twenty four and sum of divisors

**EinStone** · March 23rd 2010, 01:07 AM

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?

**chiph588@** · March 23rd 2010, 09:45 AM

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 $\displaystyle \sum_{d\mid n} d = \sum_{\substack{d\mid n\\d<\sqrt{n}}} (d+\frac{n}{d}) = \sum_{\substack{d\mid n\\d<\sqrt{n}}} 24k_d \equiv 0\mod{24}$

**EinStone** · March 23rd 2010, 03:09 PM

Nice proof, just why is this fact true?

$\sum_{d\mid n} d = \sum_{\substack{d\mid n\\d<\sqrt{n}}} (d+\frac{n}{d})$

**chiph588@** · March 23rd 2010, 04:09 PM

Originally Posted by EinStone

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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