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[PrimeNumbers] Re: Pythagorean Sets of Consecutive Primes

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  • Jens Kruse Andersen
    ... A gap cycle of 6, 8, 10 in some order can only get one longer than this before becoming inadmissible modulo 5. If we want more than 10 primes then we need
    Message 1 of 7 , Aug 31, 2011
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      David wrote:
      > Here is the first set of 9:
      >
      > {v=[
      > 206710003, 206710013, 206710019,
      > 206710027, 206710037, 206710043,
      > 206710051, 206710061, 206710067];
      > print(vector(#v-1,k,v[k+1]-v[k]));}
      >
      > [10, 6, 8, 10, 6, 8, 10, 6]

      A gap cycle of 6, 8, 10 in some order can only get one longer than
      this before becoming inadmissible modulo 5.
      If we want more than 10 primes then we need other gaps.

      Puzzle:
      If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?
      I assume so in the following.
      This means all solutions must have a gap cycle of 3 numbers.
      If 5 divides the sum of the 3 numbers then the cycle never
      becomes inadmissible modulo 5.

      (10, 24, 26) is the smallest even Pythagorean triple with sum
      divisible by 5.
      The first case of 12 consecutive primes for that cycle in some order:
      490537270893409 + d,
      for d = 0, 10, 34, 60, 70, 94, 120, 130, 154, 180, 190, 214.
      The gaps are 10, 24, 26, 10, 24, 26, 10, 24, 26, 10, 24.
      A few hundred cases of 12 non-consecutive primes were found
      during the search.

      I suspect the above is the first case of 12 consecutive primes
      with any gaps.
      A cycle with (10, 24, 26) cannot give more than 12 primes because
      it becomes inadmissible modulo 7.
      In order to get an admissible pattern with 13 primes I think we
      would need gaps of a size which makes it very hard to find
      consecutive primes.

      If the k-tuple conjecture (*) is true then there are arbitrarily
      long solutions.
      For example, (3*n#, 4*n#, 5*n#) is a Pythagorean triple for all n,
      and the gaps never become inadmissible modulo primes up to n.

      (*) There are variations of the k-tuple conjecture.
      We only need one saying that all admissible patterns have at least
      one occurrence.
      Then it follows that they also have occurrences with consecutive primes.

      --
      Jens Kruse Andersen
    • djbroadhurst
      ... Yes. Proof: Suppose the converse. Then we have 4 possibilities: 1) a^2 = b^2 + c^2, d^2 = b^2 - c^2 2) a^2 = b^2 + c^2, d^2 = c^2 - b^2 3) a^2 = b^2 - c^2,
      Message 2 of 7 , Sep 1, 2011
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        --- In primenumbers@yahoogroups.com,
        "Jens Kruse Andersen" <jens.k.a@...> wrote:

        > Puzzle:
        > If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?

        Yes.

        Proof: Suppose the converse. Then we have 4 possibilities:
        1) a^2 = b^2 + c^2, d^2 = b^2 - c^2
        2) a^2 = b^2 + c^2, d^2 = c^2 - b^2
        3) a^2 = b^2 - c^2, d^2 = b^2 + c^2
        4) a^2 = c^2 - b^2, d^2 = b^2 + c^2
        Hence there exists a pair of coprime integers (x,y)
        such that x^2+y^2 and x^2-y^2 are both odd squares.
        But each odd square is congruent to 1 modulo 8.
        Hence 2*x^2 = 2 mod 8 and x is even.
        Hence both y^2 and -y^2 are congruent to 1 modulo 4,
        which is absurd.

        David Broadhurst
      • djbroadhurst
        ... [Then I screwed up, sorry: I *should* have said:] and x is odd. I am pondering the unfortunate consequences.... David (feeling more odd then even :-)
        Message 3 of 7 , Sep 1, 2011
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          --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

          > > Puzzle:
          > > If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?
          >
          > Suppose the converse. Then we have 4 possibilities:
          > 1) a^2 = b^2 + c^2, d^2 = b^2 - c^2
          > 2) a^2 = b^2 + c^2, d^2 = c^2 - b^2
          > 3) a^2 = b^2 - c^2, d^2 = b^2 + c^2
          > 4) a^2 = c^2 - b^2, d^2 = b^2 + c^2
          > Hence there exists a pair of coprime integers (x,y)
          > such that x^2+y^2 and x^2-y^2 are both odd squares.
          > But each odd square is congruent to 1 modulo 8.
          > Hence 2*x^2 = 2 mod 8

          [Then I screwed up, sorry: I *should* have said:]

          and x is odd.

          I am pondering the unfortunate consequences....

          David (feeling more odd then even :-)
        • djbroadhurst
          ... See http://www.mathmeth.com/tom/files/pyth.pdf which claims a proof by infinite descent that x^2+y^2 and x^2-y^2 cannot both be odd squares. David (who
          Message 4 of 7 , Sep 1, 2011
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            --- In primenumbers@yahoogroups.com,
            "djbroadhurst" <d.broadhurst@...> wrote:

            > > Puzzle:
            > > If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?
            > >
            > > Suppose the converse. Then we have 4 possibilities:
            > > 1) a^2 = b^2 + c^2, d^2 = b^2 - c^2
            > > 2) a^2 = b^2 + c^2, d^2 = c^2 - b^2
            > > 3) a^2 = b^2 - c^2, d^2 = b^2 + c^2
            > > 4) a^2 = c^2 - b^2, d^2 = b^2 + c^2
            > > Hence there exists a pair of coprime integers (x,y)
            > > such that x^2+y^2 and x^2-y^2 are both odd squares.
            > > But each odd square is congruent to 1 modulo 8.
            > > Hence 2*x^2 = 2 mod 8
            >
            > [Then I screwed up, sorry

            See
            http://www.mathmeth.com/tom/files/pyth.pdf
            which claims a proof by infinite descent that
            x^2+y^2 and x^2-y^2 cannot both be odd squares.

            David (who miserably failed to find a simpler proof)
          • Jens Kruse Andersen
            ... It was apparently harder than I thought. I didn t have a proof but guessed it would be relatively easy when there is a known parametrization of all
            Message 5 of 7 , Sep 1, 2011
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              David wrote:
              > See
              > http://www.mathmeth.com/tom/files/pyth.pdf
              > which claims a proof by infinite descent that
              > x^2+y^2 and x^2-y^2 cannot both be odd squares.
              >
              > David (who miserably failed to find a simpler proof)

              It was apparently harder than I thought.
              I didn't have a proof but guessed it would be relatively easy when
              there is a known parametrization of all Pythagorean triples.
              I only did a brute force search to see there were no small solutions.

              --
              Jens Kruse Andersen
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