Loading ...
Sorry, an error occurred while loading the content.

Re: Question

Expand Messages
  • elevensmooth
    ... The parenthesized expression only need be 11 times a square. This happens whenever t is of the form 510510*u^2 + 4642*u + 10, plus several other forms. I
    Message 1 of 64 , Oct 20, 2005
    • 0 Attachment
      --- In primenumbers@yahoogroups.com, "scolnik \(fibertel\)"
      <scolnik@f...> wrote:
      >
      > because a + b*t = 11*(25631 + 46410*t) and therefore if the expression
      > between parentheses does not give an odd power of 11..

      The parenthesized expression only need be 11 times a square. This
      happens whenever t is of the form 510510*u^2 + 4642*u + 10, plus
      several other forms. I found the solution using Dario Alperns generic
      quadratic solver at

      http://www.alpertron.com.ar/QUAD.HTM

      Using "step by step" will lead to information about how to generate
      the solutions.

      William
      Poohbah of OddPerfect Number Search
      http://OddPerfect.org
    • Kermit Rose
      ... Thank you David. I had been confused by consideration of the following: if x y = z and y x, then if we set t = (y+x)/2 and s = (y-x)/2 then z = t^2 - s^2
      Message 64 of 64 , Oct 20, 2012
      • 0 Attachment
        On 10/20/2012 10:07 AM, primenumbers@yahoogroups.com wrote:
        > 2.2. Re: Question
        > Posted by: "djbroadhurst"d.broadhurst@... djbroadhurst
        > Date: Fri Oct 19, 2012 9:02 am ((PDT))
        >
        >
        >
        > --- Inprimenumbers@yahoogroups.com,
        > Kermit Rose<kermit@...> wrote:
        >
        >> >Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
        >> >
        >> >for what values of x would
        >> >x^2 + 5 x + 6 be a perfect square.
        > Set A = 1, B = 0, C = -1, D = 5, E = 0, F = 6.
        >
        > and you will get the obvious answer from Dario:
        >
        > x = -2
        > y = 0
        > and also:
        > x = -3
        > y = 0
        > Calculation time: 0h 0m 0s
        >
        > David

        Thank you David.

        I had been confused by consideration of the following:

        if x y = z and y > x,
        then if we set t = (y+x)/2 and s = (y-x)/2

        then z = t^2 - s^2
        and x = y - 2 t.

        So z = x y = (y - 2t) y = y^2 - 2 t y

        y^2 - 2 t y - z = 0

        Transforming w = y - k yields

        (w + k)^2 - 2 t (w + k) - z = 0

        w^2 + (2 k - 2 t) w + (k^2 - 2 t k - z) = 0

        which has integral solution if and only if

        (2k - 2 t)^2 + (z + 2 t k - k^2) is an integral square

        In all that derivation, I forgot that I still did not know what value t was.

        I confused myself too easily.

        Kermit








        [Non-text portions of this message have been removed]
      Your message has been successfully submitted and would be delivered to recipients shortly.