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Let $N = q^k n^2$ be an odd perfect number with special / Euler prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

In what follows, we let $$I(x) = \frac{\sigma(x)}{x}$$ denote the abundancy index of $x \in \mathbb{N}$. (The function $\sigma(x)$ is the sum of divisors of $x$.)

In this answer to a closely related question, it is proven that $$q \neq 1049$$ since $$q = 1049$$ would imply that $$105 \mid 525 = \frac{1049 + 1}{2} = \frac{q + 1}{2} \mid N,$$ contradicting the fact that $3 \cdot {5} \cdot {7} = 105 \nmid N$.

My question here is as follows:

What are the implications of $q \neq 1049$ with respect to inequalities for the abundancy indices of $q^k$ and $n^2$?

MY ATTEMPT

I have posted my own attempt as an answer to my question below.

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1 Answer 1

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Let $N = q^k n^2$ be an odd perfect number with special / Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Since $N$ is perfect and $\gcd(q,n)=1$, we have $$2 = I(N) = I(q^k n^2) = I(q^k)I(n^2).$$

In particular, since $k \geq 1$, we know that $$\frac{q+1}{q}=I(q) \leq I(q^k)$$ so that $$I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$

Likewise, from the formula for the abundancy index of a prime power, we can obtain the upper bound $$I(q^k) = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1},$$ so that we get the lower bound $$\frac{2(q-1)}{q} < \frac{2}{I(q^k)} = I(n^2).$$ Summarizing, we get the bounds $$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q - 1} < \frac{2(q-1)}{q} < I(n^2) \leq \frac{2q}{q+1}.$$

Now, since it is known that $q \neq 1049$, we have two cases to consider:

Case 1: $q < 1049$

This implies that $$\frac{1}{1049} < \frac{1}{q} \implies 1 + \frac{1}{1049} < 1 + \frac{1}{q} \implies \frac{2}{1 + \frac{1}{q}} < \frac{2}{1 + \frac{1}{1049}} \implies I(n^2) \leq \frac{2q}{q+1} < \frac{1049}{525}$$ where $$\frac{1049}{525} = 1.99\overline{809523}.$$ This further means that $$I(q^k) = \frac{2}{I(n^2)} > \frac{1050}{1049}.$$

Case 2: $q > 1049$

This yields $$\frac{1}{q} < \frac{1}{1049} \implies 1 - \frac{1}{q} > 1 - \frac{1}{1049} = \frac{1048}{1049} \implies \frac{q}{q - 1} < \frac{1049}{1048},$$ which further implies that $$I(q^k) < \frac{q}{q - 1} < \frac{1049}{1048},$$ and $$I(n^2) = \frac{2}{I(q^k)} > \frac{2\cdot(1048)}{1049)} = \frac{2096}{1049} \approx 1.998093422306959.$$

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