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If $N = qn^2$ is an odd perfect number, is it possible to have $q + 1 = \sigma(n)$?

(Note: This blog post was pulled from this [MO link].)

The title says it all.

**Question**

If $N = qn^2$ is an odd perfect number, is it possible to have $q + 1 = \sigma(n)$?

**Heuristic**

From the Descartes spoof, with quasi-Euler prime $q_1$:
$$n_1 = 3003 < \sigma(n_1) = 5376 < q_1 = 22021$$

So it appears that it might be possible to prove that $q + 1 \neq \sigma(n)$.

**Motivation**

If $q + 1 \neq \sigma(n)$, then it follows that
$$I(q) + I(n) \neq \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$
from which we obtain
$$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q},$$
since the reverse inequality
$$\frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < I(q) + I(n)$$
will violate the inequality $I(q) < \sqrt[3]{2} < I(n)$ (see this [paper]). (Note that the problematic case $n < q < \sigma(q) < \sigma(n)$ is ruled out by Brown's recently announced proof for $q < n$.)

But the inequality
$$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$
implies that the biconditional
$$q < n \Longleftrightarrow \sigma(q) < \sigma(n)$$
holds.

This biconditional is then a key ingredient in the proof of the main result in this [arXiv preprint].

The methods in that preprint are only sufficient to rule out $\sigma(q) = q + 1 = \sigma(n)$ if $3$ does not divide $n$, since we obtain
$$2.799 \approx 1 + 2^{\frac{\log(6/5)}{\log(31/25)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(5))}{\log(I(5^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$
$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$
(where the smallest prime factor $u$ of $N$ satisfies $u \geq 5$), whence we arrive at a contradiction.

**Further Considerations**

If $\sigma(q) = q + 1 = \sigma(n)$ and $3 \mid n$, then the same method does not force a contradiction, because we then have
$$2.7199 \approx 1 + 2^{\frac{\log(4/3)}{\log(13/9)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(3))}{\log(I(3^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$
$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$
where $u$ is the smallest prime factor of $N$.