# A question on logic (related to odd perfect numbers)

Let $$\sigma(x)$$ denote the sum of divisors of the positive integer $$x$$. If $$\sigma(x)=2x$$, then $$x$$ is called a perfect number.

An odd perfect number $$n$$ is said to be given in the so-called Eulerian form $$n = p^k m^2$$ where $$p$$ is the special/Euler prime satisfying $$p \equiv k \equiv 1 \pmod 4$$ and $$\gcd(p,m)=1$$. It is currently unknown whether there is an odd perfect number, despite extensive computer searches.

MOTIVATION FOR MY INQUIRY

Suppose, for the sake of our discussion, that I have the following abstract for a paper which I intend to submit to a journal (which summarizes some results about odd perfect numbers):

In this article, we consider the various possibilities for $$p$$ and $$k$$ modulo $$16$$, and show conditions under which the respective congruence classes for $$\sigma(m^2)$$ (modulo $$8$$) are attained, if $$p^k m^2$$ is an odd perfect number with special prime $$p$$. We prove that

• $$\sigma(m^2) \equiv 1 \pmod 8$$ holds only if $$p+k \equiv 2 \pmod {16}$$.
• $$\sigma(m^2) \equiv 3 \pmod 8$$ holds only if $$p-k \equiv 4 \pmod {16}$$.
• $$\sigma(m^2) \equiv 5 \pmod 8$$ holds only if $$p+k \equiv 10 \pmod {16}$$.
• $$\sigma(m^2) \equiv 7 \pmod 8$$ holds only if $$p-k \equiv 4 \pmod {16}$$.

We express $$\gcd(m^2,\sigma(m^2))$$ as a linear combination of $$m^2$$ and $$\sigma(m^2)$$. We also consider some applications under the assumption that $$\sigma(m^2)/p^k$$ is a square. Lastly, we prove a last-minute conjecture under this hypothesis.

Suppose further that the proofs of the results so presented are logically sound and correct.

QUESTIONS

Here are my questions:

(1) Does it follow that $$\sigma(m^2) \equiv 3 \pmod 8$$ and $$\sigma(m^2) \equiv 7 \pmod 8$$ are both untenable?

(2) Or does it only follow that the condition $$p - k \equiv 4 \pmod {16}$$ cannot occur?

UPDATED MARCH 01, 2020 (12:33 PM Manila time)

Apologies for the inadvertent bump! But here is the link to the preprint of the article under discussion, in case anyone is interested.

• I see no reason for $(1)$ or $(2)$ under the given informations. But clearly, if $$p-k\equiv 4\mod 16$$ should be impossible this implies that $\sigma(m^2)$ cannot be $3$ or $7$ modulo $8$ Commented Feb 29, 2020 at 7:26
• Thank you for your time, attention and comment, @Peter! Please write out your last comment as an actual answer so that I can accept it. Commented Feb 29, 2020 at 7:28

If $$p-k\equiv 4\mod 16$$ is impossible then clearly $$\sigma(m^2)\equiv 3\mod 8$$ and $$\sigma(m^2)\equiv 7\mod 8$$ are impossible as well since for those congruences $$p-k\equiv 4\mod 16$$ is necessary.