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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?

Jose Arnaldo Bebita Dris
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    This is not an answer. Just an observation. Observe that $a+\frac{bn}{σ(qk)/2}>0$ and since either a or b is positive and the other one negative, we can conclude that one of these two cases must hold: Case 1: If a is negative and b positive. In this case the following absolute values must hold $|a|<|\frac{bn}{σ(qk)/2}|$ . This will ensure that $a+\frac{bn}{σ(qk)/2}$ remains a value greater than 0. Case 2: If a is positive and b is negative then the following must hold $|a|>|\frac{bn}{σ(qk)/2}|$ . This will ensure that $a+\frac{bn}{σ(qk)/2}$ remains greater than 0. – User4576283 Apr 13 '23 at 14:36
  • Thank you very much for sharing that observation, @User4576283! =) – Jose Arnaldo Bebita Dris Apr 13 '23 at 15:58
  • I think one can use the Triangle Inequality to get improved bounds. – Jose Arnaldo Bebita Dris Apr 13 '23 at 16:00
  • In the second case, $a = 0$ cannot occur. Editing the question now to take note of this change. – Jose Arnaldo Bebita Dris Apr 14 '23 at 03:42
  • @User4576283: Please write your comment as an actual answer, including full details as needs be, so that this particular question does not remain in the unanswered queue, and furthermore, in order for myself to upvote your answer, as your approach certainly adds perspective to my method. – Jose Arnaldo Bebita Dris May 03 '23 at 15:16

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