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Math Help - For all primes greater than 3, p = 6ką1?

  1. #1
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    For all primes greater than 3, p = 6ką1?

    is it true?
    what is the proof?
    I prefer a hint if it ain't difficult to prove.
    tnx
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  2. #2
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    Quote Originally Posted by srulikbd
    is it true?
    what is the proof?
    I prefer a hint if it ain't difficult to prove.
    tnx
    Hint(?):

    Let p>3 be a prime. Consider the possible values of the remainder
    when p is divided by 6. As p>3 is prime the remainder cannot
    be divisible by any factor of 6.

    RonL
    Last edited by CaptainBlack; May 28th 2006 at 04:08 AM.
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  3. #3
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    Ok Understood

    quite easy
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  4. #4
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    Or to say it another way.
    Any number must have form,
    \begin{array}{c}6k\\6k+1\\6k+2\\6k+3\\6k+4\\6k+5
    Note that,
    \begin{array}{c}6k\\6k+2\\6k+4
    are all divisible by 2 and p>2 thus not a prime.

    And that,
    6k+3 is divisble by 3 and p>3 thus not a prime.
    Hence the only possibilites are,
    6k+1
    or,
    6k+5=6k+6-1=6(k+1)-1
    Thus,
    6k\pm 1 for some integer k.
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  5. #5
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    another question

    how do I prove that there are infinite primes that have the form xk+y, y<x, and share no common factors?
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  6. #6
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    Quote Originally Posted by srulikbd
    how do I prove that there are infinite primes that have the form xk+y, y<x, and share no common factors?
    Whoa!!!
    That is too complicated I believe the proof use's funtional analysis. It was first proven by Dirichelt (my avatar ).

    In fact in the book "Introduction to Theory of Numbers" by Hardy and Wright. Which contains elementary,algebraic and analytic number theory; proves all theorems in the book (even the prime number theorem) except this theorem! That is how complicated it is.
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  7. #7
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    cool

    coincidence
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  8. #8
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    I can present a proof to why are there infinitely many primes of form 4k+1 \mbox{ and }4k+3 if you really wish to know.
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  9. #9
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    ok

    i'll be happy!
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  10. #10
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    Quote Originally Posted by srulikbd
    i'll be happy!
    Lemma: The product of integers of the form 4k+1 is again an integer of the form 4k+1.

    Proof:: If a=4k+1 and b=4j+1 then, ab=(4k+1)(4j+1)=4(4kj+k+j)+1. Q.E.D.

    Theorem: There are infinitely many primes of form 4k+3.

    Proof: Assume there are finitely many primes of form 4k+3 call them \{ p_1,...,p_s} \}. Then form an integer,
    n=4p_1p_2\cdot ...\cdot p_s-1
    Prime factorize n as,
    n=q_1q_2\cdot ...\cdot q_r
    Note that 2\not | n because it is odd. Thus, q_i\not =2 for any 1\leq i\leq r.
    We cannot have that all the primes factors of n are of the form 4k+1 for that would imply that n has form 4k+1, which is does not because 4k-1=4(k-1)+3 by the lemma. Thus one of its factors r_i has form 4k+3. Because an odd prime has one of two forms 4k+1,4k+3. But then r_i|n thus, r_i|(4p_1p_2\cdot ...\cdot p_s-1) thus, r_i|1, because r_i is found among p_1,...,p_s thus an impossibility. Q.E.D.
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  11. #11
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    I didn't understand the end-what does | means?
    and whaat does X means? (2X...) does it means doesn't divide?
    tnx
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  12. #12
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    Quote Originally Posted by srulikbd
    I didn't understand the end-what does | means?
    and whaat does X means? (2X...) does it means doesn't divide?
    tnx
    a|b is to be read as a divides b.

    a \not| \ b is to be read as a does not divide b.

    RonL
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  13. #13
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    tnx both of u now I understand

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