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31.10.19

Expressing $\gcd(m^2, \sigma(m^2))$ as a linear combination - Part II

Let $N = p^k m^2$ be an odd perfect number with special / Euler prime $p$.

Continuing from the previous blog post titled


it is perhaps worthwhile and instructive to consider what would happen if one assumes that the Descartes-Frenicle-Sorli Conjecture that $k=1$ hold.

From the final equation in that blog post, we have
$$\gcd(m^2, \sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{2m^2}{\sigma(p^k)}$$
$$=\frac{D(m^2)}{s(p^k)}=\frac{2s(m^2)}{D(p^k)}=p\sigma(m^2) - 2(p-1){m^2},$$
where $\sigma(x)$ is the sum of divisors of $x$, $D(x)=2x-\sigma(x)$ is the deficiency of $x$, and $s(x)=\sigma(x)-x$ is the sum of the proper/aliquot divisors of $x$.

Now, assume that $k=1$.

Dividing both sides of
$$\gcd(m^2, \sigma(m^2))=p\sigma(m^2) - 2(p-1){m^2}$$
by $pm^2$, we get
$$\frac{D(m^2)}{N}=I(m^2) - 2\cdot\bigg(\frac{p-1}{p}\bigg),$$
where $I(x)=\sigma(x)/x$ is the abundancy index of $x$.

Since $k=1$, we get
$$I(p^k)=I(p)=\frac{p+1}{p}$$
from which we obtain
$$I(m^2)=\frac{2}{I(p^k)}=\frac{2p}{p+1}.$$
Finally, we derive
$$\frac{D(m^2)}{N}=\frac{2p}{p+1} - 2\cdot\bigg(\frac{p-1}{p}\bigg)=\frac{2}{p(p+1)},$$
which gives some insight as to where the expression
$$N = \frac{p(p+1)}{2}\cdot{D(m^2)}$$
comes from, when the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.