Silly question, could use some assistance

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?
 

CRGreathouse

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If you add up inverses of odd numbers, can you get an even integer?
It's an open problem. The special case n = 2 corresponds the the existence of odd perfect numbers.
 
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The special case n = 2 corresponds the the existence of odd perfect numbers.
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.
 

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.
 
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.
 
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CRGreathouse

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

My post was addressing the possibility of nonconsecutive odds.
 
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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