This question is an offshoot of this MSE post.

Let $\sigma$ be the classical sum-of-divisors function. An odd perfect number $N$ is said to be given in Eulerian form if $\sigma(N)=2N$ and $N={q^k}{n^2}$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Descartes, Frenicle, and subsequently Sorli conjectured that $k = 1$ always holds.

If the DFS conjecture on odd perfect numbers is true, then we have: $$q = \frac{\sigma(n^2)}{2n^2 - \sigma(n^2)} = -1 + 2\cdot\frac{n^2}{2n^2 - \sigma(n^2)}$$ $$\frac{q+1}{2} = \frac{n^2}{2n^2 - \sigma(n^2)}.$$

It follows that $$\sigma(n^2) = q\cdot(2n^2 - \sigma(n^2))$$ and $$n^2 = \left(\frac{q+1}{2}\right)\cdot(2n^2 - \sigma(n^2)).$$

Does it now follow that $$\gcd(n^2, \sigma(n^2)) = 2n^2 - \sigma(n^2)?$$

  • 1
    $\begingroup$ Given $\frac{q+1}2=\frac{n^2}{2n^2-\sigma(n^2)}$, the proof should be the same as in your last question. $\endgroup$ – Jaycob Coleman Mar 27 '15 at 18:33
  • $\begingroup$ Thanks for getting back to me, @JaycobColeman! =) $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 27 '15 at 20:15
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    $\begingroup$ You're welcome. $\endgroup$ – Jaycob Coleman Mar 27 '15 at 23:31

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