To prove $6|σ(6n-1) , ∀n∈ \mathbb N$

Question

Let $σ(n)$ denote the sum of all the positive divisors of $n∈ \mathbb N$. I think that $6$ divides $σ(6n-1)$ for all $n∈ \mathbb N$ , but I am not able to prove it. So, a proof of the result (if it is true and I think it is) will be much appreciated.

Answer 1

The sum of divisors is weakly multiplicative, which means that if $\operatorname{gcd}(a,b)=1$, then $\sigma(ab) = \sigma(a)\sigma(b)$. This allows to divide it into its prime-power divisors, eg $\sigma(2^a 3^b \dots) = \sigma(2^a)\sigma(3^b)\dots$.

A number of the form of $6n-1$ must have at least one prime as a divisor of this form, suppose this is $p$. It is easy to show that there must be an odd count of primes of this form, since $5^{2m}=1 \pmod 6$ So we pick one that has an odd power.

We just need to show that if $p^m = 5 \pmod 6$, then the sum of its divisors is a multiple of $6$. Since $p^{2m}=1 \pmod{6}$, it follows that at least one m for the p must be odd.

Accordingly, the sum of divisors of a power of $p$ is $1+p+p^2+p^3+\dots$ When $m$ is odd, we can divide it into $(1+p)(1+p^2+p^4+\dots)$, and the first divisor is a multiple of $6$.

Therefore the whole number must be a multiple of 6.

Aside:

The weakly multiplicative manner of $\sigma()$ is how one finds multiplicatively perfect numbers, for example. $\sigma(120) = \sigma(8) \sigma(3) \sigma(5) = 15\cdot 4 \cdot 6 = 360 = 3\times 120$. It's just about adding factors to the LHS and crossing out what's left.

Answer 2

If $n=6k-1$ then $n$ is not square. (You can easily check this fact.) So if $d\mid n$ then $d\neq n/d$. Furthermore, we get $(d,2)=(d,3)=1$. That is, $d\equiv\pm1 \pmod 6$ so we get $$ d+\frac{n}{d}=d+\frac{6k-1}{d}\equiv d-\frac{1}{d} \equiv 0 \pmod 6. $$

Therefore $6\mid d+n/d$ for all $d\mid n$. Since $n$ is not square, $$ \sigma(n)=\sum_{d\mid n,\,d<\sqrt{n}}d+\frac{n}{d}\equiv 0\pmod 6. $$