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A Property

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  • gordon_as_number
    Here is just a comment. It is possible to compile a data image as a number, as in xxx.arXiv.org/physics/0510148. That number, when factored in primes, can be
    Message 1 of 64 , Oct 20, 2005
      Here is just a comment.

      It is possible to compile a data image as a number, as in
      xxx.arXiv.org/physics/0510148. That number, when factored
      in primes, can be stored in the latter to data compress that
      image by a factor of an approximate 10^12. As the labeling
      of primes is about a digit or two below the number in accordance
      with numbers up to 10^17 (the known primes), with density of
      an approximate x/ln(x) 10^17.

      The point is, with the prime factorization of a number up to
      10^15 or so, and each prime in accordance with integers, a
      digit is lost with each prime factor. And assuming that there
      a maximum ln(N) prime factors, or rather less, most data
      compression schemes can be achieved with a factor of 10^12,
      which is actually quite amazing. (Images could be written as
      a number with approximately 10^260 digits.)

      So, depending on how fast you can factor a number with 10^100
      digits or so, compressions of 10^10 are achievable.

      Gordon
    • 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
        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








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