I've played around with some identities and came up with:
$\sigma_s(n)(1-\frac{1}{n^s}) = 0 \pmod{k}$
(for $n$ and $k$ positive integers, and $s$ an integer)
With the conditions that 1) $n$ is invertible modulo k, 2) $\phi(k)$ divides $2s$.
In other words: if the order of $k$ divides $2s$ then either $k$ divides the divisor sum $\sigma_s(n)$ or $n$ is a quadratic residue modulo $k$.
E.g. $n=6, s=5, k = 11$
$\sigma_5(6)(1 - \frac{1}{6^5}) = 11 \cdot 732 \cdot \frac{7775}{7776} \equiv 0 \pmod{11}$
Derivation:
$\sigma_s(n) = \displaystyle \sum_{d | n} d^s$
$\sigma_{-s}(n) = \displaystyle \sum_{d | n} d^{-s}$
$(\sigma_{-s}(n) ) \cdot n^s = \sigma_s(n)$
$\sigma_{s}(n) - \sigma_{-s}(n) = \displaystyle \sum_{d | n} d^{-s} (d^{2s} - 1)$
Therefore if the order divides $2s$ and $k$ doesn't divide $n$, each term of the last sum is zero modulo $k$.
I guess this is a general manipulation of: $\sigma_s(n) \equiv \sigma_{s+t}(n) \pmod {k}$ (when $\phi(k) | t$).
Finally $\sigma_{s}(n) - \sigma_{-s}(n) = \sigma_s(n) (1 - \frac{1}{n^s})$
Is this manipulation correct? Does it fall under some more general category of techniques/theorems?