<s>[INST] 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.)
 [/INST] We're looking for divisors $a, b$ such that $a \times b = n \equiv - 1 \pmod{24}$. In order for this to be possible, $a$ and $b$ must both be coprime to 24. Now note the following:
$a \times b \equiv -1 \pmod{24}$
$a^2 + a \times b \equiv a^2 -1 \pmod{24}$
$a(a + b) \equiv (a-1)(a+1) \pmod{24}$
Since $a$ is coprime to 24, and thus $a$ is odd, $(a-1), (a+1)$ must be even numbers. Furthermore, they are consecutive even numbers and thus at least one of them is also divisible by 4. In addition, as $a$ is coprime to 24, one of $(a-1), (a+1)$ must be divisible by 3 because if $a \equiv 1 \pmod{3}$ then $3|(a-1)$ and if $a \equiv 2 \mod{3}$ then $3|(a+1)$. This means that $4 \times 2 \times 3 = 24 | (a+1)(a-1)$. As $a$ is coprime to 24, this implies that $(a + b) \equiv 0 \pmod{24}$ for all $a, b$.
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