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?