<s>[INST] 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.
 [/INST] 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 $8$.
</s>