(Note: This blog post is taken verbatim from this MSE question and the answer contained therein.)
QUESTION
Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
By the definition of a perfect number, we have
$$\sigma(N) = 2N.$$
Since $\sigma()$ is a multiplicative function, and $\gcd(p,m)=1$, then we obtain
$$\sigma(p^k)\sigma(m^2)=\sigma(p^k m^2)=\sigma(N)=2N=2p^k m^2.$$
This means that
$$\frac{\sigma(p^k)}{2}\cdot\frac{\sigma(m^2)}{p^k}=m^2$$
where we note that both $\sigma(p^k)/2$ and $\sigma(m^2)/p^k$ are odd integers.
Since $m$ is odd, then $m^2 \equiv 1 \pmod 8$, from which we infer (???) that
$$(1) \hspace{0.10in} {\sigma(p^k)/2} \equiv {\sigma(m^2)/p^k} \pmod 8.$$
Here is our:
> INITIAL QUESTION: What are the implications of the assertion $(1)$?
OUR ATTEMPT
Since we have
$$\frac{\sigma(m^2)}{p^k} = \frac{m^2}{\sigma(p^k)/2}$$
then using the fact that $\sigma(p^k)/2$ is odd, we obtain
$$\left(\frac{\sigma(p^k)}{2} \equiv \frac{m^2}{\sigma(p^k)/2} \pmod 8\right)$$
$$\Rightarrow \Bigg(\left(\frac{\sigma(p^k)}{2}\right)^2 \equiv m^2 \pmod 8\Bigg)$$
$$\Rightarrow \Bigg(\left(\frac{\sigma(p^k)}{2}\right)^2 \equiv 1 \pmod 8\Bigg).$$
The congruence $y^2 \equiv 1 \pmod 8$ has the four solutions
$$y \equiv {\pm 1} \pmod 8$$
and
$$y \equiv {\pm 3} \pmod 8,$$
so we are guessing this angle leads to a dead end.
Going back to
$$\left(\frac{\sigma(p^k)}{2}\right)^2 \equiv m^2 \pmod 8,$$
it follows (???) that
$$(2) \hspace{0.10in} \frac{\sigma(p^k)}{2} \equiv {\pm m} \pmod 8.$$
>> In fact, we dare to conjecture that
$$\frac{\sigma(p^k)}{2} \equiv m \pmod 8$$
holds. Alas, we have no proof.
Here is our:
> FINAL QUESTION: We have some doubts regarding the validity of the arguments (in the sections marked with a (???)) that are used to prove Statement $(1)$ and Statement $(2)$. Do the proofs that we have presented hold water? If not, how can they be mended so as to produce a logically correct assertion?
ANSWER
This answer relies on a CAS (Computer Algebra System), notably WolframAlpha.
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Statement $(1)$ is TRUE, since given positive integers $a$ and $b$, if $ab \equiv 1 \pmod 8$, then it follows that
$$\left(a \equiv 1 \pmod 8\right) \land \left(b \equiv 1 \pmod 8\right) \Rightarrow \left(a \equiv b \equiv 1 \pmod 8\right),$$
or
$$\left(a \equiv 3 \pmod 8\right) \land \left(b \equiv 3 \pmod 8\right) \Rightarrow \left(a \equiv b \equiv 3 \pmod 8\right),$$
or
$$\left(a \equiv 5 \pmod 8\right) \land \left(b \equiv 5 \pmod 8\right) \Rightarrow \left(a \equiv b \equiv 5 \pmod 8\right),$$
or
$$\left(a \equiv 7 \pmod 8\right) \land \left(b \equiv 7 \pmod 8\right) \Rightarrow \left(a \equiv b \equiv 7 \pmod 8\right).$$
Thus, we see that in all cases,
$$ab \equiv 1 \pmod 8 \Rightarrow a \equiv b \pmod 8.$$
(Here is the WolframAlpha computational verification.)
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Statement $(2)$ is FALSE. (Here is the WolframAlpha computational verification.) In particular, given positive integers $c$ and $d$, if $c^2 \equiv d^2 \pmod 8$, it does not necessarily follow that $c \equiv {\pm d} \pmod 8$, if $cd \equiv 1 \pmod 2$.
