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A closer look at generalization of difference of squares factoring.

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  • Kermit Rose
    I discovered this simpler proof of the generalization that also gives added insight into its limitations. z = t^s - s^2 z + s^2 = t^2 z + s^2 - (t - A)^2 =
    Message 1 of 1 , Jan 11, 2008
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      I discovered this simpler proof of the generalization that also gives
      added insight
      into its limitations.

      z = t^s - s^2

      z + s^2 = t^2

      z + s^2 - (t - A)^2 = t^2 - (t - A)^2

      z + (s + A - t) * (s + t - A) = A * ( 2 * t - A )

      Z + ( A - (t - s) ) * (s + t - A) = A * (2 * t - A)

      Z - ( (t - s) - A) * (( t + s) - A) = A * (2 * t - A)


      Z - ( x - A) * ( y - A) = A * ( x + y - A)

      where x y = z

      which is the simple algebraic identity

      xy - (x - A) * (y - A) = A * ( x + y - A)


      For every value of A, there is a related factoring, but the identity
      gives no clue how
      to find any of them.

      At the moment, it looks as if finding any of them will be exactly as
      difficult as finding the
      t or s.

      Kermit Rose < kermit@... >
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