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Another equivalent to factoring

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  • Kermit Rose
    Consider the two arithmetic sequences k d2 + a2 and k + a1 where d2,a2,and a1 are integers, and k is the variable that takes on sequential values 0,1,2,3, .
    Message 1 of 1 , Jun 15 6:51 AM
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      Consider the two arithmetic sequences

      k d2 + a2
      and
      k + a1

      where d2,a2,and a1 are integers, and

      k is the variable that takes on sequential values 0,1,2,3, . . .

      We ask,

      When does the term

      (k + a1) divide ( k d2 + a2)?

      We answer as follows:


      k d2 + a2 = L ( k + a1) = L k + L a1

      L k + L a1 - k d2 = a2

      L k + L a1 - k d2 - a1 d2 = a2 - a1 d2

      (L -d2) (k + a1) = a2 - a1 d2

      We factor a2 - a1 d2 = x y

      Then

      L = d2 + x
      k = y - a1

      Conversely, if we wish to factor z = a2 - a1 d2

      and discover ( accidentally or by design ) that

      (k + a1) divides (k d2 + a2) then we know that

      (k + a1) will also divide z.


      Kermit Rose

      < kermit@... >
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