A conjecture regarding odd perfect numbers

Question

(Note: I asked this question in MSE this June 2018 but did not receive any responses there. I have therefore cross-posted it here, hoping that it gets answered.)

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 $s(z):=\sigma(z)-z$.

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 $$\frac{D(p^k)D(m^2)}{s(p^k)s(m^2)} = 2.$$ Note that we also have (Equation A) $$\frac{2D(m^2)s(m^2)}{D(p^k)s(p^k)} = \bigg(\gcd(m^2, \sigma(m^2))\bigg)^2.$$ Lastly, notice that we can easily get $$\sigma(p^k) \equiv k + 1 \equiv 2 \pmod 4$$ so that it remains to consider the possible equivalence classes for $\sigma(m^2)$ modulo $4$. Since $\sigma(m^2)$ is odd, we only need to consider two.

Here is my question:

Which equivalence class of $\sigma(m^2)$ modulo $4$ makes Equation A untenable?

I know that the answer must somehow depend on the equivalence class of $p$ and $k$ modulo $8$, but as I only know that $p \equiv k \equiv 1 \pmod 4$, I am stuck.

UPDATED September 19 2018 (Manila time) After considering various cases, I think I am able to prove that:

1. If $p \equiv k \equiv 1 \pmod 8$, then $\sigma(m^2) \equiv 3 \pmod 4$ is impossible.
2. If $p \equiv 1 \pmod 8$ and $k \equiv 5 \pmod 8$, then $\sigma(m^2) \equiv 1 \pmod 4$ is impossible.
3. If $p \equiv 5 \pmod 8$ and $k \equiv 1 \pmod 8$, then $\sigma(m^2) \equiv 1 \pmod 4$ is impossible.
4. If $p \equiv k \equiv 5 \pmod 8$, then $\sigma(m^2) \equiv 3 \pmod 4$ is impossible.

Answer 1

It is rare that complicating an expression leads to an insight. There are, of course, important counter-examples to this principle. But generally speaking, one seeks to simplify an expression, rather that add extra complications. Let me explain how I see some unnecessary complications in your post.

Let $n$ be an odd perfect number. As you mention, this means that $\sigma(n)=2n$.

Using the fact that $\sigma$ is a multiplicative function, with a very nice multiplicative definition, we can investigate how this equation behaves on prime factorizations. Since it involves the prime $2$ quite prominently, it is natural to look at this equality $2$-adically. Doing so, we discover (as Euler did) that $n$ must be of the form $n=p^k m^2$, where $p\equiv k\equiv 1\pmod{4}$ and $p$ is a prime, and $p\nmid m$.

Thus, we can remove all reference to the variable $n$, and we have $$ \sigma(p^k)\sigma(m^2)=2p^km^2. $$ We also have information about how the factors on the right side divide the factors on the left side. We have $2\mid \sigma(p^k)$, while $p^k\mid \sigma(m^2)$, and finally $m^2$ can have some of factors go into both parts.

By rewriting this equality as $$ \frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)}, $$ and asserting that both sides are odd integers, we have exactly the same information, but now it involves fractions (an unnecessary complication).

Rather than introducing fractions, it would be much cleaner to simply write $\sigma(p^k)=2t_1$, $\sigma(m^2)=p^k t_2$, and note that $m^2=t_1 t_2$. By giving names to the cofactors, you can avoid using fractions. Notice that with these identifications, your "fundamental" equality is just asserting that $t_2=t_2$.

Now, since $p\nmid m$, then $$ \gcd(\sigma(m^2),m^2)=\gcd(p^k t_2,m^2)=\gcd(t_2,m^2)=\gcd(t_2,t_1t_2)=t_2. $$ What is fundamentally happening here is we are using the fact that $N=p^k m^2$ is a partial prime factorization. The introduction of a $\gcd$ in your post seems to be an unnecessary complication. It seems to simply be a stand-in for the quantity $t_2$.

At this point, the fractions in your post get even more complicated, via the introduction of the functions $D$ and $s$. These expressions are just rewriting the given quantities as alternative expressions. This added complexity adds nothing new, since you never use (or need) any properties $D$ and $s$ except their definitions (which lets you get rid of them, in terms of more fundamental quantities).

In particular, after simplifying both sides of Equation A separately, we should get $t_2^2=t_2^2$. So, that equation, in and of itself, is a tautology in terms of the previous information. It gives no new restrictions on the congruence classes of $\sigma(m^2)$. Any restrictions on those congruence classes must come from the previous information.

This might explain why a corrigendum was necessary. It is easy to make mistakes when a situation is over-complicated by irrelevant conditions.

