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Re: [PrimeNumbers] Cubes + 1

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  • mikeoakes2@aol.com
    In a message dated 02/12/2004 14:46:19 GMT Standard Time, ... No. If q is any odd number 2, then a^q + 1 = (a+1) * (a^(q-1) - a^(q-2) + ... +1) and so has
    Message 1 of 5 , Dec 2, 2004
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      In a message dated 02/12/2004 14:46:19 GMT Standard Time,
      shawns_email_address@... writes:


      >Better yet, is any number of the form (a^q)+1 prime where q is a
      >prime other than 2?

      No.

      If q is any odd number > 2, then
      a^q + 1 = (a+1) * (a^(q-1) - a^(q-2) + ... +1)
      and so has (a+1) as a factor.

      -Mike Oakes


      [Non-text portions of this message have been removed]
    • Ray Chandler
      So after 1^q+1=2 there are no further primes for odd q. -Ray Chandler ________________________________ From: mikeoakes2@aol.com [mailto:mikeoakes2@aol.com]
      Message 2 of 5 , Dec 2, 2004
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        So after 1^q+1=2 there are no further primes for odd q.
        -Ray Chandler

        ________________________________

        From: mikeoakes2@... [mailto:mikeoakes2@...]
        Sent: Thursday, December 02, 2004 8:53 AM
        To: shawns_email_address@...
        Cc: primenumbers@yahoogroups.com
        Subject: Re: [PrimeNumbers] Cubes + 1



        In a message dated 02/12/2004 14:46:19 GMT Standard Time,
        shawns_email_address@... writes:


        >Better yet, is any number of the form (a^q)+1 prime where q is a
        >prime other than 2?

        No.

        If q is any odd number > 2, then
        a^q + 1 = (a+1) * (a^(q-1) - a^(q-2) + ... +1)
        and so has (a+1) as a factor.

        -Mike Oakes


        [Non-text portions of this message have been removed]
      • Paul Leyland
        ... We can say even more than that. All primes of the form (a^q + 1) are either 2 (i.e. a=1) or of the form (b^{2^n}+1) where b and n are integers. When
        Message 3 of 5 , Dec 2, 2004
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          On Thu, 2004-12-02 at 14:53, mikeoakes2@... wrote:
          >
          > In a message dated 02/12/2004 14:46:19 GMT Standard Time,
          > shawns_email_address@... writes:
          >
          >
          > >Better yet, is any number of the form (a^q)+1 prime where q is a
          > >prime other than 2?
          >
          > No.
          >
          > If q is any odd number > 2, then
          > a^q + 1 = (a+1) * (a^(q-1) - a^(q-2) + ... +1)
          > and so has (a+1) as a factor.

          We can say even more than that. All primes of the form (a^q + 1) are
          either 2 (i.e. a=1) or of the form (b^{2^n}+1) where b and n are
          integers. When b=2 these are the Fermat numbers which are prime for
          0<=n<=4. No others are known and no more are believed to exist.

          It's quite easy to show, and a nice simple exercise, that if x has an
          odd factor p, then a^x+1 is divisible by (a^{x/p} + 1). The case p=1
          is clearly trivial.


          Paul
        • cino hilliard
          Hi, ... By Ohakm s Razor, the long division (x^n+1)/(x+1) is the best proof that x+1 divides. However We also can also argue as follows using Baconian
          Message 4 of 5 , Dec 3, 2004
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            Hi,

            >From: Paul Leyland <pcl@...>
            >Reply-To: pcl@...
            >To: mikeoakes2@...
            >CC: shawns_email_address@..., primenumbers@yahoogroups.com
            >Subject: Re: [PrimeNumbers] Cubes + 1
            >Date: 02 Dec 2004 16:46:51 +0000

            > > In a message dated 02/12/2004 14:46:19 GMT Standard Time,
            > > shawns_email_address@... writes:
            > >
            > >
            > > >Better yet, is any number of the form (a^q)+1 prime where q is a
            > > >prime other than 2?
            > >
            > > No.
            > >
            > > If q is any odd number > 2, then
            > > a^q + 1 = (a+1) * (a^(q-1) - a^(q-2) + ... +1)
            > > and so has (a+1) as a factor.
            >
            >We can say even more than that. All primes of the form (a^q + 1) are
            >either 2 (i.e. a=1) or of the form (b^{2^n}+1) where b and n are
            >integers. When b=2 these are the Fermat numbers which are prime for
            >0<=n<=4. No others are known and no more are believed to exist.

            By Ohakm's Razor, the long division (x^n+1)/(x+1) is the best proof that x+1
            divides. However We
            also can also argue as follows using Baconian Induction for n odd x an
            integer.

            (x+1)^3 = x^3 + 1 + 3x^2 + 3x

            1. x^3 + 1 = (x+1)^3 - 3x(x+1)
            div by x+1

            (x+1)^5 = x^5 + 1 + (5!/4!1!)x^4 + (5!/3!2!)x^3 + (5!/2!3!)x^2 + (5!/1!4!)x

            2. x^5 + 1 = (x+1)^5 - 5x(x^3+1) - 10x^2(x+1)
            div by x+1

            3. x^7 + 1 = (x_1)^7 - 7x(x^5+1) - 21x^2(x^3+1) - 35x3(x+1)
            div by x+1

            4. x^9 + 1 = 9x(x^7+1) - 36x^2(x^5+1) - 84x^3(x^3+1) - 126x^4(x+1)
            div py x+1
            ...

            since binomial(n,a) = binomial(n,n-a) for a = 1,2..(n-1)/2. The subtracted
            terms with these
            coefficients can be factored to binomial(n,a)(x^(n-a) + x^a). Now these
            terms can be further
            fsactored to t_i = binomial(n,a)x^a(x^(n-2a)+1). i = 1,2..(n-1)/2

            For n = 3
            a = 1, t_1 = 3x(x+1)

            For n = 5
            a= 1, t_1 = 5x(x^3+1)
            a= 2, t_2 = 10x^2(x+1)

            For n = 7
            a=1, t_1 = 7x(x^5+1)
            a=2, t_2 = 21x^2(x^3+1)
            a=3, t_3 = 35x^3(x+1)

            For n = 9
            a=1, t_1 = 9x(x^7+1)
            a=2, t_2 = 36x^2(x^5+1)
            a=3, t_3 = 84x^3(x^3+1)
            a=4, t_4 = 126x^4(x+1)

            Continuing this process we get for any odd n, an (n-1)/2 term series to
            subrtact from (x+1)^n.
            The first (n-1)/2-1 terms of this series contains factors (x^m+1) m<n that
            have already been shown
            to be div by x+1 and the last term contains the term with factor x+1.
            obviously if x+1 divides any of
            the other facters, kx^a, in the terms, the statement still holds. Thus,
            knowing we can work backwards to the point where n=3 that x+1 divides x^n+1,
            x+1 must divide x^n+1 for all n.

            CLH
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