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Re: [PrimeNumbers] A Property

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  • Jan van Oort
    At my former employer, a French software outfit, we already had that idea. In a modified way, we incorporated it into a piece of software that compresses
    Message 1 of 64 , Oct 21, 2005
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      At my former employer, a French software outfit, we already had that idea.
      In a modified way, we incorporated it into
      a piece of software that compresses images obtained by LAR ( Large Aperture
      Radar ) reconnaissance flights from the
      French Air Force.
      Any fool can have that idea. The proof: I had it in 2000, and the French
      Air Force uses it now *sardonic grin*
      Jan van Oort

      On 10/21/05, gordon_as_number <gordon_as_number@...> wrote:
      >
      >
      >
      > Here is just a comment.
      >
      > It is possible to compile a data image as a number, as in
      > xxx.arXiv.org/physics/0510148 <http://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
      >
      >
      >
      >
      >
      > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
      > The Prime Pages : http://www.primepages.org/
      >
      >
      >
      >
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      --
      Non sunt multiplicanda entia praeter necessitatem necnon voluptatem
      ( Ockam's razor, hack # 1, beta 0.1 )


      [Non-text portions of this message have been removed]
    • 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
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        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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