## 9.6.18

### Would like to get numerical (lower [and upper?]) bounds for $p$

(This post is copied verbatim from this MSE question.)

This question is an offshoot of this earlier MSE question.

Let $\sigma(z)$ denote the sum of divisors of $z \in \mathbb{N}$, the set of positive integers.  Denote the abundancy index of $z$ by $I(z) := \sigma(z)/z$.

If $N={p^k}{m^2}$ is an odd perfect number with Euler prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then it is somewhat trivial to prove that
$$3 - \frac{p - 2}{p(p-1)} = \frac{3p^2 - 4p + 2}{p(p-1)} < I(p^k) + I(m^2)$$
and
$$I(p^k) + I(m^2) \leq \frac{3p^2 + 2p + 1}{p(p+1)} = 3 - \frac{p - 1}{p(p+1)}.$$
Now, setting $x := 3 - \bigg(I(p^k) + I(m^2)\bigg)$, we have the simultaneous inequalities
$$\frac{p-1}{p(p+1)} \leq x < \frac{p - 2}{p(p - 1)}$$
resulting in the inequalities
$$\bigg((p - 2) > xp(p-1)\bigg) \land \bigg((p - 1) \leq xp(p+1)\bigg).$$

Notice that it is known that
$$\frac{57}{20} < I(p^k)+I(m^2) < 3$$
so that we know that
$$0 < x < \frac{3}{20}.$$

We now solve the inequalities one by one.

Solution is
$$\frac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x} < p < \frac{(x+1) + \sqrt{x^2 - 6x + 1}}{2x}$$

Solution is
$$p \in \bigg(-\infty, \frac{1 - x - \sqrt{x^2 - 6x + 1}}{2x}\bigg] \bigcup \bigg[\frac{1 - x + \sqrt{x^2 - 6x + 1}}{2x},\infty\bigg)$$

Now, this is where the computations start to get messy.  Can I ask for some help?

Basically, I would like to get numerical (lower [and upper?]) bounds for $p$.

It is easily seen that
$$\frac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x} < 2, \frac{1 - x - \sqrt{x^2 - 6x + 1}}{2x} < 2$$
so your conditions reduces to
$$\frac{1 - x + \sqrt{x^2 - 6x + 1}}{2x} \leq p < \frac{1+x + \sqrt{x^2 - 6x + 1}}{2x}$$
then, one can see that the lower bound and upper bound differs by $1$ and both are unbounded in the given range of $x$. That is, there can be only one prime $p$ for any given value $x$.

Related Paper:  The abundancy index of divisors of odd perfect numbers - Part III, http://nntdm.net/volume-23-2017/number-3/53-59/

## 5.5.18

### Can the following argument be pushed to a full proof that $(p + 2)/p$ is an outlaw if $p$ is an odd prime?

(The following post is extracted verbatim from this MSE question.)

This is related to this earlier MSE question.

Let $\sigma(x)$ be the sum of the divisors of $x$, and denote the abundancy index of $x$ by $I(x):=\sigma(x)/x$.

If the equation $I(a) = b/c$ has no solution $a \in \mathbb{N}$, then the rational number $b/c$ is said to be an abundancy outlaw.

In the earlier question, it is shown that:

(a) If $I(n) = (p+2)/p$, then $n$ is an odd square.

(b) If $I(n) = (p+2)/p$, then $p \mid n$.

So now suppose to the contrary that $I(n) = (p+2)/p$.  From (a) and (b), since $p$ is a prime number, then $p^2 \mid n$.  It follows that
$$\frac{p^2 + p + 1}{p^2} = I(p^2) \leq I(n) = \frac{p+2}{p},$$
whence there is still no contradiction.

From the divisibility constraint $p^2 \mid n$, we have that $p^2 \leq n$.  We claim that $p^2 \neq n$.  Suppose to the contrary that $p^2 = n$.

Then we have
$$\frac{\sigma(n)}{n} = \frac{p+2}{p} = \frac{p(p+2)}{p^2}$$
so that
$$p^2 + p + 1 = \sigma(p^2) = \sigma(n) = p^2 + 2p,$$
which contradicts the fact that $p$ is an odd prime.

This implies that $p < \sqrt{n}$, which further means that
$$\gcd(n, \sigma(n)) = \frac{n}{p} > \sqrt{n} > {10}^8$$
where the lower bound $n > {10}^{16}$ is due to Richard F. Ryan, "Results concerning uniqueness for $\sigma(x)/x = \sigma(p^n q^m)/(p^n q^m)$ and related topics, International Math. J., 2002 , V2#5pp497-514".

Here is my question:

Can the preceding argument be pushed to a full proof that $(p + 2)/p$ is an outlaw if $p$ is an odd prime?

## 19.3.18

### If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?

(Note:  This post was copied verbatim from this MSE question.)

Let $\sigma(x)$ be the sum of the divisors of $x$.  Denote the deficiency of $x$ by $D(x) := 2x - \sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x) := \sigma(x) - x$.

