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Posers and Puzzles

Posers and Puzzles

  1. Standard member genius
    Wayward Soul
    03 Mar '05 11:35
    prove that there are infinetly many primes of the form 6k+5.
  2. 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. Donation Acolyte
    Now With Added BA
    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. 04 Mar '05 06:13 / 5 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. 04 Mar '05 09:40 / 1 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 member Bowmann
    Non-Subscriber
    22 Mar '05 23:06
    All primes above 3 are either one more or one less than a multiple of 6.
  7. 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 member Bowmann
    Non-Subscriber
    22 Mar '05 23:11
    I simply meant that what you say about twin primes would stand to reason.