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Re: Prime Numbers Conjecture

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  • Mark Underwood
    ... Along that vein, here s a more general observation: there are primes r,q such that 1*r + 2*q produces every prime 5 2*r + 3*q produces every prime 17
    Message 1 of 5 , Jul 25, 2008
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      --- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>
      wrote:
      >
      > Conjecture:
      >
      > For all prime p>=19
      >
      > we have
      >
      > p=3q+2r q and r are both primes.
      >
      >
      > Sincerely:
      >
      > Sebastián Martín Ruiz
      >

      Along that vein, here's a more general observation:

      there are primes r,q such that

      1*r + 2*q produces every prime > 5
      2*r + 3*q produces every prime > 17 (as per Sebastian's observation)
      3*r + 4*q produces every prime > 31
      4*r + 5*q produces every prime > 337
      5*r + 6*q produces every prime > 191
      6*r + 7*q produces every prime > 421
      7*r + 8*q produces every prime > 2711 (hmmm)
      8*r + 9*q produces every prime > 881
      9*r + 10*q produces every prime > 811
      10*r + 11*q produces every prime > 1979.


      Just think, somewhere, in some galaxy, a proof for this type of thing
      has already been produced. :o


      Mark

      .
    • Patrick Capelle
      ... It s not new: http://www.primepuzzles.net/conjectures/conj_047.htm http://primes.utm.edu/curios/page.php?short=19 Patrick Capelle
      Message 2 of 5 , Jul 25, 2008
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        --- Sebastian Martin wrote:
        > Conjecture:
        >  
        > For all prime p>=19
        >  
        > we have
        >  
        > p=3q+2r   q and r are both primes.
        >  
        >  
        > Sincerely:
        >  
        > Sebastián Martín Ruiz

        It's not new:
        http://www.primepuzzles.net/conjectures/conj_047.htm
        http://primes.utm.edu/curios/page.php?short=19

        Patrick Capelle
      • Bernardo Boncompagni
        I have noticed that all factors of a cyclotomic number Phi(n,b) (that is the n-th cyclotomic polynomial computed in the point b) are either a divisor of n or
        Message 3 of 5 , Jul 30, 2008
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          I have noticed that all factors of a cyclotomic number Phi(n,b) (that is
          the n-th cyclotomic polynomial computed in the point b) are either a
          divisor of n or congruent to 1 modulo n. For example:

          Phi(13,15) = 139013933454241 = 53 * 157483 * 16655159
          All 3 factors are congruent to 1 modulo 13.

          Phi(20,12) = 427016305 = 5 * 85403261
          5 is a divisor of 20 and 85403261 is congruent to 1 modulo 20.

          So I have a few questions:
          - is this a general pattern, or is it just another instance of the "law
          of small numbers"?
          - if so, what is the smallest known counterexample?
          - if not, can anyone point me to a (possibly online) demonstration?

          Thank you for your interest.

          Bernardo Boncompagni

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        • Phil Carmody
          ... There s a well-known equivalent proof for divisors of Fermat Numbers (with an extra stage at the end that you don t need). It s worth understanding that,
          Message 4 of 5 , Jul 30, 2008
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            --- On Wed, 7/30/08, Bernardo Boncompagni <RedGolpe@...> wrote:
            > I have noticed that all factors of a cyclotomic number
            > Phi(n,b) (that is
            > the n-th cyclotomic polynomial computed in the point b) are
            > either a
            > divisor of n or congruent to 1 modulo n. For example:
            >
            > Phi(13,15) = 139013933454241 = 53 * 157483 * 16655159
            > All 3 factors are congruent to 1 modulo 13.
            >
            > Phi(20,12) = 427016305 = 5 * 85403261
            > 5 is a divisor of 20 and 85403261 is congruent to 1 modulo
            > 20.
            >
            > So I have a few questions:
            > - is this a general pattern, or is it just another instance
            > of the "law
            > of small numbers"?
            > - if so, what is the smallest known counterexample?
            > - if not, can anyone point me to a (possibly online)
            > demonstration?
            >
            > Thank you for your interest.

            There's a well-known equivalent proof for divisors of Fermat Numbers (with an extra stage at the end that you don't need). It's worth understanding that, and then trying to adapt it to arbitrary cyclotomic numbers.

            Phil
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