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it's just a rough draft,...

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  • leavemsg1
    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
    Message 1 of 2 , Jun 5, 2011
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      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
    • Peter Kosinar
      ... p = 29, a = 59, b = 3033169. What s wrong in the proof? Hint: p | X*Y = (p=X or p=Y) is false. Peter
      Message 2 of 2 , 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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