Prove that if $n+1$ is divisible by $8$, then the sum of $n$'s factors is also divisible by $8$ 
Let $n$ be a positive integer. Prove that if $n+1$ is divisible by $8$, then the sum of $n$'s factors is also divisible by $8$.

If $n+1$ is divisible by $8$, then $n \equiv 7 \pmod{8}$, but I didn't see how to separate it into cases here.
 A: If $d$ is a factor of $n$, then $d \equiv \pm 1, \pm 3 \bmod 8$.
The factors of $n$ come in pairs $d, d'=n/d$, with $d \equiv -d' \bmod 8$, since $n \equiv -1 \bmod 8$.
Note that $d \ne d'$ because $n$ is not a perfect square since the squares are $\equiv 0,1 \bmod 8$.
Therefore, the sum of all factors is $\equiv 0 \bmod 8$.
A: Let $n + 1 = 8m$.
 We show first that if $d$ is a divisor of $n = 8m - 1$, then $d^2 - 1$ is divisible by $8$. Now $d$ must be odd (because $n$ is), so $d - 1$ and $d + 1$ are consecutive even numbers, so one must be a multiple of $4$, and there product $d^2 - 1$ must be a multiple of $8$.
Now $8m - 1$ cannot be a square (because squares are congruent to $0$ or $1 \bmod 8$), so its divisors come in pairs $d, \frac{8m - 1}{d}$. But $d + \frac{8m - 1}{d} = \frac{(d^2-1)+8m}{d}$ is divisible by $8$ (because $d^2 - 1$ and $8m$ are and no factor of $8$ can divide $d$). Hence the sum of all the divisors of $n - 1$ is divisible by $8$.
A: A factorization of $\,n\,$ maps mod $8\,$ to factorization of $\,-1,\,$ by $\,n\equiv -1\pmod{\! 8},\,$ i.e.
$$ dd' = n\,\Rightarrow\, dd'\equiv -1\ \ ({\rm mod}\,\ 8)$$
Further, modulo $\,8\!:\ d\mid 1\,\Rightarrow\, \color{#c00}{d^2\equiv 1},\,$ so $\,\color{#c00}d(\color{#c00}dd'\equiv -1)\,\Rightarrow\, d'\equiv -d\,\Rightarrow\, d+d'\equiv 0$
Thus the cofactors $\,d,d'$ cancel out in the sum (and $d'\not\equiv d\,$ by $\,-1$ isn't square mod $8)$
Remark $ $ Pairing via the cofactor involution is  closely related to Wilson's theorem in group form.
