(Note: I asked this question in MSE this June 2018 but did not receive any responses there. I have therefore cross-posted it here, hoping that it gets answered.)

Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the deficiency of $z$ by $D(z):=2z-\sigma(z)$, and the sum of the aliquot divisors of $z$ by $s(z):=\sigma(z)-z$.

If $n$ is odd and $\sigma(n)=2n$, then $n$ is said to be an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the form $n = p^k m^2$, where $p$ is the special / Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Starting from the fundamental equality
$$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)}$$
one can derive
$$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)} = \gcd(m^2, \sigma(m^2))$$
so that we ultimately have
$$\frac{D(m^2)}{s(p^k)} = \frac{2m^2 - \sigma(m^2)}{\sigma(p^k) - p^k} = \gcd(m^2, \sigma(m^2))$$
and
$$\frac{s(m^2)}{D(p^k)/2} = \frac{\sigma(m^2) - m^2}{p^k - \frac{\sigma(p^k)}{2}} = \gcd(m^2, \sigma(m^2)),$$
whereby we obtain
$$\frac{D(p^k)D(m^2)}{s(p^k)s(m^2)} = 2.$$
Note that we also have (**Equation A**)
$$\frac{2D(m^2)s(m^2)}{D(p^k)s(p^k)} = \bigg(\gcd(m^2, \sigma(m^2))\bigg)^2.$$
Lastly, notice that we can easily get
$$\sigma(p^k) \equiv k + 1 \equiv 2 \pmod 4$$
so that it remains to consider the possible equivalence classes for $\sigma(m^2)$ modulo $4$. Since $\sigma(m^2)$ is odd, we only need to consider two.

Here is my question:

Which equivalence class of $\sigma(m^2)$ modulo $4$ makes

Equation Auntenable?

I know that the answer must somehow depend on the equivalence class of $p$ and $k$ modulo $8$, but as I only know that $p \equiv k \equiv 1 \pmod 4$, I am stuck.

**UPDATED September 19 2018 (Manila time)** After considering various cases, I think I am able to prove that:

- If $p \equiv k \equiv 1 \pmod 8$, then $\sigma(m^2) \equiv 3 \pmod 4$ is impossible.
- If $p \equiv 1 \pmod 8$ and $k \equiv 5 \pmod 8$, then $\sigma(m^2) \equiv 1 \pmod 4$ is impossible.
- If $p \equiv 5 \pmod 8$ and $k \equiv 1 \pmod 8$, then $\sigma(m^2) \equiv 1 \pmod 4$ is impossible.
- If $p \equiv k \equiv 5 \pmod 8$, then $\sigma(m^2) \equiv 3 \pmod 4$ is impossible.