In what follows, let $a, b, c \in \mathbb{N}$ and let $x, y \in \mathbb{Z}$. Denote the abundancy index of $z \in \mathbb{N}$ by $I(z)=\sigma(z)/z$, and the deficiency of $z$ by $D(z)=2z-\sigma(z)$, where $\sigma(z)$ is the sum of the divisors of $z$.

This blog post is an offshoot of the following MSE post. Essentially, we will be using the equation

$$\mathscr{A}\text{: }ax + by= c=\gcd(a,b)=\gcd(a,c)=\gcd(c,b)$$

in what follows. Note that the values $x$ and $y$ in (Equation $\mathscr{A}$) are not unique. (As hinted by Bill Dubuque in several comments to the MSE question mentioned, this equation can be proved in many ways, one of which is via the GCD Distributive Law.)

Let $N=q^k n^2$ be an odd perfect number with Euler prime $q$. (That is, $q$ is a prime that satisfies $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Thus, $q, k, n, N \in \mathbb{N}$.)

Now, recall from this NNTDM paper that we have the equations

$$\mathscr{B}\text{: } \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2)),$$

where $\sigma(x)$ is the sum of the divisors of $x$ and $D(y)=2y-\sigma(y)$ is the deficiency of $y$.

In particular, note from (Equation $\mathscr{A}$) (and the fact that both $n^2$ and $\sigma(n^2)$ are odd) that

$$\mathscr{C}\text{: } \gcd(n^2,\sigma(n^2))=\gcd(2n^2-\sigma(n^2),n^2)=\gcd(D(n^2),n^2)$$

and

$$\mathscr{D}\text{: } \gcd(n^2,\sigma(n^2))=\gcd(2n^2 - \sigma(n^2),\sigma(n^2))=\gcd(D(n^2),\sigma(n^2)).$$

But notice that (from (Equation $\mathscr{B}$) and (Equation $\mathscr{C}$)) we have

$$\frac{2n^2}{\sigma(q^k)}=\frac{\sigma(n^2)}{q^k}=\gcd(n^2,\sigma(n^2))=\gcd(D(n^2),n^2),$$

and that (from (Equation $\mathscr{B}$) and (Equation $\mathscr{D}$)) we also have

$$\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2))=\gcd(D(n^2),\sigma(n^2)).$$

These last two equations imply that

$$\frac{\sigma(n^2)}{q^k} \text{ divides } D(n^2)$$

and

$$\frac{D(n^2)}{\sigma(q^{k-1})} \text{ divides } \sigma(n^2).$$

It follows from the last two divisibility constraints that

$$\frac{\sigma(n^2)}{{q^k}\sigma(q^{k-1})} \mid \frac{D(n^2)}{\sigma(q^{k-1})} \mid \sigma(n^2)$$

and also that

$$\frac{D(n^2)}{{q^k}\sigma(q^{k-1})} \mid \frac{\sigma(n^2)}{q^k} \mid D(n^2).$$

We therefore conclude that $\sigma(q^{k-1})$ divides $\sigma(n^2)$ and that $q^k$ divides $D(n^2)$.

This last divisibility constraint appears to result in a contradiction, as it implies that

$$\frac{D(n^2)}{q^k}=\frac{2n^2 - \sigma(n^2)}{q^k}$$

is an integer, which further means that

$$\frac{2n^2}{q^k}$$

must also be an integer, since $\sigma(n^2)/q^k$ is an integer. This contradicts $\gcd(q,n)=\gcd(q^k,n^2)=1$.

However, I think I may have used the assumption $k=1$ implicitly in the above proof.

__Update (September 20, 2017, 7:00 AM [Manila time])__

The assumption $k=1$ was not implicitly used in the argument above. It turned out that

$$\frac{D(n^2)}{{q^k}\sigma(q^{k-1})} \not\in \mathbb{N}.$$

From (Equation $\mathscr{B}$), we also obtain

$$\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2)),$$

from which we have

$$\frac{\sigma(n^2)-n^2}{q^k - \sigma(q^k)/2}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2)).$$

But we can simplify the first fraction in the last equation as follows

$$\frac{\sigma(n^2)-n^2}{q^k - \sigma(q^k)/2} = \frac{\sigma(n^2)-n^2}{D(q^k)/2} = 2\cdot\frac{\sigma(n^2)-n^2}{D(q^k)}.$$

In particular, by using (Equation $\mathscr{B}$) again, we get

$$D(q^k)D(n^2) = 2(\sigma(n^2) - n^2)\sigma(q^{k-1}).$$

This implies that the deficiency function $D$ is not multiplicative since, in particular, $\gcd(q^k, n^2) = 1$ but

$$D(q^k n^2) = 0 < 2(\sigma(n^2) - n^2)\sigma(q^{k-1}) = D(q^k)D(n^2).$$

(In general, if $\gcd(X,Y)=1$, then the deficiency function satisfies $D(XY) \leq D(X)D(Y)$. See this article for a proof.)

Note that

$$D(q^k n^2) = 0 < 2(\sigma(n^2) - n^2)\sigma(q^{k-1}) = D(q^k)D(n^2).$$

(In general, if $\gcd(X,Y)=1$, then the deficiency function satisfies $D(XY) \leq D(X)D(Y)$. See this article for a proof.)

Note that

$$\sigma(n^2) - n^2$$

is also called the sum of the aliquot parts of $n^2$.

Here are some relevant OEIS hyperlinks for the number sequences used in this blog post:

OEIS sequence A033879 - Deficiency of $z$, or $2z - \sigma(z)$

OEIS sequence A001065 - Sum of proper divisors (or aliquot parts) of n: $\sigma(z) - z$

Note the following (trivial!) relationships between these two sequences:

$$A033879 + 2\cdot(A001065) = \sigma(z)$$

$$A033879 + A001065 = z$$

Lastly, notice that the inequality $1 < I(z) < 2$ follows immediately from

$$\frac{A033879 + A001065}{A033879 + A001065} < \frac{\sigma(z)}{z} := \frac{A033879 + 2\cdot(A001065)}{A033879 + A001065} < \frac{2\cdot(A033879) + 2\cdot(A001065)}{A033879 + A001065}.$$

Lastly, notice that the inequality $1 < I(z) < 2$ follows immediately from

$$\frac{A033879 + A001065}{A033879 + A001065} < \frac{\sigma(z)}{z} := \frac{A033879 + 2\cdot(A001065)}{A033879 + A001065} < \frac{2\cdot(A033879) + 2\cdot(A001065)}{A033879 + A001065}.$$

__Comments from the readers of this blog are most welcome. Please feel free to shoot me an e-mail.__