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question about a proof

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  • David Litchfield
    Hello all and happy New Year, Could someone tell me, please, if this stands up ? All odd numbers of the form (a^2 + b^2) / 2 with both a and b being odd, that
    Message 1 of 4 , Jan 2, 2003
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      Hello all and happy New Year,
      Could someone tell me, please, if this "stands up"?

      All odd numbers of the form (a^2 + b^2) / 2 with both a and b being odd,
      that is those numbers that lie equidistant between two odd perfect squares,
      are the sum of two other perfect squares, one odd and the other even and of
      the form ((a+b)/2)^2 and ((a-b)/2)^2

      n = (a^2 + b^2) / 2 = x^2 + y^2

      let x = (a+b)/2
      let y = (a-b)/2

      so that

      n = (a^2 + b^2) / 2

      can be rewritten as

      n = ((x + y)^2 + (x - y)^2) / 2

      2n = (x + y)^2 + (x - y)^2

      2n = (x + y) * (x + y) + (x - y) * (x - y)

      2n = x^2 + xy + xy + y^2 + x^2 -xy -xy + y^2

      2n = 2x^2 + 2y^2

      n = x^2 + y^2

      Cheers,
      David Litchfield
    • Phil Carmody
      ... (03:22) gp (a^2+b^2)/2 - (((a+b)/2)^2+((a-b)/2)^2) %1 = 0 i.e. if n=(a^2+b^2)/2, then n=(((a+b)/2)^2+((a-b)/2)^2) Phil ===== The answer to life s mystery
      Message 2 of 4 , Jan 2, 2003
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        --- David Litchfield <Mnemonix@...> wrote:
        > Hello all and happy New Year,
        > Could someone tell me, please, if this "stands up"?
        >
        > All odd numbers of the form (a^2 + b^2) / 2 with both a and b being odd,
        > that is those numbers that lie equidistant between two odd perfect squares,
        > are the sum of two other perfect squares, one odd and the other even and of
        > the form ((a+b)/2)^2 and ((a-b)/2)^2

        (03:22) gp > (a^2+b^2)/2 - (((a+b)/2)^2+((a-b)/2)^2)
        %1 = 0

        i.e. if n=(a^2+b^2)/2, then n=(((a+b)/2)^2+((a-b)/2)^2)


        Phil


        =====
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        Sex and death. -- Ian 'Lemmy' Kilminster

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      • Phil Carmody
        ... It means I m running The PARI Group s gp package at 3:22am. It s free, it s incredibly powerful, and it answers these kinds of questions without even
        Message 3 of 4 , Jan 2, 2003
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          --- David Litchfield <mnemonix@...> wrote:
          > Cheers for the reply, Phil. Could you explain one thing:
          > > (03:22) gp >
          > I'm not sure exactly what this means.

          It means I'm running The PARI Group's "gp" package at 3:22am.

          It's free, it's incredibly powerful, and it answers these kinds of questions
          without even breaking a sweat.

          > Further, I take it all I had need to have done was
          >
          > > i.e. if n=(a^2+b^2)/2, then n=(((a+b)/2)^2+((a-b)/2)^2)

          I was simply rewording the identity
          (a^2+b^2)/2-(((a+b)/2)^2+((a-b)/2)^2) = 0
          into a form more similar to your original question.

          Phil


          =====
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          Sex and death. -- Ian 'Lemmy' Kilminster

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        • Jon Perry
          So we can write: (a+b)/2)^2+((a-b)/2)^2 =a^2/4 + ab/2 + b^2/4 + a^2/4 - ab/2 + b^2/4 =a^2/2 + b^2/2 Jon Perry perry@globalnet.co.uk
          Message 4 of 4 , Jan 3, 2003
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            So we can write:

            (a+b)/2)^2+((a-b)/2)^2

            =a^2/4 + ab/2 + b^2/4 + a^2/4 - ab/2 + b^2/4

            =a^2/2 + b^2/2

            Jon Perry
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