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Math Help - help me to prove this conjecture

  1. #1
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    help me to prove this conjecture

    Conjecture: If p1 * p2 * p3 * ... pn + 1 is prime for every positive integer n, where p1,p2,...,pn are the n smallest prime number

    So far, I have tried this

    Proof (direct) :

    Let pj be a prime number which is greater than 1

    (p1*p2*p3*p4*...pn) % pj = 0
    1 is divisible by pj


    Therefore this conjecture is true?
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  2. #2
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    Quote Originally Posted by Discrete View Post
    Conjecture: If p1 * p2 * p3 * ... pn + 1 is prime for every positive integer n, where p1,p2,...,pn are the n smallest prime number

    So far, I have tried this

    Proof (direct) :

    Let pj be a prime number which is greater than 1

    (p1*p2*p3*p4*...pn) % pj = 0
    1 is divisible by pj


    Therefore this conjecture is true?
    2*3*5*7*11*13 + 1= 30031 = 59*509


    RonL
    Last edited by CaptainBlack; July 15th 2007 at 10:35 AM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    2*3*5*7*11*13 = 30030 = 59*509


    RonL
    what do you mean
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  4. #4
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    2*3*5*7*11*13 = 30030 = 59*509


    RonL
    I think Discrete mean the following:
    p_1\cdot p_2\cdot \ldots \cdot p_n+1 is prime, where p_1,p_2,\ldots , p_n are the first n primes.
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  5. #5
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    Quote Originally Posted by red_dog View Post
    I think Discrete mean the following:
    p_1\cdot p_2\cdot \ldots \cdot p_n+1 is prime, where p_1,p_2,\ldots , p_n are the first n primes.


    that is right red dog, that's the question and is my proof right?
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  6. #6
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    Quote Originally Posted by Discrete
    Therefore this conjecture is true?
    No, it just shows that the number is divisible by a different prime.
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  7. #7
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    Quote Originally Posted by red_dog View Post
    I think Discrete mean the following:
    p_1\cdot p_2\cdot \ldots \cdot p_n+1 is prime, where p_1,p_2,\ldots , p_n are the first n primes.
    There was a typo I missied the +1.

    2*3*5*7*11*13 + 1 = 30031 = 59*509

    I make that one more than the product of the first 6 primes is composite.
    One counter example is sufficent to refute a conjecture.

    RonL
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  8. #8
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    but how do I want to proof this by using direct, contraposition or contradiction proof?
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  9. #9
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    Quote Originally Posted by Discrete View Post
    but how do I want to proof this by using direct, contraposition or contradiction proof?
    Proof by contradiction:

    Suppose the conjecture true, show the counter example, which gives
    a contradiction, hence the supposition is false.

    RonL
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  10. #10
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    Quote Originally Posted by CaptainBlack View Post
    Proof by contradiction:

    Suppose the conjecture true, show the counter example, which gives
    a contradiction, hence the supposition is false.

    RonL

    yes I know this is the way of how to proof it by contradiction but then for the case in this question, how do I deal with it?
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  11. #11
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    Quote Originally Posted by Discrete View Post
    yes I know this is the way of how to proof it by contradiction but then for the case in this question, how do I deal with it?
    Your conjecture is false!

    RonL
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  12. #12
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    Is it possible to use the proof that I have mentioned in my first post?? 1 is not divisible by any prime numbers, so therefore any prime number which is added by 1 becomes no prime number, right??
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  13. #13
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    Quote Originally Posted by Discrete View Post
    Is it possible to use the proof that I have mentioned in my first post?? 1 is not divisible by any prime numbers, so therefore any prime number which is added by 1 becomes no prime number, right??
    Not by any prime in your list used in constructing the number, yes. If that's what you are trying to prove then the proof is correct. As in CaptainBlack's example
    2 \cdot 3\cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1 = 30031 = 59 \cdot 509

    Note that 30031 is not divisible by any of 2, 3, 5, 7, 11, or 13, but it is divisible by the primes 59 and 509.

    -Dan
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  14. #14
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    Quote Originally Posted by Discrete View Post
    Is it possible to use the proof that I have mentioned in my first post?? 1 is not divisible by any prime numbers, so therefore any prime number which is added by 1 becomes no prime number, right??
    As every prime p not equal to 2 is odd, p+1 is even and so not a prime.

    But 2 is prime and 2+1=3 which is also prime.

    RonL

    [I must say that I am lost, what is the conjecture - or whatever - that is under discussion here?]
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  15. #15
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    This is how I understand this user.

    Let p_1,...,p_n be the first n primes.

    Then, m=p_1p_2...p_n+1 is not divisible by any of these p_i. (Remember Euclid's Proof).

    So, (and that is what he claims) it must mean that m is prime itself because it is not divisible by any of those primes.

    That is his conjecture.

    But that is false because we only showed that m is divisible by another prime. Either itself, if is it, or a completely different prime if it is not.
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