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An amusing error - somewhat OT

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  • j_m_berg
    I ve been tinkering with Fermat Number factoring for several months. Given my lack of math background and my preference to explore on my own, I ve been having
    Message 1 of 1 , May 2, 2003
      I've been tinkering with Fermat Number factoring for several months.
      Given my lack of math background and my preference to explore on my
      own, I've been having fun rediscovering many things that are rather
      simple (to most people here).

      One thing I've been playing with is a simple transformation of the
      standard equation.

      F = (A*2^M+1)*(B*2^M+1)
      F = A*B*2^(2*M)+A*2^M+B*2^M+1
      (F-1)/2^M = (A*B*2^M) + (A+B)

      A couple of days ago, I suddenly realized that if I could deduce what
      value to subtract from (F-1)/2, I could simply factor the result to
      determine the product of A*B. So I started to explore using small
      values of A and B.

      That's when I made the fatal error of transposing from
      thinking "subtract (A+B)" and turned it to thinking "mod(A+B)". Given
      small values of A and B, the product of A*B was small enough that
      using mod(A+B) would extract the sum of A+B, leaving just the product
      of A*B*2^M.

      This excited me since I knew that K values typically are not prime
      and consist of realitively small factors. I immediately setup a large
      test using the unresolved portion of F12 to determine the
      factorization of (F-1)/2^M mod(A+B). But it didn't take me too long
      to realize that this wasn't giving me valid results.

      At that point I went to bed and tried to ignore it long enough to get
      some sleep. Then while drinking my morning coffee the next day, my
      error suddenly tapped me on my shoulder. That's when I realized that
      I was reducing the value by mod(A+B) instead of subtracting A+B. Talk
      about a letdown!

      So, I'm back to dabbling and tinkering with Fermat Number factoring.
      My hind brain keeps telling me there is a way to reduce Fermat
      Numbers. Given the long history of research on the subject and my
      limited skills, I doubt there'll be any breakthrough. But you know
      what? It sure is fun to try!

      And no, I'm not certain why I'm sharing this. But it seemed amusing
      to me and consequently I thought I'd share it for it's limited humor
      quotient. :)

      Jay Berg
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