. #4  
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    Actually, the wiki article seems quite clear: Every even number greater than 2 can be expressed as the sum of two primes.

    All the other forumulations are equivalent to this:

    1. Apart from 2, there are no even prime numbers, so stating 'odd' is unnecessary - they both have to be odd to sum to an even number.

    2. ll even numbers are integers, so the owrds integer and number are interchangeable in this context.

    3. The first even number greater than 2 is 4, so that makes no difference - you can state it in terms of >2 or >=4.

    All versions are correct, and mathematicians simply state the equivalent that is most useful at the time for them.
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  6. #5  
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    Thank you, sunshinewarrior. My real difficulty was trying to pinpoint the one considered the Goldbach conjecture in modern times, and I was, to say the least, confused between the different versions of the conjecture.

    Once again, thank you. However, I have asked another question in my second post on this thread, regarding a possible way to answer it. I have no intention of burdening you, but could you possibly see your way to answering if it is a valid method?
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  7. #6  
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    Quote Originally Posted by Liongold
    Also, if we could prove that the sum of any two odd primes is equivalent to the sum of any two even numbers and then show that all even numbers are the sum of two even numbers, can we then say that all even numbers are the sum of two odd primes (including one, of course)?

    Of course, I know virtually no maths at all, so I could be talking bollocks.
    All even numbers bar 2 are by definition the sum of two even numbers. They are also, by definition, the sum of two odd numbers as well.

    The key problem is trying to get a combination of odd numbers, each of which is prime.

    If you consider the sequence of prime numbers, starting from 2 upwards, you will notice that it is, in its specifics, unpredictable. Given a particular prime number 'p', you cannot tell how many integers are between it and the next prime number on the sequence.

    Gauss and others did great work on primes, but a lot of it was statistical - figuring out on average the number of primes less than a given integer, for instance.

    Now if you took any tow primes (apart from 2), their sum must necessarily be an even number, since those primes will be odd numbers themselves. If their sum is an even number then it will necessarily also be the sum of two even numbers.

    Unfortunately, this will not prove anything, because the proof you want works in the opposite direction: you want to show that these sums of two primes cover each and every even number greater than 2. And you cannot simply keep adding primes, as you know - because for a proof to be rigorous, it must be demonstrated to hold good for every conceivable even number (that is, it must go to infinity) and we cannot count that high. So it has to be a formal proof.

    If you have such an idea, great. But as stated, I think your current idea needs work. I hope you can see why.

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  8. #7  
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    Quote Originally Posted by Liongold
    Thank you, sunshinewarrior. My real difficulty was trying to pinpoint the one considered the Goldbach conjecture in modern times, and I was, to say the least, confused between the different versions of the conjecture.

    Once again, thank you. However, I have asked another question in my second post on this thread, regarding a possible way to answer it. I have no intention of burdening you, but could you possibly see your way to answering if it is a valid method?
    You won't solve the Goldbach conjecture that way. It is trivial to show that any even number greater than 2 is itself the sum of two even numbers. You do that as follows. 4 = 2+2. If N is even and N>4 then N = 2 + (N-2).

    But the statement that "that the sum of any two odd primes is equivalent to the sum of any two even numbers" doesn't seem to make sense since it requires that given any two odd primes p and q that p+q=m+n for any two even numbers m and n. If instead you meant to say that "the sum of any two even numbers is the sum of two odd primes" then the proof of that statement would indeed prove the Goldbach conjecture. However, since any even number 4 or greater is, as noted above trivially the sum of two even numbers, this is not a real simplification so you are simply back to the original, very hard to prove Goldbach conjecture.
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    But the statement that "that the sum of any two odd primes is equivalent to the sum of any two even numbers" doesn't seem to make sense since it requires that given any two odd primes p and q that p+q=m+n for any two even numbers m and n.
    Well, it is possible to do so.

    Firstly, take any two odd primes.

    We know that the sum of any two odd numbers is always even, and that all primes barring two are odd. This implies that the sum of any two odd primes is always equal to an even number.

    Wouldn't this mean that p+q = an even number = m+n?

    I value your opinion, so could you please tell me if I have made a istake in my reasoning so far? Thank you.

    However, since any even number 4 or greater is, as noted above trivially the sum of two even numbers, this is not a real simplification so you are simply back to the original, very hard to prove Goldbach conjecture.
    Thank you very much, DrRocket.

