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adjacents and primes

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  • Mark Underwood
    Hi all Have been looking at a^r + b^s = prime, where a,b are adjacent integers as are r,s. Occassionally there is a glimmer of order, like for instance 20^3 +
    Message 1 of 6 , Dec 17, 2005
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      Hi all

      Have been looking at a^r + b^s = prime,
      where a,b are adjacent integers as are r,s.

      Occassionally there is a glimmer of order, like for instance

      20^3 + 21^4 = prime
      21^3 + 22^4 = prime
      22^3 + 23^4 = prime
      23^3 + 24^4 = prime
      24^3 + 25^4 = prime.

      Is there any longer? (I don't know, haven't looked into it.)

      Anyways I wonder, given any integer a >1 is there always at least one
      adjacent integer b such that a^r + b^s is prime, where r,s are
      adjacent integers no more than a?

      For instance:

      26: 26^2 + 27^3 = prime
      29: 29^10 + 28^9 = prime
      35: 35^2 + 34^3 = prime

      Mark
    • Werner D. Sand
      Hi Mark, odd prime numbers have the final number 1,3,7 or 9. The expression n^3 + (n+1)^4 is, as easily to show, always odd and not divisible by 5, and thus
      Message 2 of 6 , Dec 18, 2005
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        Hi Mark,

        odd prime numbers have the final number 1,3,7 or 9. The expression
        n^3 + (n+1)^4 is, as easily to show, always odd and not divisible by
        5, and thus has likewise the final numbers 1,3,7 or 9. n and (n+1)
        are coprime, their powers too, and the sum of the powers is thus not
        divisible by the prime factors of n and (n+1). All this creates a
        certain predisposition for primality. But only a certain.

        Werner
      • Kermit Rose
        From: Werner D. Sand Date: 12/18/05 13:52:38 To: primenumbers@yahoogroups.com Subject: [PrimeNumbers] re:adjacents and primes Hi Mark, odd prime numbers have
        Message 3 of 6 , Dec 18, 2005
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          From: Werner D. Sand
          Date: 12/18/05 13:52:38
          To: primenumbers@yahoogroups.com
          Subject: [PrimeNumbers] re:adjacents and primes

          Hi Mark,

          odd prime numbers have the final number 1,3,7 or 9. The expression
          n^3 + (n+1)^4 is, as easily to show, always odd and not divisible by
          5, and thus has likewise the final numbers 1,3,7 or 9. n and (n+1)
          are coprime, their powers too, and the sum of the powers is thus not
          divisible by the prime factors of n and (n+1). All this creates a
          certain predisposition for primality. But only a certain.

          Werner




          From kermit@...

          In fact n^3 + (n+1)^4

          is never divisible by 3, by 5, or by 31, and maybe not divisible by some
          few primes greater than 31.

          Note that 31 is equal to 1 mod 3 and equal to 1 mod 5.


          n^3 + (n+1)^4 can be divisible by 7, 11, 13, 17, 23 and 29.

          [Non-text portions of this message have been removed]
        • Kermit Rose
          From: Mark Underwood Date: 12/17/05 13:21:44 To: primenumbers@yahoogroups.com Subject: [PrimeNumbers] adjacents and primes Hi all Have been looking at a^r +
          Message 4 of 6 , Dec 18, 2005
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            From: Mark Underwood
            Date: 12/17/05 13:21:44
            To: primenumbers@yahoogroups.com
            Subject: [PrimeNumbers] adjacents and primes

            Hi all

            Have been looking at a^r + b^s = prime,
            where a,b are adjacent integers as are r,s.


            From kermmit@...

            This is equivalent to investigating

            whether or not

            a^r + (a+1)^(r+1) is a prime polynomial.

            Generally prime polynomials produce prime integers in the pattern
            you have observed.

