# Thread: twenty four and sum of divisors

1. ## twenty four and sum of divisors

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

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

Maybe that helps?

2. Suppose $\displaystyle d\mid n \implies \exists \; a\in\mathbb{N} \text{ s.t. } ad=n\implies ad\equiv -1 \mod{24}$.

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

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

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

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

3. Nice proof, just why is this fact true?

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

4. Originally Posted by EinStone
Nice proof, just why is this fact true?

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

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