    If you have such an idea, great. But as stated, I think your current idea needs work. I hope you can see why.
    Thank you, sunshinewarrior, or would you prefer to be called shank? Anyway, thank you.
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  10. #9  
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    Quote Originally Posted by Liongold
    Wouldn't this mean that p+q = an even number = m+n?
    Yes. But...

    The two are different things. What you have for now is the simple result that the sum of two primes will be equal to some combination of even numbers, but that there is always such a combination does not mean that the set of all these combinations is equal to the set of all possible combinations. It is the latter that you have to prove, and which no mathematician has been able to do so far.


    Quote Originally Posted by Liongold
    Thank you, sunshinewarrior, or would you prefer to be called shank? Anyway, thank you.
    shanks is simpler, and preferable I suppose; I simply made the mistake of registering myself here as sunshine warrior (another nick), and since I detest multiple registrations I let it be...
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  11. #10  
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    Yes. But...

    The two are different things. What you have for now is the simple result that the sum of two primes will be equal to some combination of even numbers, but that there is always such a combination does not mean that the set of all these combinations is equal to the set of all possible combinations. It is the latter that you have to prove, and which no mathematician has been able to do so far.
    Ah, now I understand. Thanks, shanks.
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  12. #11  
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    Quote Originally Posted by Liongold
    But the statement that "that the sum of any two odd primes is equivalent to the sum of any two even numbers" doesn't seem to make sense since it requires that given any two odd primes p and q that p+q=m+n for any two even numbers m and n.
    Well, it is possible to do so.

    Firstly, take any two odd primes.

    We know that the sum of any two odd numbers is always even, and that all primes barring two are odd. This implies that the sum of any two odd primes is always equal to an even number.

    Wouldn't this mean that p+q = an even number = m+n?

    I value your opinion, so could you please tell me if I have made a istake in my reasoning so far? Thank you.
    Your problem is with the quantifiers. If you take two odd primes, then indeed the sum is even. And any even number 4 or greater is the sum of two even numbers, which is nothing more than N = 2 + (N-2). So you can say that the sum of ANY two odd primes greater than 1 can be written as the sum of two even numbers. But you cannot say tha the sum of two odd primes is the sum of ANY two even numbers. For instance 3+5 is not 6 + 12. Another way to say this is that given two odd primes greater than 1, p and q there exist two positive even numbers m and n such that p+q=m+n.
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  13. #12  
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    Thank you, DrRocket. I see I'll have to revise my method so far. Thank all of you yet again.
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  14. #13 Re: The Goldbach Conjecture 
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    Quote Originally Posted by Liongold
    The Goldbach Conjecture states that every even integer greater than 4 is the sum of two odd prime numbers.

    In other versions, I have seen the word odd excluded, and sometimes the number 4 itself has undergone changes.

    What exactly is the correct version of the Goldbach Conjecture? Having checked Wikipedia, I have become confused as to what the modern version of the conjecture is.
    This stated it nicely with an example.

    http://en.wikipedia.org/wiki/Goldbach%27s_conjecture

    They say any even integer over two, "can be stated as the sum of two prime numbers". They do not have to be the same numbers.

    That must be wild as the numbers get larger. Could make a cool computer code if you numbered the primes in an index. Ha-ha.





    Sincerely,


    William McCormick
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  15. #14 Re: The Goldbach Conjecture 
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    Quote Originally Posted by William McCormick
    if you numbered the primes in an index.
    This suggests you think the primes are countable. They might be; can you prove it?
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    Not holding my breath here!

    Anyway, this discussion got me thinking back to a thm I once had to prove and found hard (though it is actually elementary):

    The ring of integers modulo n is a field iff n is prime

    The proof I offered was said to be "not exactly wrong, but extremely inelegant". See if you guys can do better; I'll start us off....

    Suppose we denote by the set of integers exactly divisible by some . Let us further use the notation for the quotient set as , this being the ring of integers mod .

    Note that is an ideal in . This fact is needed, so we need to show this is so.

    Note also that is the additive identity for . This fact is also needed, so we need to show this too.

    Now - since a field differs from a ring by having a multiplicative inverse for every element, it will suffice to show that the thm as stated will be true when has a multiplicative inverse for every element only when is prime.

