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PATCH: 2^n+1 = np?

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  • Andrey Kulsha
    Hello! Sorry, there was a little error. At first there was a proof for such fact that p is composite; then I obtained yet stronger result (namely 3|p), but I
    Message 1 of 1 , May 30, 2001
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      Hello!

      Sorry, there was a little error. At first there was a proof
      for such fact that p is composite; then I obtained yet
      stronger result (namely 3|p), but I didn't update the
      proof... :-)

      ***
      >Proof:
      >
      >Let p=(2^n+1)/n is prime greater than 3. n must be odd, so
      >2^n+1 is divisible by 3; p is prime, hence, n is divisible
      >by 3.
      ***

      You should read:

      ***
      Proof:

      Let p=(2^n+1)/n isn't divisible by 3. n must be odd, so
      2^n+1 is divisible by 3; hence, n is divisible by 3.
      ***

      Best wishes,

      Andrey
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