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Math Help - twenty four and sum of divisors

  1. #1
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    twenty four and sum of divisors

    If n ≡ −1 (mod 24), then 24 divides \sigma (n) = \sum_{d|n}d

    I have proven already that if  n = \prod_{j=1}^s p_j^{a_j} is the prime factorization of n, then
    \sigma (n) = \prod_{j=1}^s \frac{p_j^{a_j+1}-1}{p_j-1}.

    Maybe that helps?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Suppose  d\mid n \implies \exists \; a\in\mathbb{N} \text{ s.t. } ad=n\implies ad\equiv -1 \mod{24}.

    Finding an inverse modulo  24 is special because if it exists,  d^{-1}\equiv d\mod{24} .

    Therefore  ad\equiv-1\mod{24}\implies a\equiv -d^{-1}\equiv -d \mod{24} \implies a=24k-d .

    Last but not least you need to show  n is not square. I'll let you take a stab at that. (Hint: Consider  n modulo  4 .)

    So we have then that  \displaystyle \sum_{d\mid n} d = \sum_{\substack{d\mid n\\d<\sqrt{n}}} (d+\frac{n}{d}) = \sum_{\substack{d\mid n\\d<\sqrt{n}}} 24k_d \equiv 0\mod{24}
    Last edited by chiph588@; December 12th 2010 at 07:53 PM.
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  3. #3
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    Nice proof, just why is this fact true?

     \sum_{d\mid n} d = \sum_{\substack{d\mid n\\d<\sqrt{n}}}  (d+\frac{n}{d})
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by EinStone View Post
    Nice proof, just why is this fact true?

     \sum_{d\mid n} d = \sum_{\substack{d\mid n\\d<\sqrt{n}}}  (d+\frac{n}{d})
    Well if  d\mid n and  d < \sqrt{n} , then  \frac{n}{d} > \sqrt{n} .

    So when we pair up  d and  \frac{n}{d} we have to make sure not to use the same  d twice, so we stop at  \sqrt{n} since by now every  d dividing  n has been accounted for.
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