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Math Help - Proving step to full proof

  1. #1
    Senior Member I-Think's Avatar
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    Proving step to full proof

    I was doing a proof to prove that there exists arbitrarily long strings of composite numbers.
    In more technical language, for every m\in {N}, there exists a n such that
    n, n+1, n+2, n+3,..., n+m is composite

    to do so, I let n=m!,(for m>3) then we get the sequence:
    m!, m!+1, m!+2, m!+3,..., m!+m

    Every element of this sequence is obviously divisible by some integer, except m!+1.

    Now I must prove that m!+1 is always an composite. I have no idea how to do this. I tried induction and contradiction.
    Help please
    Last edited by CaptainBlack; October 1st 2010 at 11:04 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by I-Think View Post
    I was doing a proof to prove that there exists arbitrarily long strings of composite numbers.
    In more technical language, for every m\element{N}, there exists a n such that
    n, n+1, n+2, n+3,..., n+m is prime

    to do so, I let n=m!,(for m>3) then we get the sequence:
    m!, m!+1, m!+2, m!+3,..., m!+m

    Every element of this sequence is obviously divisible by some integer, except m!+1.

    Now I must prove that m!+1 is always an integer. I have no idea how to do this. I tried induction and contradiction.
    Help please
    There are so many typos in this I think you should correct them before expecting a reply.

    Also this is number theory.

    To show that for any n\in \mathbb{N} there is a set of  n consecutive composites choose $$ m so that m=n+1.

    Then m!+2, m!+3,.. m!+m are all composite, and there are  m-1=n of these.

    CB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by I-Think View Post

    Now I must prove that m!+1 is always an composite. I have no idea how to do this. I tried induction and contradiction.
    Help please
    You can't since if $$ m=2 then m!+1=5 is prime

    CB
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  4. #4
    Senior Member I-Think's Avatar
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    I apologize for the typos. I also apologize for posting this in the wrong forum but I did not expect the problem to involve Number Theory, as it is a topic I had not yet covered.

    And as for proving that m!+1is always composite, I had stated it was over the range m>3
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  5. #5
    Member Traveller's Avatar
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    Let n = (m+2)! + 2

    P.S: m! + 1 is composite when m = p-1 for some prime p > 3.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by i-think View Post
    i apologize for the typos. I also apologize for posting this in the wrong forum but i did not expect the problem to involve number theory, as it is a topic i had not yet covered.

    And as for proving that m!+1is always composite, i had stated it was over the range m>3
    ok 11!+1

    cb
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  7. #7
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    Infact following should work

    n = (m+1)! + 2
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