    See if your proof is more "elegant" than mine.
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  17. #16  
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    Quote Originally Posted by Guitarist
    Not holding my breath here!

    Anyway, this discussion got me thinking back to a thm I once had to prove and found hard (though it is actually elementary):

    The ring of integers modulo n is a field iff n is prime

    The proof I offered was said to be "not exactly wrong, but extremely inelegant". See if you guys can do better; I'll start us off....

    Suppose we denote by the set of integers exactly divisible by some . Let us further use the notation for the quotient set as , this being the ring of integers mod .

    Note that is an ideal in . This fact is needed, so we need to show this is so.

    Note also that is the additive identity for . This fact is also needed, so we need to show this too.

    Now - since a field differs from a ring by having a multiplicative inverse for every element, it will suffice to show that the thm as stated will be true when has a multiplicative inverse for every element only when is prime.

    See if your proof is more "elegant" than mine.
    If n is not prime then n=mk for some m and some k but mk=0 mod n whence m and k are divisors of zero and Zn is therefore not a field.

    Suppose that n is prime and m<n, m not 0. Consider all elements of Zn of the form in Zn this is finite, and since n is prime is is never 0 mod n by the unique factorization therem. So there is a k and l<k so that then

    This seems to be a bit off topic in a thread aabout the Goldbach conjecture unless I am missing something.
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  18. #17  
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    Quote Originally Posted by DrRocket
    Suppose that n is prime and m<n, m not 0. Consider all elements of Zn of the form in Zn this is finite, and since n is prime is is never 0 mod n by the unitque factorization therem. So there is a k and l<k so that them
    You could just consider modulo . For any ,



    ________ (since and are coprime)

    ________ (since and is prime)

    ________

    Hence the list modulo is a permutation of modulo ; therefore such that .
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  19. #18 Re: The Goldbach Conjecture 
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    Quote Originally Posted by Guitarist
    Quote Originally Posted by William McCormick
    if you numbered the primes in an index.
    This suggests you think the primes are countable. They might be; can you prove it?
    I do not know that you can create a formula to deduce where they will or will not occur in the future. There are patterns however they seem random, or appear to alter, and I believe there will be exceptions as the numbers get higher.

    But you could easily enough setup a computer routine to kick out every prime number from zero to any usable number rather easily and quickly. By actually testing the number by all existing numbers below it. Creating a list of numbers that only create remainders or decimal points. And just number them or code them into a table.

    I don't know that it would be useful though. It would probably be more effort then it would be worth. And I know there are codes that are very brief already. Using a similar index table.


    Sincerely,


    William McCormick
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  20. #19 Re: The Goldbach Conjecture 
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    Quote Originally Posted by William McCormick
    Quote Originally Posted by Guitarist
    Quote Originally Posted by William McCormick
    if you numbered the primes in an index.
    This suggests you think the primes are countable. They might be; can you prove it?
    I do not know that you can create a formula to deduce where they will or will not occur in the future. There are patterns however they seem random, or appear to alter, and I believe there will be exceptions as the numbers get higher.

    But you could easily enough setup a computer routine to kick out every prime number from zero to any usable number rather easily and quickly. By actually testing the number by all existing numbers below it. Creating a list of numbers that only create remainders or decimal points. And just number them or code them into a table.

    I don't know that it would be useful though. It would probably be more effort then it would be worth. And I know there are codes that are very brief already. Using a similar index table.


    Sincerely,


    William McCormick
    Actually picking out prime numbers is not that easy, and the difficulty in factoring large numbers, even with computers, and determining if they are prime is basis of a great many codes.
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  21. #20  
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    Quote Originally Posted by DrRocket
    This seems to be a bit off topic in a thread aabout the Goldbach conjecture unless I am missing something.
    Yes it is. But I doubt the G. conjecture is likely to be solved via chat-room exchanges, but I could easily be wrong.

    But...

    I was getting a little depressed about the low volume of traffic in this sub (why depressed? I ask myself; no answer), so I thought I would try and kick-start using something vaguely relevant, rather than launching into yet one more of my interminable monologues,

    But then again - if there are no members here who want to chew the mathematical fat just for the fun of it, then, hey, wtf!
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  22. #21  
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    The Goldbach conjecture is pretty difficult to prove indeed... ...but always up for a challenge (although not having the capability to chew or digest any of Guitarist's mathematical fat :wink.

