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Re: [PrimeNumbers] Question why are 2 and 5 unlike any of the other primes?

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  • mikeoakes2@aol.com
    In a message dated 28/02/04 00:28:21 GMT Standard Time, tftn@earthlink.net ... Your conjecture doesn t stand scrutiny. 2^(2^n)-1 = (2^(2^(n-1)))^2-1 which has
    Message 1 of 2 , Feb 28, 2004
      In a message dated 28/02/04 00:28:21 GMT Standard Time, tftn@...
      writes:


      > In my work with Primes (where 1 isn't taken as the zeroth Prime),
      > I've found that the subset {2,5} has unique properties.
      > Conjecture:
      > Primes of the sort should form 8 unique sets if I'm right about this:
      > 2^n+/-1
      > 2^(2^n)+/-1
      > 5^n+/-2
      > 5^(5^n)+/-2
      > and 4 new types like:
      > 5^(2^n)+/-2
      > 2^(5^n)+/-1
      > are also indicated.
      >

      Your conjecture doesn't stand scrutiny.

      2^(2^n)-1 = (2^(2^(n-1)))^2-1
      which has (2^(2^(n-1)))+/-1 as algebraic factors and so is never prime.

      2^(5^n)+/-1 = (2^(5^(n-1)))^5+/-1
      which has (2^(5^(n-1)))+/-1 as a factor and so is never prime.

      -Mike Oakes

      To Mr Bagula: as already requested, please do not post identically to so many
      groups (4 in this case); I already sent the above reply yesterday to
      numbertheory, and it now has to go to primenumbers as well; people get fed up with
      seeing the same post several times.

      To the primenumbers moderator: since he ignored your request, cannot you
      please from now on suppress these multiple postings?



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