            [Non-text portions of this message have been removed]
          • Jens Kruse Andersen
            ... 7 divides a^3 + (a+1)^4 for a==5 (mod 7). 6 consecutive primes is then the best possible for that formula. It first occurs starting at a = 841369395. a^5 +
            Message 5 of 6 , Jan 3, 2006
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              December 17 Mark Underwood wrote:

              > Have been looking at a^r + b^s = prime,
              > where a,b are adjacent integers as are r,s.
              >
              > Occassionally there is a glimmer of order, like for instance
              >
              > 20^3 + 21^4 = prime
              > 21^3 + 22^4 = prime
              > 22^3 + 23^4 = prime
              > 23^3 + 24^4 = prime
              > 24^3 + 25^4 = prime.
              >
              > Is there any longer? (I don't know, haven't looked into it.)

              7 divides a^3 + (a+1)^4 for a==5 (mod 7).
              6 consecutive primes is then the best possible for that formula.
              It first occurs starting at a = 841369395.

              a^5 + (a+1)^4 never has a factor below 17.
              17 divides for both a==1 and a==4 (mod 17).
              The formula can avoid small factors for 13 consecutive a.
              The first case of 6 primes starts at a = 1469217.
              The first 7 is:

              188934537^5 + 188934538^4
              188934538^5 + 188934539^4
              188934539^5 + 188934540^4
              188934540^5 + 188934541^4
              188934541^5 + 188934542^4
              188934542^5 + 188934543^4
              188934543^5 + 188934544^4

              > Anyways I wonder, given any integer a >1 is there always at least one
              > adjacent integer b such that a^r + b^s is prime, where r,s are
              > adjacent integers no more than a?
              >
              > For instance:
              >
              > 26: 26^2 + 27^3 = prime
              > 29: 29^10 + 28^9 = prime
              > 35: 35^2 + 34^3 = prime

              I don't allow exponent 0 (a^0=1).

              If exponent 1 is allowed:
              The smallest a>1 for which a^r + b^s is never prime for
              b = a+/-1, s = r+/-1, and r,s <= a, is a = 13361.
              This assumes prp's for smaller a are really primes.

              The cases where prp's above 3000 digits were needed:
              1252^1155 + 1253^1154 (3578 digits)
              3319^992 + 3318^991 (3493 digits)
              9818^765 + 9819^764 (3054 digits)
              9819^764 + 9818^765 (3054 digits, same prp as for a=9818)
              10127^888 + 10126^887 (3557 digits)
              11051^1176 + 11050^1177 (4760 digits)

              PrimeForm/GW made prp tests. 13361^13361 has 55126 digits.
              I stopped there but guess there are primes for larger exponents.

              If exponent 1 is not allowed then the prime 3743^1 + 3744^2 is
              skipped and the smallest a>2 is a = 3743.
              For a=2, there are no allowed r,s values at all.

              --
              Jens Kruse Andersen
            • Mark Underwood
              Jens I don t know how you can go up that high, good lord! Amazing. Thank you for finding (at over 55,000 digits!) what I could never have hoped to. kind
              Message 6 of 6 , Jan 3, 2006
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                Jens I don't know how you can go up that high, good lord! Amazing.
                Thank you for finding (at over 55,000 digits!) what I could never
                have hoped to.

                kind regards,
                Mark




                --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
                <jens.k.a@g...> wrote:
                >
                (snip)

                > The smallest a>1 for which a^r + b^s is never prime for
                > b = a+/-1, s = r+/-1, and r,s <= a, is a = 13361.
                > This assumes prp's for smaller a are really primes.
                >
                > The cases where prp's above 3000 digits were needed:
                > 1252^1155 + 1253^1154 (3578 digits)
                > 3319^992 + 3318^991 (3493 digits)
                > 9818^765 + 9819^764 (3054 digits)
                > 9819^764 + 9818^765 (3054 digits, same prp as for a=9818)
                > 10127^888 + 10126^887 (3557 digits)
                > 11051^1176 + 11050^1177 (4760 digits)
                >
                > PrimeForm/GW made prp tests. 13361^13361 has 55126 digits.
                > I stopped there but guess there are primes for larger exponents.
                >
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