Here is my question:

If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?
$$D(q^k)D(n^2)=2s(q^k)s(n^2)$$

I only know that $k=1$ is true if and only if (one of) the following conditions hold:

(1) $\sigma(n^2)/q \mid n^2$

(2) $D(n^2) \mid n^2$

(3) $\gcd(n^2, \sigma(n^2)) = D(n^2)$

Source of Equation

From the fundamental equation
$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \gcd(n^2, \sigma(n^2))$$
we obtain
$$\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k} = \frac{D(n^2)}{s(q^k)} = \gcd(n^2, \sigma(n^2))$$
and
$$\frac{\sigma(n^2) - n^2}{\frac{2q^k - \sigma(q^k)}{2}} = \frac{2s(n^2)}{D(q^k)} = \gcd(n^2, \sigma(n^2)),$$
using the identity
$$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B}.$$

Reference

## 5.3.18

### If $\frac{σ(x)}{x}=\frac{p+2}p$ where $p$ is an odd prime, does it follow that $x$ is an odd square?

(Note:  The following proof was copied verbatim from the answer of MSE user Alex Francisco.)

First, note that for any coprime $a, b \in \mathbb{N}_+$, there is$$I(ab) = I(a) I(b).$$

Suppose there is an even number $n = 2^k \cdot l$, where $k \geq 1$ and $l$ odd, such that$$I(n) = \frac{p + 2}{p}.$$

Case 1: $k \geq 2$. Then$$\frac{p + 2}{p} = I(2^k \cdot l) = I(2^k) I(l) \geq I(2^k) = \frac{2^{k + 1} - 1}{2^k} = 2 - \frac{1}{2^k} \geq \frac{7}{4},$$
which implies $4(p + 2) \geq 7p$, contradictory to $p \geq 3$.

Case 2: $k = 1$ and $p \geq 5$. Then, analogously,$$\frac{p + 2}{p} \geq 2 - \frac{1}{2^k} = \frac{3}{2},$$
which implies $2(p + 2) \geq 3p$, contradictory to $p \geq 5$.

Case 3: $k = 1$ and $p = 3$. Then$$\frac{5}{3} = I(2l) = \frac{3}{2} I(l) \Longrightarrow I(l) = \frac{10}{9}.$$

If $l$ has an odd prime factor $q < 10$, suppose $q^m \mathbin{\|} l$, then$$\frac{10}{9} = I(l) = I(q^m) I\left( \frac{l}{q^m} \right) \geq I(q^m) = \frac{q^{m + 1} - 1}{q^m (q - 1)}\\ = 1 + \frac{1}{q - 1}\frac{q^m - 1}{q^m} \geq 1 + \frac{1}{q - 1}\frac{q - 1}{q} = 1 + \frac{1}{q} > \frac{10}{9},$$
a contradiction. Now suppose the prime factorization of $l$ is$$l = \prod_{i = 1}^s p_i^{a_i},$$
then $p_i \geq 11$. Because$$\frac{10}{9} = I(l) = \prod_{i = 1}^s \frac{1 + p_i + \cdots + p_i^{a_i}}{p_i^{a_i}},$$
then$$9 \prod_{i = 1}^s \sum_{j = 0}^{a_i} p_i^j = 10 \prod_{i = 1}^s p_i^{a_i} \Longrightarrow \left. 9 \,\middle|\, \prod_{i = 1}^s p_i^{a_i} \right.,$$
contradictory to $p_i \geq 11$.

Therefore, there does not exist an even positive integer $n$ such that$$I(n) = \frac{p + 2}{p}.$$

## 20.9.17

### Some remarks on the Descartes-Frenicle-Sorli conjecture on odd perfect numbers

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

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

## 12.9.17

### On computing $\gcd\left(\sigma(q^k),\sigma(n^2)\right)$ when $q^k n^2$ is an odd perfect number with Euler prime $q$

In this blog post, we compute
$$\gcd\left(\sigma(q^k),\sigma(n^2)\right)$$
when $N = q^k n^2$ is an odd perfect number with Euler prime $q$.

From this paper in NNTDM, we have the equation
$$i(q) := \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right).$$

In particular, we know that the index $i(q)$ is an integer greater than $5$ [Dris, Luca (2016)] (a copy is available in arXiv).

We now attempt to compute an expression for $\gcd\left(\sigma(q^k),\sigma(n^2)\right)$ in terms of $i(q)$.

First, since we have
$$\sigma(q^k)\sigma(n^2) = \sigma({q^k}{n^2}) = \sigma(N) = 2N = 2{q^k}{n^2}$$
we obtain
$$\sigma(q^k) = \frac{2 q^k n^2}{\sigma(n^2)} = \frac{2n^2}{\sigma(n^2)/q^k} = \frac{2n^2}{i(q)}$$
and
$$\sigma(n^2) = \frac{2 q^k n^2}{\sigma(q^k)} = {q^k}\cdot\bigg(\frac{2n^2}{\sigma(q^k)}\bigg) = {q^k}{i(q)},$$
so that we get
$$\gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg).$$

Now, since $\gcd(q, n) = \gcd(q^k, 2n^2) = 1$ and $i(q)$ is odd, we get
$$\gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

Hence, we conclude that
$$G := \gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$

I posed the following question in MathOverflow:

I seem to recall that somebody (was it Pomerance [?] et. al) proved that
$$G \neq 1.$$  Does anybody here happen to know a reference?  Additionally, does
$G \neq 1$ imply that $G = i(q)$?