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](https://en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers). 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](http://emis.impa.br/EMIS/journals/INTEGERS/papers/n39/n39.pdf) (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](https://mathoverflow.net/questions/310919) 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:**

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> 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?
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> 
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**Reference:**
[The Abundancy index of divisors of odd perfect numbers – Part III](http://nntdm.net/papers/nntdm-23/NNTDM-23-3-053-059.pdf)