On a GCD approach to odd perfect numbers

Question

Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Let $\sigma(z)$ denote the classical sum of divisors of the positive integer $z$.

Define the GCDs $$G = \gcd(\sigma(p^k),\sigma(m^2))$$ $$H = \gcd(m,\sigma(m^2))$$ and $$I = \gcd(m^2,\sigma(m^2)).$$

Recall that we must have $$\frac{\sigma(p^k)}{2}\cdot\frac{\sigma(m^2)}{p^k} = m^2,$$ and that $$I = \gcd(m^2,\sigma(m^2)) = \frac{\sigma(m^2)}{p^k}.$$

One can show that $$G = \gcd(\sigma(p^k)/2,\sigma(m^2)/p^k) = \frac{\left(\gcd(\sigma(p^k)/2, m)\right)^2}{\sigma(p^k)/2}$$ and $$H = \gcd(m,\sigma(m^2)/p^k) = \frac{m}{\sigma(p^k)/2}\cdot\gcd(\sigma(p^k)/2,m),$$ whence we obtain $$G \times I = H^2.$$ (Notice further that the divisibility constraints $G \mid H$ and $H \mid I$ must hold.)

Trivially, we must have $H \mid m$ and $G \mid \sigma(p^k)/2$.

This means that there exist positive integers $x$ and $y$ such that $$m = Hx$$ and $$\sigma(p^k)/2 = Gy.$$

We obtain $$m^2 = H^2 x^2$$ $$\sigma(p^k)/2 = Gy$$ $$\frac{m^2}{\sigma(p^k)/2} = I = \frac{H^2}{G}\cdot\frac{x^2}{y},$$ from which it follows that $$y = x^2.$$

It can be shown that $$x = \frac{\sigma(p^k)/2}{\gcd(\sigma(p^k)/2,m)}.$$

Here is my question:

Is it possible to rule out $x > 1$?

Notice that I am trying to prove that the divisibility constraint $\sigma(p^k)/2 \mid m$ holds, which is true if and only if $x = 1$.

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Updated (January 03, 2024)

I am getting $$\sigma(p^k)/2 = Gx^2$$ $$m = Hx$$ $$GJ = H$$ where $$J = \frac{m}{\gcd(\sigma(p^k)/2,m)},$$ so that
$$\frac{\sigma(p^k)}{2x^2} = G$$ $$\frac{m}{x} = H$$ $$\frac{\sigma(p^k)}{2x^2}\cdot{J} = \frac{m}{x}$$ and therefore $$J = \frac{m}{\sigma(p^k)/2}\cdot\frac{\sigma(p^k)/2}{\gcd(\sigma(p^k)/2,m)}.$$ I am not too sure whether this, indeed, forces $\sigma(p^k)/2 \mid m$. Alas, this is where I get stuck!

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Last Update (January 03, 2024)

Subtracting, I obtain $$H - G = \frac{\gcd(\sigma(p^k)/2, m)}{\sigma(p^k)/2}\left(m - \gcd(\sigma(p^k)/2, m)\right).$$

I do think this implies $\sigma(p^k)/2 \mid m$, since either we have $$\sigma(p^k)/2 \mid \gcd(\sigma(p^k)/2, m)$$ which forces $\sigma(p^k)/2 \mid m$, by the definition of GCD; or $$\sigma(p^k)/2 \mid \left(m - \gcd(\sigma(p^k)/2, m)\right)$$ which still forces $\sigma(p^k)/2 \mid m$.

Note: This argument presupposes that $\sigma(p^k)/2 = q$ is prime. But then, if such is the case, then $\sigma(p^k)/2$ is squarefree, by which the required conclusion then follows from the divisibility constraint $\sigma(p^k)/2 \mid m^2$.

Answer 1

As I mentioned in another answer, using GCDs and fractions is usually a bad idea. Here is how I would interpret all of this material.

Let $N$ be an odd perfect number. Write its prime factorization as $N=\prod_{i=1}^{k}p_i^{e_i}$. We may assume $p_1\equiv e_1\equiv 1\pmod{4}$, as usual, and all the other exponents are even. Fix $m:=\prod_{i=2}^{k}p_i^{e_i/2}$.

From $\sigma(N)=2N$ and the fact that for any prime $p$ we have $p\nmid \sigma(p^e)$, we see that if we write $\sigma(m^2)=\prod_{i=1}^{k}p_i^{f_i}$, then $f_1=e_1$.

Set $s:=\prod_{i=2}^{k}p_i^{f_i}$, so that $\sigma(m^2)=p_1^{e_1}s$. Also, fix $t:=m^2/s = \prod_{i=2}^{k}p_i^{e_i-f_i}$. Notice that $\sigma(p_1^{e_1})=2t$.

Now, you defined the quantity $G$, which is $$ G:=\gcd(\sigma(p_1^{e_1}),\sigma(m^2))=\gcd(2t,p_1^{e_1}s)=\gcd(s,t)=\prod_{i=2}^{k}p_i^{\min(e_i-f_i,f_i)}. $$ You also defined the quantity $H$, which is $$ H:=\gcd(m,\sigma(m^2))=\gcd(m,p_1^{e_1}s)=\gcd(m,s)=\prod_{i=2}^{k}p_i^{\min(e_i/2,e_i-f_i)}. $$ Finally, you defined the quantity $I$, which is $$ I:=\gcd(m^2,\sigma(m^2))=\gcd(m^2,p_1^{e_1}s)=\gcd(st,s)=s=\prod_{i=2}^{k}p_i^{f_i}. $$

Thus, these quantities, and every other equality in your post boils down to trivial identities involving minimums in terms of $e_i$ and $f_i$. They provide no new information. For example, the quantity $x$ you define is $$ x:=m/H = \prod_{i=2}^{k}p_i^{\max(0,e_i/2-f_i)}. $$ Your question about whether we can rule out $x>1$, essentially asks whether we can rule out $f_i<e_i/2$. (And now, one should ask: Can you rule that out in the spoof odd perfect numbers? If not, then to rule it out in actual odd perfect numbers one would need to make essential use of "true" primality (and not merely "spoof" primality) in some way. It seems like all of the GCD manipulation is compatible with spoof factorizations, if we replace GCD with its spoof-like counterpart.)