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Thread: 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 $\displaystyle p>3$ be a prime. Consider the possible values of the remainder
    when $\displaystyle p$ is divided by $\displaystyle 6$. As $\displaystyle p>3$ is prime the remainder cannot
    be divisible by any factor of $\displaystyle 6$.

    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,
    $\displaystyle \begin{array}{c}6k\\6k+1\\6k+2\\6k+3\\6k+4\\6k+5$
    Note that,
    $\displaystyle \begin{array}{c}6k\\6k+2\\6k+4$
    are all divisible by 2 and $\displaystyle p>2$ thus not a prime.

    And that,
    $\displaystyle 6k+3$ is divisble by 3 and $\displaystyle p>3$ thus not a prime.
    Hence the only possibilites are,
    $\displaystyle 6k+1$
    or,
    $\displaystyle 6k+5=6k+6-1=6(k+1)-1$
    Thus,
    $\displaystyle 6k\pm 1$ for some integer $\displaystyle k$.
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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 $\displaystyle 4k+1 \mbox{ and }4k+3$ 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 $\displaystyle 4k+1$ is again an integer of the form $\displaystyle 4k+1$.

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

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

    Proof: Assume there are finitely many primes of form $\displaystyle 4k+3$ call them $\displaystyle \{ p_1,...,p_s} \}$. Then form an integer,
    $\displaystyle n=4p_1p_2\cdot ...\cdot p_s-1$
    Prime factorize $\displaystyle n$ as,
    $\displaystyle n=q_1q_2\cdot ...\cdot q_r$
    Note that $\displaystyle 2\not | n$ because it is odd. Thus, $\displaystyle q_i\not =2$ for any $\displaystyle 1\leq i\leq r$.
    We cannot have that all the primes factors of $\displaystyle n$ are of the form $\displaystyle 4k+1$ for that would imply that $\displaystyle n$ has form $\displaystyle 4k+1$, which is does not because $\displaystyle 4k-1=4(k-1)+3$ by the lemma. Thus one of its factors $\displaystyle r_i$ has form $\displaystyle 4k+3$. Because an odd prime has one of two forms $\displaystyle 4k+1,4k+3$. But then $\displaystyle r_i|n$ thus, $\displaystyle r_i|(4p_1p_2\cdot ...\cdot p_s-1)$ thus, $\displaystyle r_i|1$, because $\displaystyle r_i$ is found among $\displaystyle p_1,...,p_s$ 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
    $\displaystyle a|b$ is to be read as $\displaystyle a$ divides $\displaystyle b$.

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

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

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