Proving $n \mid \sigma(n^2)$

Question

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 $$\gcd\left(\sigma(q^k)/2,i(q^k)\right)\cdot{i(q^k)} = \left(\gcd(n,i(q^k))\right)^2 \tag{*}$$ holds, where $$i(q^k) = \gcd(n^2,\sigma(n^2)) = \frac{\sigma(N/q^k)}{q^k} = \frac{n^2}{\sigma(q^k)/2}$$ is the index of $N$ at the prime-power $q^k$.

Note that it is also known that $i(q^k) \geq 3$ (a result which has since been improved by several authors), and this implies from Equation $(*)$ that $$\gcd(n,i(q^k)) = 1$$ is untenable.

Here, we will attempt to prove the following:

CLAIM: $\gcd(n,i(q^k)) = n$

MY ATTEMPT AT A PROOF OF THE CLAIM

Suppose to the contrary that $$\gcd(n,i(q^k)) \neq n.$$

Since $\gcd(n,i(q^k)) \mid n$ holds in general (by the definitional property of GCD), then it follows that $$\gcd(n,i(q^k)) < n.$$

But by Bezout's Identity, we have $$\gcd(n,i(q^k)) = an + bi(q^k)$$ where $a,b \in \mathbb{Z}$.

However, we also have $$i(q^k) = \frac{n^2}{\sigma(q^k)/2}.$$

Hence, $$3 \leq an + \frac{bn^2}{\sigma(q^k)/2}= an + bi(q^k) = \gcd(n,i(q^k)) < n.$$

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:

• $a < 0 < b$
• $b < 0 < a$.

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?

Answer 1

This is a partial answer, which proves that $n \nmid \sigma(n^2)$ in fact holds.

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Define the GCDs $$I_{n^2} = \gcd\left(n^2,\sigma(n^2)\right) = \gcd\left(n^2,I_{n^2}\right) = \frac{n^2}{\sigma(q^k)/2}$$ $$G_q = \gcd\left(\sigma(q^k)/2,I_{n^2}\right)$$ $$H_n = \gcd\left(n,I_{n^2}\right).$$

We have the identity $${G_q}\cdot{I_{n^2}}=\left(H_n\right)^2.$$

Since $I_{n^2} \geq 3$ (Dris - JIS, September 2012), it follows that $H_n \neq 1$.

Consequently, it follows that $H_n \geq 3$.

Suppose to the contrary that $n \mid \sigma(n^2)$. This is equivalent to the condition $\sigma(q^k)/2 \mid n$, since we can write $$\frac{\sigma(n^2)}{n}=\frac{q^k n}{\sigma(q^k)/2}.$$

We then have $$H_n = n.$$

By Bezout's Identity, we have $$H_n = \gcd\left(n,I_{n^2}\right) = an + bI_{n^2}$$ where $a,b \in \mathbb{Z}$.

However, we also have $$I_{n^2} = \frac{n^2}{\sigma(q^k)/2}.$$

Dividing both sides of the equation $$H_n = an + b\cdot\left(\frac{n^2}{\sigma(q^k)/2}\right)$$ by $n$, we then obtain $$1 = \frac{H_n}{n} = a + b\cdot\left(\frac{n}{\sigma(q^k)/2}\right)$$ so that $$1 - a = b\cdot\left(\frac{n}{\sigma(q^k)/2}\right).$$

In particular, since it is known that $\sigma(q^k)/2 \neq n$ (see this MSE answer), then $\sigma(q^k)/2 \mid n$ implies that $$\frac{n}{\sigma(q^k)/2} > 1$$

(There is an error beginning this portion of the proof, as the succeeding part only works when $b$ is positive. This literally means that we have, so far, only been able to rule out $a < 0 < b$.)

so that $$1 - a > b$$ or $$a + b < 1.$$

At the same time, we know (from the original post) that either $$a < 0 < b$$ or $$b < 0 < a.$$

If one could show that $$0 < a + b$$ unconditionally holds, then together with $$a + b < 1$$ this would contradict the fact that $a, b \in \mathbb{Z}$.

As pointed out by Tony Kimani Kuria in a comment:

When we have $$b < 0 < a$$ then the following must hold: $$\left|a\right| > \left|\frac{bn}{\sigma(q^k)/2}\right| > \left|b\right|.$$

So we have a contradiction in this case, as we have $$a = \left|a\right| > \left|b\right| = -b$$ so that $0 < a + b$.

By the contrapositive, when $$\left|b\right| > \left|a\right|$$ then the following must hold: $$a < 0 < b.$$ Note by this comment that the condition $$\left|a\right| = \left|b\right|$$ cannot hold. We obtain $$b = \left|b\right| > \left|a\right| = -a$$ so that $0 < a + b$, which together with $a + b < 1$ contradicts $a, b \in \mathbb{Z}$.