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Prime Numbers Conjecture

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  • Sebastian Martin
    Conjecture:   For all prime p =19   we have   p=3q+2r   q and r are both primes.     Sincerely:   Sebastián Martín Ruiz
    Message 1 of 5 , Jul 25 2:34 AM
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      Conjecture:
       
      For all prime p>=19
       
      we have
       
      p=3q+2r   q and r are both primes.
       
       
      Sincerely:
       
      Sebastián Martín Ruiz


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      [Non-text portions of this message have been removed]
    • 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 2 of 5 , Jul 25 7:51 AM
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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 3 of 5 , Jul 25 11:05 AM
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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 4 of 5 , Jul 30 10:19 AM
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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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            the land and the missionaries had the bible.
            They taught how to pray with our eyes closed.
            When we opened them, they had the land and we
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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 5 of 5 , Jul 30 1:35 PM
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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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