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Re: Factorizing technique

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  • ben tyers
    The method that I am using to obtain is unique to best the best of my knowledge. The basic arguement is: D=Ax+Bx-AB+x^2 where D is the number to be factored
    Message 1 of 1 , Sep 13, 2003
      The method that I am using to obtain is unique to best the best of my
      knowledge.
      The basic arguement is:

      D=Ax+Bx-AB+x^2
      where D is the number to be factored example : D=1709023

      where x is the aproxiamte root of d x=1307
      rounded to next integer

      where A+/-x & B+/-x are the Factors of D A=+116

      B=-106

      by putting these numbers back into the formula,
      1709023=(116*1307) - (106*1307) - (116*106) 1307*1307
      1709023= 151612 - 138542 - 12296 + 1708249
      obviously x+A and X+B are the prime factors, in this case 1423 and 1201

      In the above example A is +, and B is -, However the way in which I obtain
      these values from D via the above formula can lead to the opposite. The
      beauty of this formula is that it doesn't seek either or A and B, but is
      happy with any.
      The above has been put as simply as possible to explain the basis of my
      workings

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