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RE: [PrimeNumbers] An observation on Fermat factorizations.

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  • Mc Laughlin, James
    Hello Ed, It is fairly obvious. If F_n = 2^(2^n)+1 = (2^(n+2)k+1)*(2^(n+2)l+1) = 2^(2n+4)k*l + 2^(n+2)(k+l) +1 then 2^(2^n-n-2)=2^(n+2)k*l + (k+l) , so 2^(n+2)
    Message 1 of 6 , Jul 28, 2005
      Hello Ed,

      It is fairly obvious.

      If
      F_n = 2^(2^n)+1 = (2^(n+2)k+1)*(2^(n+2)l+1) = 2^(2n+4)k*l + 2^(n+2)(k+l) +1

      then

      2^(2^n-n-2)=2^(n+2)k*l + (k+l) ,

      so 2^(n+2) | (k+l).


      Jimmy Mc Laughlin.

      http://www.trincoll.edu/~jmclaugh/


      ________________________________

      From: primenumbers@yahoogroups.com on behalf of ed pegg
      Sent: Thu 7/28/2005 12:58 PM
      To: primenumbers@yahoogroups.com; durman@...
      Subject: [PrimeNumbers] An observation on Fermat factorizations.


      Frits Staudt sent me an interesting observation on Fermat factorizations.
      A data source on them: http://www.prothsearch.net/fermat.html

      He added up the k's, and noticed the sum divisible by higher powers of 2.

      F5 is 5 2^7+1 * 52347 2^7+1
      Observation: 5+52347 = 52352 = 409 2^7

      F6 is 1071 2^8+1 * 262814145745 2^8+1
      Observation: 1071 + 262814145745 = 1026617761 2^8

      F7 Observation: 116503103764643 + 11141971095088142685
      = 21761889840218569 2^9

      F8 Observation: 604944512477 +
      45635566267264637582599393652151804972681268330878021767715 =
      22282991341437811319628610181714748521817025552481917129 2^11

      Is this obvious/interesting?

      Ed Pegg Jr



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