    Was wondering whether there could be a clue in the intrinsic structure of the primes that are used in the sums `to get even´. Started with the following simple premise:

    All primes > 3 can be expressed in sums of smaller primes by .

    Take f.i. the sum of two random primes:

    7+11 = (5+2) + (7+2+2) = ((3+2)+2) + ((5+2)+2+2) = ((3+2)+2)+(((3+2)+2)+2+2) = 6 + 6*2 = 18

    Pretty straight forward results of course, but this last element in the formula could be an indication that all even numbers between 6 and 18 could be created by summing the prime numbers encountered in the process. Even when a constraint is added:

    with and

    Example:
    06 = 3+3
    08 = 5+3
    10 = 5+5
    12 = 7+5
    14 = 7+7
    16 = 5+11
    18 = 7+11

    It seems to work. So let's test it on another interval and now without counting down all the way to 3, e.g.:

    13+23 = (11+2) + (19+2+2) = 30+3*2 = 36

    30 = 11+19
    32 = 13+19
    34 = 11+23
    36 = 13+23

    or:

    43+61 = (41+2) + (59+2) = 100+2*2 = 104

    100 = 41+59
    102 = 43+59
    102 = 41+61
    104 = 43+61

    or:

    97+13 = (89+ 4*2) + (11+2) = 100 + 5*2 = 110

    100 = 89+11
    102 = 89+13
    104 = ?
    106 = ?
    108 = 97+11
    110 = 97+13

    Here it fails since I probably have more 2's (and even numbers in the interval) than available primes. How to get rid of the 2's and reduce the interval? This can only be done by subtracting 2's from the next prime that should be closer by than the previous one.

    97+13 = (89+ 4*2) + (17- 2*2) = 106 + 2*2 = 110

    102 = 89+13
    106 = 89+17
    110 = 97+13
    114 = 97+17

    But then the even numbers start to hop by 4 rather than 2.

    Fail sofar...

    Maybe a proof should start with:

    Suppose there exists a set of even numbers on the interval [n,n+k] that can not be created by the sum of two primes, then this would imply that...

    smart text that therefore no primes are allowed to occur in ...

    ...and hopefully a strong contradiction somewhere around here. 8)
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  23. #22  
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    Quote Originally Posted by Guitarist
    But then again - if there are no members here who want to chew the mathematical fat just for the fun of it, then, hey, wtf!
    I guess we could all wait for William McCormick to produce the proof that you requested. And wait. And wait. ................zzzzzzzzzzz
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  24. #23 Re: The Goldbach Conjecture 
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    Quote Originally Posted by DrRocket

    Actually picking out prime numbers is not that easy, and the difficulty in factoring large numbers, even with computers, and determining if they are prime is basis of a great many codes.
    You could knock out all even numbers. You could knock out all numbers ending in five. Then just run what is left, by dividing every number half as large as themselves into themselves.

    That would not take long. It could be setup in assembly language to run like a program that calculates square roots.

    But to be honest by the time you do all that the overhead is going to be the same as or more.

    Because it will not even remove one decimal place in many cases and add one in others.

    Sincerely,


    William McCormick
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  25. #24  
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    Quote Originally Posted by accountabled
    But then the even numbers start to hop by 4 rather than 2.

    Fail so far...
    Hmm... recall that if is prime, then so is , also if is even, then so is . So no failure then, right? (you might like to show the truth of these two assertions)

    But the point is, you could keep going as you did until hell froze over, and still not have a proof. Who is to say that the (hell frozen + 1)th case won't break the pattern?

    Quote Originally Posted by DrRocket
    I guess we could all wait for William McCormick to produce the proof that you requested.
    Well, the problem was rather difficult (wink), don't be too hard on him!
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    Another attempt:

    Suppose Levy's conjecture is true i.e.

    Any odd number (i.e. prime and non-prime) has the shape

    e.g.
    7=2*2+3
    9=2*3+3
    11=2*3+5
    13=2*5+3
    15=2*5+5
    17=2*5+7
    ...

    Since it's pretty easy to proof that all even numbers () can be created by the sum of two odd numbers, it also means that based on the premise every even number can be expressed as the sum of

    That's still 4 prime numbers in the sum and we need only two for the Goldbach conjecture.

    Of course it's also true that if:



    and:



    then:



    but also that there must be a and a :



    so,





    so for all :



    (pretty obvious result of course, but just wanted to go through the logic).

