Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
Since the divisor sum $\sigma$ is multiplicative and $N = q^k n^2$ is perfect (with $\gcd(q,n)=1$), we obtain $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2.$$
This implies that $$\frac{\sigma(n^2)}{n}=\frac{q^k n}{\sigma(q^k)/2}. \tag{*}$$
If $n \mid \sigma(n^2)$, then the LHS of Equation (*) is an integer. Hence, the RHS is also an integer. Consequently, we have $$\dfrac{\sigma(q^k)}{2} \mid q^k n.$$ But we know that $$\gcd\Bigg(q^k,\dfrac{\sigma(q^k)}{2}\Bigg) = \gcd(q^k,\sigma(q^k)) = 1.$$ It follows that $\sigma(q^k)/2 \mid n$.
Conversely, if $\sigma(q^k)/2 \mid n$, then the RHS of Equation (*) is an integer. Hence, the LHS is also an integer. Hence, we conclude that $$n \mid \sigma(n^2).$$
We can now state the following proposition:
THEOREM: If $q^k n^2$ is an odd perfect number with special prime $q$, then $\sigma(q^k)/2 \mid n$ if and only if $n \mid \sigma(n^2)$.
Here is my:
INQUIRY: Is this argument logically sound? If this is not correct, how can it be mended so as to produce a valid proof?
