Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e., $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$). ($\sigma(x)$ gives the sum of the divisors of $x$, and $y$ is perfect if $\sigma(y) = 2y$.)
In this paper, it is shown that the inequality $n < q$ is sufficient for Sorli's conjecture that $k = 1$ to hold.
Somebody pointed out to me that it is obvious that the following implication is true:
$$\sigma(n) \leq q^k \Longrightarrow k = 1$$
Is it indeed so? It's not obvious to me for two reasons:
$$n < q^k \Longrightarrow \left\{n < q \Longleftrightarrow k = 1\right\}$$
$$k > 1 \Longrightarrow q < n$$
That is, in order to prove the supposedly obvious claim, it would suffice to show that the condition
$$q < n < q^k$$
never holds.
After getting back to my correspondent, the claim that
$$\sigma(n) \leq q^k \Longrightarrow k = 1,$$
where $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, has been retracted.