    Back to two primes (if Levy is correct).
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    Quote Originally Posted by accountabled
    Another attempt:

    Suppose Levy's conjecture is true i.e.

    Any odd number (i.e. prime and non-prime) has the shape

    e.g.
    7=2*2+3
    9=2*3+3
    11=2*3+5
    13=2*5+3
    15=2*5+5
    17=2*5+7
    ...

    Since it's pretty easy to proof that all even numbers () can be created by the sum of two odd numbers, it also means that based on the premise every even number can be expressed as the sum of

    That's still 4 prime numbers in the sum and we need only two for the Goldbach conjecture.

    Of course it's also true that if:



    and:



    then:



    but also that there must be a and a :



    so,





    so for all :



    (pretty obvious result of course, but just wanted to go through the logic).

    Back to two primes (if Levy is correct).
    You may have just solved the mystery of the universe, but I do not follow the variables you use. Ha-ha.

    Could you explain it with words?

    Sincerely,


    William McCormick
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    William, it only solves world hunger sofar... :-D

    My reasoning in words is as follows:

    Assuming that Levy's conjecture, that any odd number equal or larger than 7 can be produced by the sum of "twice a prime (P) plus a prime", is indeed correct, I try to assess what the implications could be.

    Here's just one. Since all all odd numbers (O) > 7 also comprise of all the prime numbers, then also all odd prime numbers (except 2,3 and 5) could be produced by the sum of "twice a prime plus a prime".

    Since Goldbach conjectured that all even numbers (E) larger or equal than 4 are the sum of two primes, we can also write this equation as (twice a prime + a prime) + (twice a prime + a prime). This gives you the Goldbach conjecture decomposed into 4 prime numbers (of which some could be the same of course).

    My first assumption is that if you assume you can make all odd numbers from "twice a prime plus a prime", then it should be easy to prove that the sum of two these odd numbers can produce any even number (i.e. the conclusion is that with 4 primes you can produce all even numbers).

    But that's still four primes and I would like to get down to two. I then tried to find a way to express the last two primes (indexed by k,l) in the "4-prime Goldbach equation" into the first two primes (indexed by i,j).

    One connection I tried is that there must be i,j's and k,l's that produce odd numbers that are two apart. It should be true for all odd numbers produced by "primes i,j + 2". that there is an equal odd number produced by primes k,l. I therefore can express the i,j's in the k,l's and don't need the k,l's (although some doubt about the correctness of this step is slumbering in my head...).

    Example:

    2*13 + 7 = 33
    2*2 + 31 = 33+2

    2*13+ 7 + 2 = 2*2 + 31

    Other ways I thought off with potential to reduce the 4 primes to less?

    1. Maybe a trick with and . If we combine these in the Goldbach conjecture then the sum becomes . Doesn't produce all even numbers of course, but maybe we could relate it back to the k,l primes.

    2. Maybe there's a trick with and with the Goldbach sum becoming . Again not sure if leads anywhere.

    Boarding on a trip to Rio now, so some more time to think about all of this. I don't preclude that the conclusion could be that my reasoning contains some major flaws...

    But hey, progress can only be made through a process of creative destruction. 8)

    P.S.
    Although my biggest nightmare is of course to find a rock solid prove of the Goldbach conjecture at 30000 feet above the ocean, but then without any internet connection and suddenly the pilot announcing some serious engine problems... I vaguely recall a similar story about a postcard, a stormy boat trip and an alleged proof of the Riemann hypothesis?
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  29. #28  
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    Quote Originally Posted by accountabled
    William, it only solves world hunger sofar... :-D

    My reasoning in words is as follows:

    Assuming that Levy's conjecture, that any odd number equal or larger than 7 can be produced by the sum of "twice a prime (P) plus a prime", is indeed correct, I try to assess what the implications could be.

    Here's just one. Since all all odd numbers (O) > 7 also comprise all the prime numbers, then also all odd prime numbers (except 2,3 and 5) could be produced by the sum of "twice a prime plus a prime".

    Since Goldbach conjectured that all even numbers (E) larger than or equal to 4 are the sum of two primes, we can also write this equation as (twice a prime + a prime) + (twice a prime + a prime). This gives you the Goldbach conjecture decomposed into 4 prime numbers (of which some could be the same of course).

