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?

__POSTED ANSWER__

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 $q \geq 97$.

$$\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2}$$

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 $q \geq 97$.