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Thread: infinitely many primes of the form 6k + 5 and 6K + 1

  1. #1
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    infinitely many primes of the form 6k + 5 and 6K + 1

    hey the question is

    show that there are infinitely many primes of the form 6k + 5. does the method work for 6k + 1.

    my answer so far is

    suppose there are finite primes of the form 6k + 5
    order them such: p(1) < p(2) <....< p(n)
    let R = 6(p(1)p(2)...p(n)) + 5
    R can't be prime, if it is R > p(n)
    R can't be composite as any division will give a remainder of 5
    therefore there are infinitely many primes

    i think there's something not quite right with it and i can't use Dirichlet's Theorem

    kudos for any help!
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  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
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    409

    good work

    You have hit the nail on the head. Your proof is correct. Here are some clarifications:

    Let $\displaystyle p_1<p_2<p_3<...<p_n$ be a finite list of primes of the form $\displaystyle 6k+5$. Let $\displaystyle R=6p_1p_2p_3...p_n+5$. Since $\displaystyle R>p_n$, $\displaystyle R$ cannot be prime, so it is composite. When divided by any $\displaystyle p_i$, it leaves a remainder $\displaystyle 5$, therefore $\displaystyle R$ is not divisible by any prime of the form $\displaystyle 6k+5$. So $\displaystyle R$ must be the product of primes only of the form $\displaystyle 6k+1$. But if you take the product of numbers of the form $\displaystyle 6k+1$, the product is also of the form $\displaystyle 6k+1$. So $\displaystyle R$ must leave a remainder $\displaystyle 1$ when divided by $\displaystyle 6$. But by construction, $\displaystyle R$ leaves a remainder $\displaystyle 5$ when divided by $\displaystyle 6$. We have reached a contradiction, therefore our premise is false: there must be an infinite number of primes of the form $\displaystyle 6k+5$.

    This argument does not work for the $\displaystyle 6k+1$ case. Can you figure out why?
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