    My first assumption is that if you assume you can make all odd numbers from "twice a prime plus a prime", then it should be easy to prove that the sum of two these odd numbers can produce any even number (i.e. the conclusion is that with 4 primes you can produce all even numbers).

    But that's still four primes and I would like to get down to two. I then tried to find a way to express the last two primes (indexed by k,l) in the "4-prime Goldbach equation" into the first two primes (indexed by i,j).

    One connection I tried is that there must be i,j's and k,l's that produce odd numbers that are two apart. It should be true for all odd numbers produced by "primes i,j + 2". that there is an equal odd number produced by primes k,l. I therefore can express the i,j's in the k,l's and don't need the k,l's (although some doubt about the correctness of this step is slumbering in my head...).

    Example:

    2*13 + 7 = 33
    2*2 + 31 = 33+2

    2*13+ 7 + 2 = 2*2 + 31

    Other ways I thought off with potential to reduce the 4 primes to less?

    1. Maybe a trick with and . If we combine these in the Goldbach conjecture then the sum becomes . Doesn't produce all even numbers of course, but maybe we could relate it back to the k,l primes.

    2. Maybe there's a trick with and with the Goldbach sum becoming . Again not sure if leads anywhere.

    Boarding on a trip to Rio now, so some more time to think about all of this. I don't preclude that the conclusion could be that my reasoning contains some major flaws...

    But hey, progress can only be made through a process of creative destruction. 8)

    P.S.
    Although my biggest nightmare is of course to find a rock solid prove of the Goldbach conjecture at 30000 feet above the ocean, but then without any internet connection and suddenly the pilot announcing some serious engine problems... I vaguely recall a similar story about a postcard, a stormy boat trip and an alleged proof of the Riemann hypothesis?
    A friend of mine has one of those unsinkable boats. I will come out and get you. He uses it to crush ice and open up canals.


    It will take a while for that to go into my mind as well. I like that you look at it differently then I do. It makes the world seem like a bigger place. Thanks.

    I was taught that two was not a prime number because it could be divided by one or in half.

    If you multiply two by 0.5 you get a whole number. That is not true of a prime.

    I just saw what you did with the (P) and (O) to show they are variables. Thanks.


    Sincerely,


    William McCormick
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  30. #29  
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    Quote Originally Posted by Guitarist

    (...)

    Hmm... recall that if is prime, then so is , also if is even, then so is . So no failure then, right? (you might like to show the truth of these two assertions)

    (...)
    Guitarist,

    Building on your point. To my knowledge, Goldbach (nor Euler) have actually stated that the two primes in the sum should be positive integers. So, we might as well rewrite the conjecture as:

    "All positive even integers can be expressed as the difference between two primes."

    The gap between two subsequent primes is always even and can be made arbitrarily large. This is based on the fact that for :



    must all be composite (Havil 2003).

    So what's wrong with:

    2 = 5 + -3
    4 = 11 + -7
    6 = 29 + -23
    8 = 97 + -89
    etc.

    Of course the faculty based proof above, that any consecutive block of n composites will occur, doesn't imply that all even gaps will also occur between two subsequent primes. But in 2004, Ben Green and Terence Tao showed in the Green-Tao theorem that there are arbitrarily long arithmetic progressions of prime numbers. It is therefore always possible to find a progression of prime numbers of equal spacing and any length.

    Or is this reasoning incorrect?
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  31. #30  
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    At the risk of landing on everybody's IGNORE list... 8)

    Still pondering about the Goldbach conjecture. Had another thought that I've outlined on the following diagram:




    I started to experiment with a few numbers and then wrote a VBA program to see the results for more n's and got the following results:




    This is of course in line with the expectations, since the Goldbach conjecture has been empirically tested up to a very large number and therefore an even number that can not be produced by the sum of two primes was not to be expected. However, based on the trends in the graph I wonder whether there is a general rule to be derived that says f.i.:

    All prime numbers below can never be sufficient to produce all their associated equidistant composites between and so that both together sum up to n.

    There is an approximation for the number for prime numbers below which is . But then the number of combinations (including to multiply them all together to produce the matching composites in the domain is quite a bit more tricky. I did some further tests on the number of odd primes that are actually used in the production of the composites (circles around 70%), but that didn't lead to any deeper insights.

    Any thoughts/hints/tips ?
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