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RE Greatest Factor Less Than Root

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  • Jose Ramón Brox
    From: John W. Nicholson ... If N is even, factor with 2 until you get N odd. Once N is odd, you can write N as the difference of
    Message 1 of 1 , Jul 10, 2008
      From: "John W. Nicholson" <reddwarf2956@...>

      >Let N be a integer. How does one go about finding the largest factor,
      >c, less than the square root of a number, N^(1/2)?

      If N is even, factor with 2 until you get N' odd.

      Once N' is odd, you can write N' as the difference of two squares:

      N' = a^2 - b^2 =(a+b)(a-b)

      Then, execute the following algorithm:

      Start with a=ceil(Sqrt(N')).
      Check if b= Sqrt(a^2-N') is an integer
      if it is, END LOOP
      if it isn't, then increase a, a=a+1
      Largest factor of N' less than its square root = (a-b)
      (Largest factor of N' bigger than its square root = (a+b) )

      And so, you just have to remember the 2^r factor that we extracted earlier and calculate
      your result properly for N=2^r·N'.

      If you are interested in improving the algorithm, I recommend you look for some info about
      "Fermat's factorization" and other methods like the "quadratic sieve"

      Jose Brox
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