Further to this MSE question, I would like to pose a follow-up inquiry:
If $n \in \mathbb{N}$ and $(\sigma(n) - n) \mid (n - 1)$, does it follow that $n$ and $\sigma(n)$ would have to be coprime, so that $n$ is a solitary number (by Greening's Theorem)?
user128932 commented in the linked MSE question that:
If $nA - \sigma(n)B = 1$ with $A = B + 1$, then the above divisibility relation would work.
Substituting $A = B + 1$, I get:
$$n(B + 1) - \sigma(n)B = 1$$
so that
$$nB + n - \sigma(n)B = 1.$$
Consequently:
$$n - 1 = B(\sigma(n) - n),$$
which gives:
$$(\sigma(n) - n) \mid (n - 1).$$
Does it indeed work?
I hope I'm not missing something.
If $\sigma(n)$ and $n$ had a common factor $d>1$, then we could write
$d(\sigma(n)/d-n/d)\mid n-1$, but $\gcd(d,n-1)=1$, a contradiction.
Edit:
In general for $k,a\in\Bbb Z^+$, $$\sigma_k(n)-n^k\mid a\implies\gcd(\sigma_k(n),n^k)\mid a,$$ so if $a$ is coprime to $\sigma_k(n)$ or $n^k$, then they must be coprime to each other.