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 $\sigma(x)=\sigma_1(x)$.
It is known that the identity $$\gcd\left(\sigma(q^k)/2,i(q)\right)\cdot{i(q)} = \left(\gcd(n,i(q))\right)^2 \tag{*}$$ holds, where $$i(q) = \gcd(n^2,\sigma(n^2)) = \frac{\sigma(N/q^k)}{q^k} = \frac{n^2}{\sigma(q^k)/2}$$ is the index of $N$ at the prime $q$.
Note that it is also known that $i(q) \geq 3$ (a result which has since been improved by several authors), and this implies from Equation $(*)$ that $$\gcd(n,i(q)) = 1$$ is untenable.
Here, we will attempt to prove the following:
CLAIM: $\gcd(n,i(q)) = n$
MY ATTEMPT AT A PROOF OF THE CLAIM
Suppose to the contrary that $$\gcd(n,i(q)) \neq n.$$
Since $\gcd(n,i(q)) \mid n$ holds in general (by the definitional property of GCD), then it follows that $$\gcd(n,i(q)) < n.$$
But by Bezout's Identity, we have $$\gcd(n,i(q)) = an + bi(q)$$ where $a,b \in \mathbb{Z}$.
However, we also have $$i(q) = \frac{n^2}{\sigma(q^k)/2}.$$
Hence, $$3 \leq an + \frac{bn^2}{\sigma(q^k)/2}= an + bi(q) = \gcd(n,i(q)) < n.$$
Dividing both sides by $n$ yields $$0 < \frac{3}{n} \leq a + \frac{bn}{\sigma(q^k)/2} < 1.$$
Note that $ab > 0$ cannot happen, as $a < 0$ and $b < 0$ would contradict the lower bound, while $a > 0$ and $b > 0$ would contradict the upper bound. Hence, either of the following cases must occur:
- $a \leq 0 < b$
- $b < 0 < a$.
Here is my:
QUESTION: As I am relatively a beginner in the nuances of Bezout's Identity, particularly how it is applied to this problem, I was hoping somebody with that brilliant insight and greater experience may help with ruling out the remaining two cases. Do you see a way to do that?