Showing $24|(n+1)\implies24|\sigma(n)$ Question:

Show that if $n$ is a positive integer such that $24$ divides into $n + 1$, then $24$ divides the sum of all divisors of $n$ (denoted in number theory by $\sigma(n)$ or $\sigma_1(n)$). 

For example if $n = 95$, then $n + 1 = 96 = 4 \times 24$ and the sum of the divisors of $n$ is $$1 + 5 + 19 + 95 = 120 = 5 \times 24.$$ (Note that the number $n$ is included among its divisors.)

We use a pairing argument, working first modulo $3$ and then modulo $8$.
We are told that $n\equiv -1\pmod{24}$. It follows that $n\equiv -1\pmod{3}$. Split the set of divisors of $n$ into unordered pairs $\{a,b\}$ such that $ab=n$. (Since $n\equiv -1\pmod 3$, the number $n$ is not a perfect square, so every divisor of $n$ is taken care of.)
For any pair $\{a,b\}$ with $ab=n$, one of $a$ and $b$ is congruent to $1$ modulo $3$, and the other is congruent to $-1$. So all pair sums are congruent to $0$. Therefore the sum of all pair sums, that is, the sum of the divisors, is congruent to  $0$ modulo $3$.
The same idea works modulo $8$. We are told that $n\equiv -1\pmod{8}$. If $\{a,b\}$ is an unordered pair with $ab=n$, then either (i) One of $a$ and $b$ is congruent to $1$ modulo $8$, and the other is congruent to $-1$, or (ii) One of $a$ and $b$ is congruent to $3$ modulo $8$, and the other is congruent to $-3$. In either case the sum is congruent to $0$ modulo $8$.
Thus the sum of the divisors of $n$ is divisible by $3$ and by $8$, and therefore by $24$.
