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odd perfect numbers

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  • Kermit Rose
    An odd perfect number, N, would have to be of the form N = p1^a1 p2^a2 . . . p_k^a_k where each of the p s are odd primes. The sum of the divisors of N,
    Message 1 of 1 , Sep 27, 2007
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      An odd perfect number, N, would have to be of the form

      N = p1^a1 p2^a2 . . . p_k^a_k

      where each of the p's are odd primes.

      The sum of the divisors of N, ( including N) is the product

      S = (1 + p1 + p1^2 + . . . + p1^a1) ( 1 + p2 + p2^2 + . . . + p2^a2) .
      . . (1 + p_k + p_k ^2 + . . . + p_k ^a_k )


      Thus an odd perfect number must satisfy the requirement

      2 p1^a1 p2^a2 . . . p_k^a_k = (1 + p1 + p1^2 + . . . + p1^a1) ( 1 +
      p2 + p2^2 + . . . + p2^a2) . . . (1 + p_k + p_k ^2 + . . . + p_k ^a_k )

      where each p1, p2, . . . p_k is odd and greater than 2.


      Thus (1 + p1 + p1^2 + . . . + p1^a1) ( 1 + p2 + p2^2 + . . . + p2^a2)
      . . . (1 + p_k + p_k ^2 + . . . + p_k ^a_k ) / [ p1^a1 p2^a2 . . .
      p_k^a_k ] = 2


      Is it possible for (1 + p1 + p1^2 + . . . + p1^a1) ( 1 + p2 + p2^2 + .
      . . + p2^a2) . . . (1 + p_k + p_k ^2 + . . . + p_k ^a_k ) / [
      p1^a1 p2^a2 . . . p_k^a_k ]

      to equal 2 when all the p1, p2, ... p_k are odd and greater than 2?

      Since 2 = 2 mod 4, we see that the numerator must not = 0 mod 4, and so
      exactly one of the a1, a2, ...a_k must be odd, and all the rest must be
      even.

      This shows that an odd perfect number must be an odd power of a prime
      times a square.

      I'm sure that many more elementary observations have been made about
      requirements of odd perfect numbers.

      I anticipate that the proof that odd perfect numbers do not exist would
      follow closely the proof that even perfect numbers are a prime times a
      power of 2.

      I have not yet done a literature research on it. Does anyone here know
      off the top of their head other tidbits about odd perfect numbers?




      Kermit < kermit@... >
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