Let $\sigma(M)$ denote the sum of divisors of the positive integer $M$. If $N$ is odd and $\sigma(N)=2N$, then $N$ is called an odd perfect number. Euler showed 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 $\gcd(q,n)=1$.
Since $q$ is prime and $q \equiv k \equiv 1 \pmod 4$, then we have the divisibility constraint $$q+1 = \sigma(q) \mid \sigma(q^k) \mid \sigma(N) = 2N$$ from which it follows that $(q+1)/2 \mid N = q^k n^2$, which implies that $(q+1)/2 \mid n^2$.
Likewise, since $\gcd(q^k,\sigma(q^k))=1$, then it can be shown that $$q \mid q^k \mid \sigma(n^2).$$
Here is my question:
If $q^k n^2$ is an odd perfect number with special prime $q$, then does $(q+1)/2 \mid \sigma(n^2)$ hold?
MY ATTEMPT
Let $N = q^k n^2$ be an odd perfect number with special prime $q$.
Assume to the contrary that $(q+1)/2 \mid \sigma(n^2)$. Since $(q+1)/2 \mid n^2$, this assumption implies that $$\frac{q+1}{2} \mid \gcd(n^2, \sigma(n^2)).$$
It is known that $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)}=\gcd(n^2, \sigma(n^2)),$$ where $D(x) = 2x-\sigma(x)$ is the deficiency and $s(x) = \sigma(x)-x$ is the sum of the aliquot divisors of the positive integer $x$.
Alas, this is where I get stuck.