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

24586Question

Expand Messages
  • Kermit Rose
    Oct 20, 2012
      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]
    • Show all 64 messages in this topic