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Prime Numbers Proof

  1. Feb 11, 2005 #1
    I've been going over this on paper for a while now, so I was wondering if maybe you guys would be able to point me in the right direction. We're supposed to prove that if every even natural number greater than 2 is the sum of two primes, than every odd number greater than 5 is the sum of 3 primes. Any ideas?
    Last edited: Feb 11, 2005
  2. jcsd
  3. Feb 11, 2005 #2


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    distinct primes? because if not, just add 3 to every even number.
  4. Feb 11, 2005 #3
    Yes a prime is defined as a number only divisible by itself and 1.
  5. Feb 11, 2005 #4


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    i meant, do all the primes have to be different? I would guess not, since 4, for example, only has 2+2. So if duplicates are allowed, the problem is pretty simple. just add 3.
  6. Feb 11, 2005 #5
    The problem doesn't say whether all primes have to be different. HOwever, your solutoin will not work. An exam[ple of why is in the case of x=6, 6+3 = 9, and 9 is not a prime number.
  7. Feb 11, 2005 #6


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    Take n to be an odd integer. Then n-3 is an even number, and by the premise, it can be expressed as the sum of two primes, p and q. So n = p+q+3 is one way of expressing n as the sum of three primes.
  8. Feb 11, 2005 #7
    Here is some thought:

    Any two odd numbers added together is always even, correct?

    Any odd and even numbers added together is always odd.

    I dont know if that will help but...?

    Were there any other constraints placed on the criteria of proof?
    For example, when you add the two or three prime numbers do they have to be distinct or can you add the same one twice or three times?
  9. Feb 11, 2005 #8
    Your logic makes sense StatusX. Is there anything else to your thought?
  10. Feb 11, 2005 #9
    My first thought was this one:

    every odd number can be represented by 2n+1 with [itex] n \in N [/itex].
    So 2n is a sum of two primes. Then 2n+1 is a sum of three numbers. The question is whether 1 is a prime number, and unfortunately not
    So my proof doesn't work.

    Try 2n+1 = 2(n-1+1)+1 = 2(n-1)+3.
    So 2(n-1) is an even number. We know that even numbers can be represented by a sum of 2 prime numbers, then 2(n-1)+3 is a sum of
    three prime numbers

    (I hope my proof is correct)
  11. Feb 11, 2005 #10
    what do you mean? sure 9 is not a prime, but 9 is odd.... and you wanna prove all odd is sum of 3 prime......

    just take any even number and add 3
  12. Feb 11, 2005 #11


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    He's solved the entire problem for you. What more can you want ?
    Last edited: Feb 12, 2005
  13. Feb 12, 2005 #12


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    Gokul43201 --
    The integer "3" is not relevant to the "premise of question". The latter imposed conditions only on EVEN integers greater than 2, the first of which is 4, and then asserted the composition of ODD integers greater than 5. In any event, StatusX's proof stands as the first definitive proof provided in this thread.

  14. Feb 12, 2005 #13


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    Guess I was terribly sleepy, when I read that. The first time I read it, I missed the "if" before the condition. Then I realized that the condition was essentially assuming Goldbach but for some reason, a few minutes later, I forgot that the numbers were supposed to be even !! :eek:

    Anyway, I'm now fully awake. :bugeye:
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