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Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.

My question is as is in the title:

If $p^k m^2$ is an odd perfect number with special prime $p$, is it possible to have $p = k$?

We know that $$p = k \implies \sigma(p^k)/2 \text{ is not squarefree}$$ and that $$p = k \implies p \text{ is bounded.}$$

Alas, this is where we get stuck!

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  • $\begingroup$ Conjecturally, the Descartes-Frenicle-Sorli Conjecture predicts that $k=1$. If this conjecture is true, then it would disprove $p = k$. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Feb 9 at 7:43
  • $\begingroup$ Additionally, note that Chen and Luo proved that $$\sigma(m^2) \equiv 1 \pmod 4 \iff p \equiv k \pmod 8$$ and $$\sigma(m^2) \equiv 3 \pmod 4 \iff p \equiv (k + 4) \pmod 8.$$ This implies that, if $p = k$, then $\sigma(m^2) \equiv 3 \pmod 4$ is untenable. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Feb 9 at 7:50
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    $\begingroup$ This question seems to be very difficult. From the existence of spoofs where the power on the spoof special prime can be nearly arbitrary, we know that there is no purely combinatorial restriction on $k$. Some congruence restrictions (as you mention) can be determined on the cofactor $m$, but not enough to rule out the possibility that $p=k$. An answer could only be found using primality in some essential way. $\endgroup$
    – Pace Nielsen
    Commented Feb 9 at 17:05
  • $\begingroup$ Thank you for your comment, @PaceNielsen! I think that qualifies as an answer. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Feb 11 at 4:11

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