Posers and Puzzles

Posers and Puzzles

  1. Standard membergenius
    Wayward Soul
    Your Blackened Sky
    Joined
    12 Mar '02
    Moves
    15128
    03 Mar '05 11:35
    prove that there are infinetly many primes of the form 6k+5.
  2. my head
    Joined
    03 Oct '03
    Moves
    671
    03 Mar '05 12:39
    Originally posted by genius
    prove that there are infinetly many primes of the form 6k+5.
    the proff that there are infinite primes is established:
    one more or less than any multiple of number b is garenteed not to be a multiple of b.
    for any prime p, p!+1 is garenteed to be a larger prime, because p! is a multiple of all numbers p or less, and so p!+1 is not a multiple of any of them. if p!+1 is not prime, then it must have prime multipels greater than p. thus there are infinete primes. p!-1 can also be used.

    let us rephrase the term 6k+5 to better fit this argument: 2(3)k-1.
    if k is alwas prime, then we have numbers 2(3)k which all have only three prime multipls.

    i'll continue this after class.
  3. DonationAcolyte
    Now With Added BA
    Loughborough
    Joined
    04 Jul '02
    Moves
    3790
    03 Mar '05 21:23
    Originally posted by genius
    prove that there are infinetly many primes of the form 6k+5.
    Off the top of my head:

    Claim: If x is a natural number such that x = -1 mod 6, then x has a prime factor which is congruent to -1 mod 6.

    Proof: x is clearly not a multiple of 2 or 3, so its prime factors are all congruent to either 1 or -1 mod 6. But if they were all congruent to 1 mod 6, x would also be congruent to 1 mod 6.

    Now suppose that p1,p2,...,pn is a complete list of primes which are congruent to -1 mod 6. Then (p1p2...pn)^2 - 2 is a natural number which
    a) is congruent to -1 mod 6
    b) doesn't have any of p1,p2,...,pn as a prime factor, contradicting the claim.

    Hence there are infinitely many primes of the form 6k+5, QED.


    Apparently, it is known that every arithmetic sequence either contains infinitely many primes, or every term in the sequence is divisible by a particular prime. I have no idea how this was proved, but I get the impression that it's quite a hard result.
  4. Joined
    29 Feb '04
    Moves
    22
    04 Mar '05 06:135 edits
    Originally posted by genius
    prove that there are infinetly many primes of the form 6k+5.
    Well, all natural numbers are of the form: 6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5

    Clearly 6k cannot be prime (divisible by 6), neither can 6k+2 (divisible by 2), 6k+3 (divisible by 3), 6k+4 (divisible by 2).

    So the only "prime candidates" are 6k+1 and 6k+5. note that 6k+1 is one after a multiple of 6 and 6k+5 is one before a multiple of 6.

    Hence, if we have two consecutive odds that are prime, they will be either side of a multiple of 6.
    That is, all twin primes are of the form 6n +/- 1
    And the Twin Primes Conjecture remains famously unproven.

    http://mathworld.wolfram.com/TwinPrimes.html

    .
  5. Joined
    29 Feb '04
    Moves
    22
    04 Mar '05 09:401 edit
    Sorry, I posted before noticing Acolyte's proof.
    Anyway, it's quite interesing that all twin primes must be of the form 6n +/- 1
  6. Standard memberBowmann
    Non-Subscriber
    RHP IQ
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    17 Mar '05
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    1345
    22 Mar '05 23:06
    All primes above 3 are either one more or one less than a multiple of 6.
  7. Joined
    29 Feb '04
    Moves
    22
    22 Mar '05 23:09
    Originally posted by Bowmann
    All primes above 3 are either one more or one less than a multiple of 6.
    Been there, proved that.
  8. Standard memberBowmann
    Non-Subscriber
    RHP IQ
    Joined
    17 Mar '05
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    1345
    22 Mar '05 23:11
    I simply meant that what you say about twin primes would stand to reason.
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