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(Preamble: This question is tangentially related to this earlier one.)

Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the deficiency of $z$ by $D(z):=2z-\sigma(z)$, and the sum of the aliquot divisors of $z$ by $s(z):=\sigma(z)-z$. Finally, let the abundancy index of $z$ be denoted by $I(z):=\sigma(z)/z$.

If $n$ is odd and $\sigma(n)=2n$, then $n$ is said to be an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the form $n = p^k m^2$, where $p$ is the special / Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Starting from the fundamental equality $$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)}$$ one can derive $$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)} = \gcd(m^2, \sigma(m^2))$$ so that we ultimately have $$\frac{D(m^2)}{s(p^k)} = \frac{2m^2 - \sigma(m^2)}{\sigma(p^k) - p^k} = \gcd(m^2, \sigma(m^2))$$ and $$\frac{s(m^2)}{D(p^k)/2} = \frac{\sigma(m^2) - m^2}{p^k - \frac{\sigma(p^k)}{2}} = \gcd(m^2, \sigma(m^2)),$$ whereby we obtain $$\frac{D(p^k)D(m^2)}{s(p^k)s(m^2)} = 2.$$

We focus on what we can derive from $$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)} = \frac{D(m^2)}{s(p^k)} = \gcd(m^2,\sigma(m^2)).$$ We obtain $$2m^2 - \sigma(m^2) = D(m^2) = s(p^k)\gcd(m^2,\sigma(m^2)) = (\sigma(p^k) - p^k)\gcd(m^2,\sigma(m^2)) = \sigma(p^k)\frac{\sigma(m^2)}{p^k} - {p^k}\frac{2m^2}{\sigma(p^k)} = I(p^k)\sigma(m^2) - \frac{2m^2}{I(p^k)}.$$

Thus, we get $$\gcd(m^2,\sigma(m^2)) = \frac{D(m^2)}{s(p^k)} = \frac{I(p^k)\sigma(m^2) - \frac{2m^2}{I(p^k)}}{s(p^k)}.$$

We therefore have $$\gcd(m^2,\sigma(m^2)) = \frac{I(p^k)}{s(p^k)}\sigma(m^2) - \frac{1}{I(p^k)s(p^k)}(2m^2).$$

Here is my question:

Is it possible to express $$\gcd(m^2,\sigma(m^2)) = \frac{I(p^k)}{s(p^k)}\sigma(m^2) - \frac{1}{I(p^k)s(p^k)}(2m^2)$$ as an integral linear combination of $m^2$ and $\sigma(m^2)$ (in terms, of course, of $p$ and $k$)?

Sanity Check

When $k=1$, I have $$\gcd(m^2,\sigma(m^2)) = D(m^2) = 2m^2 - \sigma(m^2),$$ since $s(p^k)=1$ when $k=1$.

When $k=1$, I obtain $$\frac{I(p^k)}{s(p^k)}\sigma(m^2) - \frac{1}{I(p^k)s(p^k)}(2m^2) = I(p)\sigma(m^2) - \frac{1}{I(p)}(2m^2) = \frac{p+1}{p}\sigma(m^2) - \frac{2p}{p+1}(m^2).$$ Since $p^k m^2 = pm^2$ is assumed to be a(n) (odd) perfect number, then $I(p)I(m^2) = 2$, from which we get $$I(p) = \frac{2}{I(m^2)} \text{ and } I(m^2) = \frac{2}{I(p)}.$$ Hence, $$\frac{p+1}{p}\sigma(m^2) - \frac{2p}{p+1}(m^2)$$ simplifies to $$\frac{2}{I(m^2)}\sigma(m^2) - I(m^2){m^2} = 2m^2 - \sigma(m^2).$$

Jose Arnaldo Bebita Dris
  • 8,471
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    It seems that the approach used in sanity check only works for $k=1$. For general $k$, it only simplifies the expression to $\dfrac{2m^2-σ(m^2)}{s(p^k)}$. – Ѕᴀᴀᴅ Jun 26 '18 at 15:10
  • @AlexFrancisco - Yes, quite indeed. I was wondering whether anybody here has some other insight(s) as to how to simplify $$\gcd(m^2,\sigma(m^2)) = \frac{I(p^k)}{s(p^k)}\sigma(m^2) - \frac{1}{I(p^k)s(p^k)}(2m^2).$$ – Jose Arnaldo Bebita Dris Jun 27 '18 at 07:33
  • In particular, I would be interested in an expression that rewrites $\gcd(m^2,\sigma(m^2))$ as an integral linear combination of $\sigma(m^2)$ and $m^2$ (in terms, of course, of $p$ and $k$). – Jose Arnaldo Bebita Dris Jun 27 '18 at 08:44

1 Answers1

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It turns out that it is possible to express $\gcd(m^2, \sigma(m^2))$ as an integral linear combination of $m^2$ and $\sigma(m^2)$, in terms of $p$ alone.

To begin with, write $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{\sigma(p^{k-1})}=\frac{(2m^2 - \sigma(m^2))(p-1)}{p^k - 1}.$$

Now, using the identity $$\frac{A}{B}=\frac{C}{D}=\frac{A-C}{B-D},$$ where $B \neq 0$, $D \neq 0$, and $B \neq D$, we obtain $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)-(2m^2 - \sigma(m^2))(p-1)}{p^k - (p^k - 1)},$$ from which we get $$\gcd(m^2,\sigma(m^2))=\sigma(m^2)-(2m^2 - \sigma(m^2))(p-1)=2m^2 - p(2m^2 - \sigma(m^2)) = 2m^2 - pD(m^2),$$ or equivalently, $$\gcd(m^2,\sigma(m^2))=2(1 - p)m^2 + p\sigma(m^2).$$

Jose Arnaldo Bebita Dris
  • 8,471