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22752Re: [PrimeNumbers] it's just a rough draft,...

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  • Peter Kosinar
    Jun 5, 2011
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      Bill said:
      > let (2^p +1)/3 = m, and let p be prime.
      >
      > assume that (2^p +1)/3 = a*b, then 2^p +1 = 3ab, and 2^p
      > = 3ab -1; 2^(p-1)= (3ab-1)/2, and using algebra 2^(p-1)-1
      > = (3ab-3)/2 and since p | [2^(p-1)-1] then p | [(3ab-3)/2]
      > such that either p=3 or p=[(ab-1)/2]; 2^p = 2^((ab-1)/2)
      > and 2^p +/-1 = 2^((ab-1)/2) +/- 1 such that ab | 2^p +/- 1
      > since ab | 2^((ab-1)/2) +/- 1 from Euler's criteria, and
      > that's impossible, so 'm' must be prime; nice!

      p = 29, a = 59, b = 3033169. What's wrong in the proof?
      Hint: "p | X*Y => (p=X or p=Y)" is false.

      Peter
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