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25339Re: Yet another factoring puzzle

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  • djbroadhurst
    Aug 24, 2013
      --- In primenumbers@yahoogroups.com, "Kevin Acres" <research@...> wrote:

      > > Definition: Let F(n) = ((5^n-9)/4)^2-5 for integer n > 0.
      > > Exercise 2: For odd n > 1, prove that F(n)/4 is composite.
      >
      > I did work out an algorithm to derive one divisor of F(n)/4 for odd n, but
      > that's probably a long way from proving it composite.

      What is needed is a formula: F(n)=L(n)*M(n). Then it suffices
      to show that each of L(n) and M(n) has an odd prime divisor
      for odd n > 1. Here is a clue:

      4*F(11) = 24414063^2 - 15625^2

      David
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