The journey of a thousand miles begins with a single step.
(Preamble: This blog post is lifted verbatim from https://arnienumbers101.blogspot.com/2023/05/a-journey-of-thousand-miles-begins-with.html .) https://rasputinnumbers.blogspot.com/2023/05/some-new-results-on-odd-perfect-numbers.html https://arnienumbers.blogspot.com/2023/05/dris-conjecture-holds-if-and-only-if.html In what follows, we will denote the classical sum of divisors of the positive integer $x$ by $$\sigma(x)=\sigma_1(x).$$ We will also denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. We start with a minor technical lemma: Lemma 1: If $N = p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then $I(m^2) < \zeta(2)$ if and only if $p = 5$ and $k \neq 1$. Proof: Suppose that $I(m^2) < \zeta(2)$. Since $2(p - 1)/p < I(m^2)$, then it follows that $$\frac{2(p - 1)}{p} < \zeta(2).$$ Hence, we derive $$p < \frac{12}{12 - {\pi}^2} < 6,$$ from which we infer that $p = 5$ (since $p$ is a