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Let $n = p^k m^2$ be an odd perfect number given in Eulerian form (i.e. $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$). Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.

This question (and the answer contained therein) presented a proof of the following proposition, the first conditional on the hypothesis that $\sigma(m^2)/p^k$ is a square, and the second one was unconditional:

If $n = p^k m^2$ is an odd perfect number with special prime $p$, then $m^2 - p^k$ is not a square.

In both proofs, the equation $p^k + 1 = 2m$ was obtained, and was shown to lead to a contradiction.

Here are my questions:

(1) Does the contradiction imply that the inequation $p^k + 1 \neq 2m$ holds for all odd perfect numbers $n = p^k m^2$, in general?

(2) If the answer to Question (1) is YES, which of the following inequalities hold in general for an odd perfect number?

(a) $p^k + 1 < 2m$; or

(b) $p^k + 1 > 2m$?

MY ATTEMPT

I figured this much from this closely related MathOverflow question, where we tried to disprove the Dris Conjecture (that $p^k < m$) and therefore the Descartes-Frenicle-Sorli Conjecture (that $k=1$) in one fell swoop. However, we were unsuccessful, as Pascal Ochem pointed out that our method required that $p^k \geq 2m - 1$.

Jose Arnaldo Bebita Dris
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    Would the quick downvoter care to explain his/her downvote please? Some form of feedback, hopefully constructive, would go a long way towards improving future questions/posts of mine. – Jose Arnaldo Bebita Dris May 09 '21 at 08:24

1 Answers1

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This is a partial answer, which proves that $p^k + 1 \neq 2m$ holds in general for all odd perfect numbers.

Let $n = p^k m^2$ be an odd perfect number with special prime $p$.

Suppose to the contrary that $p^k + 1 = 2m$.

Then $m^2 - p^k = m^2 - (2m - 1) = m^2 - 2m + 1 = (m - 1)^2$, which contradicts the proposition in the OP.

Jose Arnaldo Bebita Dris
  • 8,471