Let $N = q^k n^2$ be a hypothetical odd perfect number given in Eulerian form. (That is, $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.)
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
This post attempts to prove that the inequality $I(n) > 3/2$ holds unconditionally, using a variation on the factor chain method.
Assume to the contrary that $I(n) \leq 3/2$. As shown in this answer to a closely related question, $I(n) \neq 3/2$. Furthermore, it is also shown there that $I(n) < 3/2$ implies $q = 5$.
We get the series of implications: $$I(n) < \frac{3}{2} \implies q = 5 \implies 3 = \frac{q + 1}{2} \mid \frac{\sigma(q^k)}{2} \mid n^2 \implies 3 \mid n \text{ (since } 3 \text{ is squarefree) }$$ $$\implies \frac{4}{3} = I(3) \leq I(n) < \frac{3}{2} \text{ (no contradiction) }$$ $$3 \mid n^2 \implies 3^2 \mid n^2 \implies \sigma(3^2) \mid \sigma(n^2) \mid N = 5^k n^2$$ $$\implies 13 \mid n^2 \implies 13 \mid n \text{ (since } 13 \text{ is squarefree) }$$ $$\implies \frac{4}{3}\cdot\frac{14}{13} = I(3)I(13) \leq I(n) < \frac{3}{2} \text{ (no contradiction) }$$ $$13 \mid n^2 \implies 13^2 \mid n^2 \implies \sigma(13^2) \mid \sigma(n^2) \mid N = 5^k n^2 \implies 183 = 3\cdot{61} \mid n^2$$ $$\implies 61 \mid n^2 \implies 61 \mid n \text{ (since } 61 \text{ is squarefree) }$$ $$\implies \frac{4}{3}\cdot\frac{14}{13}\cdot\frac{62}{61} = I(3)I(13)I(61) \leq I(n) < \frac{3}{2} \text{ (no contradiction) }$$ $$61 \mid n^2 \implies 61^2 \mid n^2 \implies \sigma(61^2) \mid \sigma(n^2) \mid N = 5^k n^2 \implies 3783 = 3\cdot{13}\cdot{97} \mid n^2$$ $$\implies 97 \mid n^2 \implies 97 \mid n \text{ (since } 97 \text{ is squarefree) }$$ $$\implies \frac{4}{3}\cdot\frac{14}{13}\cdot\frac{62}{61}\cdot\frac{98}{97} = I(3)I(13)I(61)I(97) \leq I(n) < \frac{3}{2} \text{ (no contradiction) }$$ $$97 \mid n^2 \implies 97^2 \mid n^2 \implies \sigma(97^2) \mid \sigma(n^2) \mid N = 5^k n^2 \implies 9507 = 3\cdot{3169} \mid n^2$$ $$\implies 3169 \mid n^2 \implies 3169 \mid n \text{ (since } 3169 \text{ is squarefree) }$$ $$\implies I(3)I(13)I(61)I(97)I(3169) \leq I(n) < \frac{3}{2} \text{ (no contradiction) }$$ $$3169 \mid n^2 \implies 3169^2 \mid n^2 \implies \sigma(3169^2) \mid \sigma(n^2) \mid N = 5^k n^2$$ $$\implies 10045731 = 3\cdot{3348577} \mid n^2$$ $$\implies 3348577 \mid n^2 \implies 3348577 \mid n \text{ (since } 3348577 \text{ is squarefree) }$$ $$\implies I(3)I(13)I(61)I(97)I(3169)I(3348577) \leq I(n) < \frac{3}{2} \text{ (no contradiction) }$$ $$3348577 \mid n^2 \implies 3348577^2 \mid n^2 \implies \sigma(3348577^2) \mid \sigma(n^2) \mid N = 5^k n^2$$ $$\implies 11212971273507 = 3\cdot{3737657091169} \mid n^2$$ $$\implies 3737657091169 \mid n^2 \implies 3737657091169 \mid n \text{ (since } 3737657091169 \text{ is squarefree) }$$ $$\implies I(3)I(13)I(61)I(97)I(3169)I(3348577)I(3737657091169) \leq I(n) < \frac{3}{2} \text{ (no contradiction) }$$ $$3737657091169 \mid n^2 \implies 3737657091169^2 \mid n^2 \implies \sigma(3737657091169^2) \mid \sigma(n^2) \mid N = 5^k n^2$$ $$\implies 13970080531169648034877731 = 3\cdot{181}\cdot{26042690887}\cdot{987900542491} \mid n^2$$ $$\implies 181 \mid n^2 \implies 181 \mid n \text{ (since } 181 \text{ is squarefree) }$$ $$\implies 26042690887 \mid n^2 \implies 26042690887 \mid n \text{ (since } 26042690887 \text{ is squarefree) }$$ $$\implies 987900542491 \mid n^2 \implies 987900542491 \mid n \text{ (since } 987900542491 \text{ is squarefree) }$$ $$\implies I(3)I(13)I(61)I(97)I(3169)I(3348577)I(3737657091169)I(181)I(26042690887)I(987900542491) \leq I(n) < \frac{3}{2} \text{ (no contradiction) }$$ $$181 \mid n^2 \implies 181^2 \mid n^2 \implies \sigma(181^2) \mid \sigma(n^2) \mid N = 5^k n^2 \implies 32943 = 3\cdot{79}\cdot{139} \mid n^2$$ $$\implies 79 \mid n^2 \implies 79 \mid n \text{ (since } 79 \text{ is squarefree) }$$ $$\implies 139 \mid n^2 \implies 139 \mid n \text{ (since } 139 \text{ is squarefree) }$$ But this implies that $$I(3)I(13)I(61)I(97)I(3169)I(3348577)I(3737657091169)I(181)I(26042690887)I(987900542491)I(79)I(139) \leq I(n) < \frac{3}{2},$$ where we have the approximation $$I(3)I(13)I(61)I(97)I(3169)I(3348577)I(3737657091169)I(181)I(26042690887)I(987900542491)I(79)I(139) \approx 1.512675290866383731626612,$$ which is a contradiction.
We therefore conclude that the inequality $$I(n) > \frac{3}{2}$$ must hold unconditionally.
Here are my inquiries:
QUESTIONS: Is this argument sound? If not, can it be mended so as to produce a valid proof?
