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.


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.


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.

  • 1
    $\begingroup$ 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 $8$ $\endgroup$
    – Peter
    Commented Feb 29, 2020 at 7:26
  • $\begingroup$ 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. $\endgroup$ Commented Feb 29, 2020 at 7:28

1 Answer 1


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.


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