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Math Help - 3 consecutive prime numbers.

  1. #1
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    3 consecutive prime numbers.

    Of the form n, n+2, n+4

    We have 3, 5, 7

    29, 31 ,33 etc

    I guess they must be infinite? How is it proved.

    I realise the final digit has to be in the pattern 9,1,3 or 7,9,1 but what next?
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  2. #2
    Member grgrsanjay's Avatar
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    33 is not prime i guess
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  3. #3
    Member grgrsanjay's Avatar
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    i don't think any other numbers exist
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  4. #4
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    It can easily be proved that if you have 3 integers of the form n, n+2, n+4 then one of them must be divisible by 3 (try it yourself). Thus no such prime triplets exist, other than 3,5,7.
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  5. #5
    Member grgrsanjay's Avatar
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    yea at least one should be divisible by 3

    i guess this thread is closed
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  6. #6
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    If you assume n=3k then case closed
    n=3k+1 then n+2=3(k+1)
    n=3k+2 then n+4=3(k+2)

    So, that proves it?

    How would you formally write this assuming nothing excepting basic axioms or addition etc.

    Are there any other proofs?
    Is there any way of obtaining every single proof of a question like this in general? Suppose you found 10 proofs out of 10, how would you prove or ascertain that there is no eleventh proof? Or is this idiotic speculation?

    If you add the three you get 3(n+2), is it true that the sum of 3 different integers, if divisible by three, then at least one of the three must also be divisible by three? How is this proved, if so?
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  7. #7
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    Quote Originally Posted by berachia View Post
    If you assume n=3k then case closed
    n=3k+1 then n+2=3(k+1)
    n=3k+2 then n+4=3(k+2)

    So, that proves it?

    How would you formally write this assuming nothing excepting basic axioms or addition etc.

    Are there any other proofs?
    Is there any way of obtaining every single proof of a question like this in general? Suppose you found 10 proofs out of 10, how would you prove or ascertain that there is no eleventh proof? Or is this idiotic speculation?

    If you add the three you get 3(n+2), is it true that the sum of 3 different integers, if divisible by three, then at least one of the three must also be divisible by three? How is this proved, if so?
    You did formally write the proof. You split it into all possible cases and showed that for each case, one of those integers is divisible by 3.

    I don't know about your second question, but about your last - what about 5 + 5 + 5?
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  8. #8
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by berachia View Post
    Are there any other proofs?
    Is there any way of obtaining every single proof of a question like this in general? Suppose you found 10 proofs out of 10, how would you prove or ascertain that there is no eleventh proof? Or is this idiotic speculation?
    I don't know much about proof theory, but at the very least we'd need to decide on some criteria for setting up an equivalence relation between proofs; for example, merely changing variable names should not be considered making a new proof. Your question could touch on the field of computer proof checking and Godel's famous theorems.
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