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A proposition for a better factorisation algorithm

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  • bhelmes_1
    A beautiful day, a very brief proposition for a better factorisation algorithm, based on the basic ideas of Pollard p-1 algorithm 1. Instead of using N, take C
    Message 1 of 1 , Apr 6 9:15 AM
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      A beautiful day,

      a very brief proposition for a better factorisation algorithm,
      based on the basic ideas of Pollard p-1 algorithm

      1. Instead of using N, take C

      a) Instead of an exponentation by primes, use only an exponentation by N

      b) Choose natural numbers a(1) and b(1) not equal 0 for the start point

      c) Calculate a(n+1)+b(n+1)=(a(n)+b(n)I)^exponent mod f
      where f is the composite number

      d) calculate res=gcd (a(n+1), f) and res=gcd (b(n+1), f) not every time, but after 1000 exponentation for example.

      res gives the searched factor of f

      e) You can store the value af a(1000*k) and b(1000*k)
      if b(1000*(k+1))=0 you can recalculate the sequence
      and make each time a gcd by every exponentation

      For the second part of Pollard p-1 you could use a lattice instead of searching linear which can be done on a cluster

      2. Instead of using N, use the field of adjoined square
      a+b*sqrt (A) where A is a non quadratic residuum mod f

      same algorithm as before

      3. Instead of using N, use the field of complex adjoined square
      a+b*sqrt(A)I where A is a non quadratic residuum mod f

      same algorithm as before

      You can execute 1.,2. and 3. parallel.

      Nice greetings from the primes
      Bernhard
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