If $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, then it is somewhat trivial to prove that
$$I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < I(n),$$
which led Dris to conjecture that $q^k < n$. Brown, Starni, and Dris has shown that $q^k < n$ follows from the truth of the Descartes-Frenicle-Sorli Conjecture that $k=1$. In fact, it is known (by work of Dris) that
$$k = 1 \implies \sigma(q^k) < n,$$
where $\sigma(x)$ is the sum of divisors of the positive integer $x$. The problem is to prove $q^k < n$ unconditionally. (In other words, it would suffice to prove the implication $k > 1 \implies q^k < n$.)

Article
On a Conjecture of Dris Regarding Odd Perfect Numbers

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