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  • paulmillscv
    Hello, Thanks for your response to the last question. For those of you who cannot wait until Xmas here is another question. There may be others as well. Find
    Message 1 of 6 , Oct 31, 2003
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      Hello,
      Thanks for your response to the last question. For those of
      you who cannot wait until Xmas here is another question. There may be
      others as well.

      Find the least 3 integer cubes so that X^3 = b, Y^3 = a + b and
      Z^3 = 2a + b. I.e. 3 cubes in arithmetic progression. If that is
      too hard , try 2 cubes in AP (!) or too easy, try 4 cubes in AP etc.

      Regards,
      Paul Mills
      Kenilworth,
      England.
    • Lawrence Hon
      -1 , 0 , 1
      Message 2 of 6 , Oct 31, 2003
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        -1 , 0 , 1

        On Fri, 31 Oct 2003, paulmillscv wrote:

        > Hello,
        >       Thanks for your response to the last question. For those of
        > you who cannot wait until Xmas here is another question. There may be
        > others as well.
        >
        > Find the least  3 integer  cubes so that X^3 = b,  Y^3 = a + b and
        > Z^3 = 2a + b.  I.e. 3 cubes in arithmetic progression.  If that is
        > too hard , try 2 cubes in AP (!) or too easy, try 4 cubes in AP etc.
        >
        > Regards,
        > Paul Mills
        > Kenilworth,
        > England.
        >
        >
        >
        >
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      • Andrew Swallow
        ... Then of course, any positive integer n gives -n,0,n as an answer. Is this a clever solution type question? Or a boring computer search type
        Message 3 of 6 , Oct 31, 2003
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          --- In primenumbers@yahoogroups.com, Lawrence Hon <lh2072@c...>
          wrote:
          >
          > -1 , 0 , 1

          Then of course, any positive integer n gives -n,0,n as an answer.

          Is this a 'clever solution' type question? Or a boring 'computer
          search' type question...? Cuz looking at the 'clever solution'
          approach, it doesn't seem to go too far...

          Andy
        • Chris Caldwell
          ... Am I misreading this? Seems to be you are asking for solutions to 2Y^3 = X^3 + Z^3 If we let r be a solution to r^2 + r + 1 = 0, then we have the
          Message 4 of 6 , Oct 31, 2003
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            At 12:11 PM 10/31/2003 +0000, you wrote:
            >Hello,
            > Thanks for your response to the last question. For those of
            >you who cannot wait until Xmas here is another question. There may be
            >others as well.
            >
            >Find the least 3 integer cubes so that X^3 = b, Y^3 = a + b and
            >Z^3 = 2a + b. I.e. 3 cubes in arithmetic progression. If that is
            >too hard , try 2 cubes in AP (!) or too easy, try 4 cubes in AP etc.

            Am I misreading this? Seems to be you are asking for solutions to

            2Y^3 = X^3 + Z^3

            If we let r be a solution to r^2 + r + 1 = 0, then we have the solutions

            Y = 0, X = r^n*Z where n = 0, 1 or 2

            and

            X = r^n*Y = r^m*Z

            but sadly that is it in regular integers or Q[r] (Phil's field). The same
            is true (sans the last solution) if we replace 2 by any prime that is
            congruent to 2 or 5 modulo 9 (or those primes squared).


            Short answer, Lawrence was right: -1,0,1,

            Chris
          • Mike Oakes
            ... of ... be ... etc. ... solutions ... The same ... is ... I expect you re right, Chris (if you correct your typo and say X = - r^n*Z ), and both the
            Message 5 of 6 , Nov 1, 2003
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              --- In primenumbers@yahoogroups.com, Chris Caldwell <caldwell@u...>
              wrote:
              > At 12:11 PM 10/31/2003 +0000, you wrote:
              > >Hello,
              > > Thanks for your response to the last question. For those
              of
              > >you who cannot wait until Xmas here is another question. There may
              be
              > >others as well.
              > >
              > >Find the least 3 integer cubes so that X^3 = b, Y^3 = a + b and
              > >Z^3 = 2a + b. I.e. 3 cubes in arithmetic progression. If that is
              > >too hard , try 2 cubes in AP (!) or too easy, try 4 cubes in AP
              etc.
              >
              > Am I misreading this? Seems to be you are asking for solutions to
              >
              > 2Y^3 = X^3 + Z^3
              >
              > If we let r be a solution to r^2 + r + 1 = 0, then we have the
              solutions
              >
              > Y = 0, X = r^n*Z where n = 0, 1 or 2
              >
              > and
              >
              > X = r^n*Y = r^m*Z
              >
              > but sadly that is it in regular integers or Q[r] (Phil's field).
              The same
              > is true (sans the last solution) if we replace 2 by any prime that
              is
              > congruent to 2 or 5 modulo 9 (or those primes squared).
              >
              >
              > Short answer, Lawrence was right: -1,0,1,
              >
              > Chris

              I expect you're right, Chris (if you correct your typo and say X = -
              r^n*Z ), and both the equation and your assertions have a familiar
              ring to them, but I can neither prove that no other solutions exist
              nor lay hands on a relevant reference.
              Help us out, please!

              Mike
            • Chris Caldwell
              ... The wonderful text: L. J. Mordell Diophantine Equations Academic Press, 1969, page 124-130 includes things like: Theorem 3: Let a = p or p^2 where p =
              Message 6 of 6 , Nov 1, 2003
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                At 01:12 PM 11/1/2003 +0000, you wrote:
                >I expect you're right, Chris (if you correct your typo and say X = -
                >r^n*Z ), and both the equation and your assertions have a familiar
                >ring to them, but I can neither prove that no other solutions exist
                >nor lay hands on a relevant reference.
                >Help us out, please!

                The wonderful text: L. J. Mordell "Diophantine Equations"
                Academic Press, 1969, page 124-130 includes things like:

                Theorem 3: Let a = p or p^2 where p = 2, 5 (mod 9) is a prime,
                and let e be a unit in Q(r) (r^2 + r + 1 = 0). Then the equation

                x^3 + y^3 + eaz^3 = 0

                has no solutions (x,y,z,e) in Q(r) except

                z = 0, x = -y, -ry, -r^2y

                unless a = 2, when the equation also has the solutions

                x^3 = y^3 = -e z^3, e = +/-1.

                Proof is about one page and is done by choosing x, y, z
                pairwise relatively prime and N(xyz) minimal. Then do
                descent...

                But we are wandering away from primes...

                Chris
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