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22751it's just a rough draft,...

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  • leavemsg1
    Jun 5, 2011
      but I think it's fine.

      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!

      I have it posted on my website... www.oddperfectnumbers.com
      along with some other proofs. Enjoy!

      Bill Bouris
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