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Generalization of difference of squares factoring.

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  • Kermit Rose
    Generalization of difference of squares factoring. One method of searching for factors is to find A such that A^2 - z = c^2. Then having found A and C, we
    Message 1 of 1 , Jan 10, 2008
      Generalization of difference of squares factoring.


      One method of searching for factors is to find A such that

      A^2 - z = c^2.

      Then having found A and C, we know that

      z = (A - C) * (A + C)



      The generalization of this is based on

      z = (A - C) * (A + C + j) if and only if

      A * (A + j) - z
      = A * (A + j) - (A - C) * (A + C + j)
      = A^2 + A j - [ A^2 + A C + A j - A C - C^2 - C j]
      = A^2 + A j - [A^2 + A j - C^2 - C j]
      = C^2 + C j
      = C * (C + j)

      Thus
      We look for integers Z + C * (C + j) that can be factored as
      A * ( A + j).


      If we find a [ Z + C * (C + j) ] that factors as A * (A + j),

      we know that

      z = (A - C) * (A + C + j)


      Now we have the following questions, which will probably be a matter of
      experimenting,

      is


      what is the best order in which to search for the A, C, and j?

      and

      will this generalization ever do better than the restricted search for

      A^2 - z = C^2?




      Kermit Rose < kermit@... >
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