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Let $\sigma(x)$ be the sum of divisors of the positive integer $x$. If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number. The question of existence of odd perfect numbers is the longest unsolved problem of mathematics.

Euler proved that an odd perfect number, if one exists, must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Broughan, Delbourgo, and Zhou prove in IMPROVING THE CHEN AND CHEN RESULT FOR ODD PERFECT NUMBERS (Lemma 8, page 7) that if $\sigma(n^2)/q^k$ is a square, then the Descartes-Frenicle-Sorli conjecture that $k=1$ holds.

So now suppose that $\sigma(n^2)/q^k$ is a square. This implies that $k=1$, and also that $\sigma(n^2) \equiv 1 \pmod 4$, since $\sigma(n^2)/q^k$ is odd and $q \equiv k \equiv 1 \pmod 4$.

The congruence $\sigma(n^2) \equiv 1 \pmod 4$ then implies that $q \equiv k \pmod 8$. (See this MO post for the details.) Substituting $k=1$, we obtain $$q \equiv 1 \pmod 8.$$

This implies that the lowest possible value for the special prime $q$ is $17$. (That is, this argument breaks the barriers at $q=5$ and $q=13$, under the assumption that $\sigma(n^2)/q^k$ is a square.) Note that, if $q=17$, then $(q+1)/2 = 3^2 \mid n^2$.

Here is my question:

Can we push the lowest possible value from $q \geq 17$, to say, $q \geq 41$ or even $q \geq 97$, using the ideas in this post, and possibly more?

Reference: The Abundancy index of divisors of odd perfect numbers – Part III

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  • $\begingroup$ Isn't that more suitable for MO? $\endgroup$ – enedil Mar 10 at 11:06
  • $\begingroup$ @enedil, I agree. I just wanted to see what the folks here have to say regarding this inquiry before I cross-post it over at MO (after some time, of course). $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 10 at 11:09
  • $\begingroup$ @enedil, I have just posted an answer below a few minutes ago. Note the simplicity in the argument. It is for exactly this same reason that I refrain from asking my questions in MO, because they tend to be not so well-received there (as opposed to here at MSE). $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 10 at 11:40
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If $(q+1)/2$ is an odd square, then $(q+1)/2 \equiv 1 \pmod 8$, so that $q \equiv 1 \pmod {16}$. This rules out $41$, $73$, and $89$.

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

Note that if $$\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2}$$ is a square, then $k=1$ and $\sigma(q^k)/2 = (q+1)/2$ is also a square.

The possible values for the special prime satisfying $q < 100$ and $q \equiv 1 \pmod 8$ are $17$, $41$, $73$, $89$, and $97$.

For each of these values: $$\frac{q_1 + 1}{2} = \frac{17 + 1}{2} = 9 = 3^2$$ $$\frac{q_2 + 1}{2} = \frac{41 + 1}{2} = 21 \text{ which is not a square.}$$ $$\frac{q_3 + 1}{2} = \frac{73 + 1}{2} = 37 \text{ which is not a square.}$$ $$\frac{q_4 + 1}{2} = \frac{89 + 1}{2} = 45 \text{ which is not a square.}$$ $$\frac{q_5 + 1}{2} = \frac{97 + 1}{2} = 49 = 7^2$$

Thus, if $\sigma(n^2)/q^k$ is a square and we could rule out $q=17$, it would follow that $q \geq 97$.

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  • $\begingroup$ Blimey! I missed the possibility that $q=89$. Editing my answer now to reflect this update. $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 10 at 11:57

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