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Proof about primes.

  1. Jan 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Except 2 and 3 , prove that their an infinite amount of primes
    of the form 6m+1 and 6m+5 for some integer m
    It says to use Euclid's method but replace the +1 with a -1.
    3. The attempt at a solution
    Would I just multiply some of these forms together and subtract 1
    [itex] (6m+1)....(6n+1)-1=x [/itex]
    If I divide my new number x by (6m+1) it wont divide it evenly.
    this doesn't seems like it proves it, am I on the right track?
     
  2. jcsd
  3. Jan 11, 2013 #2

    haruspex

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    I wouldn't think so. You would only be able to show that x is not divisible by primes of the form 6m+1. You won't be able to show it is not divisible by other primes.
     
  4. Jan 11, 2013 #3

    mfb

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    If you multiply k primes of the form (6m+1) and l primes of the form (6m+5), you get a product of the form (6m+1) for even l and (6m+5) for odd l. This allows to find new primes, assuming a finite number of primes of those types. I am not sure how to get an infinite number of (6m+1)-primes, but it is certainly possible for (6m+5)-primes.
     
  5. Jan 11, 2013 #4
    ok thanks for the replies Ill work on it.
     
  6. Jan 11, 2013 #5

    haruspex

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    Suppose p to be the largest prime of a given one of those two types, and construct the product of all primes (except a certain few) up to p.
     
  7. Jan 14, 2013 #6
    So I make a product of all primes of the form 6m+1 and then add one
    to it. Im still not really sure exactly what you guys are telling me, Ill keep thinking about it.
    another thing i noticed is that all of the odd numbers greater than 5 are of the form
    6m+1 or 6m+3 or 6m+5, but we know that 6m+3=3(2m+1) so it never produces primes.
    so we know because their are an infinite amount of odd primes that
    either or both 6m+1 or 6m+3 contain an infinite amount of primes.
     
    Last edited: Jan 14, 2013
  8. Jan 15, 2013 #7

    haruspex

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    That's not what I wrote: "the product of all primes (except a certain few) up to p."
     
  9. Jan 15, 2013 #8

    mfb

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    1 is a bad thing to add, as the result will be divisible by 2.
    Right
     
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