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Silly question, could use some assistance

  1. Feb 27, 2009 #1
    If you add up inverses of odd numbers, can you get an even integer?
    Of course, I mean you only use each odd integer once... so 1/3 * 6 doesn't work.

    I remember seeming to think that if this could be proved impossible, one could provide an elementary proof of Fermat's last theorem. Undoubtedly false then as it is now, but...

    Thoughts? Is this a rabbit hole? Or can you prove / disprove the adding up inverses of odd integers thing?
     
  2. jcsd
  3. Feb 27, 2009 #2

    CRGreathouse

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    It's an open problem. The special case n = 2 corresponds the the existence of odd perfect numbers.
     
  4. Feb 27, 2009 #3
    Could you elaborate on that? I don't see how

    [tex]\frac{1}{2n+1}+\frac{1}{2m+1}=2k[/tex]

    could have any nontrivial solution.
     
  5. Feb 27, 2009 #4

    CRGreathouse

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    Sorry for not being clear. Exhibiting an odd perfect number would give a set of odd integers whose inverses would sum to 2. I didn't mean for there to be only two terms. In fact, the number of terms for this special case would be at least 3^8 (75 - 16 + 1) = 393,660.
     
  6. Feb 27, 2009 #5
    It is known--and found in elementary books--that the sum of successive integers:

    [tex] \sum_{n=1}^{n=k}\frac{1}{n}[/tex] is never an integer for k greater than 1.
     
    Last edited: Feb 27, 2009
  7. Feb 27, 2009 #6

    CRGreathouse

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    I believe that's even known starting from an arbitrary positive integer, not just 1.

    My post was addressing the possibility of nonconsecutive odds.
     
  8. Feb 27, 2009 #7
    Oh yes, I don't require that they be consecutive. Sorry for the confusion.

    I'm frustrated that I can't remember how this was connected to FLT. I sincerly doubt the relationship was even correct, much less that it would have been an iff chain the likes of which could be used to prove anything.

    Good times...
     
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