Moreover, as useful test cases you might consider spoof OPNs. If you only want to focus on congruence conditions that apply to minor modifications of $\sigma$, then any result you prove should equally apply to spoofs. For example, the theorem by Chen and Luo (you cited) seems to apply equally well to spoofs. In particular, if you believe that you have shown that the special prime must be $5$ (which would be an enormous advance in odd perfect numbers), you should do what's called a "sanity check", where you test your theorem again one of the spoof OPNs, to see whether or not you made an error (and also to see if you really do use more than mere congruence considerations).

Answer 2

The following assertion appears in Theorem 3.3 (page 7, equations (5) to (6)) of Odd multiperfect numbers by Shi-Chao Chen and Hao Luo:

Let $n=\pi^{\alpha} M^2$ be an odd $2$-perfect number, with $\pi$ prime, $\gcd(\pi,M)=1$, and $\pi \equiv \alpha \equiv 1 \pmod 4$. Then \begin{align*} \sigma(M^2) \equiv 1 \pmod 4 &\iff \pi \equiv \alpha &\pmod 8, \\ \sigma(M^2) \equiv 3 \pmod 4 &\iff \pi \equiv \alpha + 4 &\pmod 8. \\ \end{align*}

Answer 3

Further to this recent corrigendum in NNTDM, which corrects an oversight in Some modular considerations regarding odd perfect numbers – Part II, we realized that we do in fact have the following biconditionals:

In this article, we consider the various possibilities for $p$ and $k$ modulo $16$, and show conditions under which the respective congruence classes for $\sigma(m^2)$ (modulo $8$) are attained, if $p^k m^2$ is an odd perfect number with special prime $p$. We can prove that:

• $\sigma(m^2) \equiv 1 \pmod 8$ holds if and only if $p + k \equiv 2 \pmod {16}$
• $\sigma(m^2) \equiv 3 \pmod 8$ holds if and only if $p - k \equiv {12} \pmod {16}$
• $\sigma(m^2) \equiv 5 \pmod 8$ holds if and only if $p + k \equiv {10} \pmod {16}$
• $\sigma(m^2) \equiv 7 \pmod 8$ holds if and only if $p - k \equiv 4 \pmod {16}$.

If $p=5$, then these logical equivalences take the form

• $\sigma(m^2) \equiv 1 \pmod 8$ holds if and only if $k \equiv {13} \pmod {16}$
• $\sigma(m^2) \equiv 3 \pmod 8$ holds if and only if $k \equiv 9 \pmod {16}$
• $\sigma(m^2) \equiv 5 \pmod 8$ holds if and only if $k \equiv 5 \pmod {16}$
• $\sigma(m^2) \equiv 7 \pmod 8$ holds if and only if $k \equiv 1 \pmod {16}$.

If $k=1$, then these logical equivalences take the form

• $\sigma(m^2) \equiv 1 \pmod 8$ holds if and only if $p \equiv 1 \pmod {16}$
• $\sigma(m^2) \equiv 3 \pmod 8$ holds if and only if $p \equiv 13 \pmod {16}$
• $\sigma(m^2) \equiv 5 \pmod 8$ holds if and only if $p \equiv 9 \pmod {16}$
• $\sigma(m^2) \equiv 7 \pmod 8$ holds if and only if $p \equiv 5 \pmod {16}$.

In particular, we obtain the following assertion:

PROPOSITION. Let $p^k m^2$ be an odd perfect number with special prime $p$. If the conjunction $p=5$ and $k=1$ is true, then the congruence $\sigma(m^2) \equiv 7 \pmod 8$ holds.

Answer 4

This answer is a direct response to Professor Pace Nielsen's suggestion to perform a "sanity check" and complements this other answer by providing a proof of the following biconditionals:

If $p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then the following equivalences hold:

• $\sigma(m^2) \equiv 1 \pmod 8 \iff p + k \equiv 2 \pmod {16}$
• $\sigma(m^2) \equiv 3 \pmod 8 \iff p - k \equiv 12 \pmod {16}$
• $\sigma(m^2) \equiv 5 \pmod 8 \iff p + k \equiv 10 \pmod {16}$
• $\sigma(m^2) \equiv 7 \pmod 8 \iff p - k \equiv 4 \pmod {16}$.

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Illustration of proof for one biconditional (as the other proof arguments are more or less similar):

Assume that $\sigma(m^2) \equiv 1 \pmod 8$. Note that this implies that $\sigma(m^2) \equiv 1 \pmod 4$. By Chen and Luo $\ldots$. Please do not downvote, this post is currently under construction. I will return to it again tomorrow morning.

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Professor Pace Nielsen writes in a comment that (and I quote, "verbatim"):

"I agree that this quoted result passes all sorts of sanity checks. We have $$\left(\frac{\sigma(p^k)}{2p^k}\right)\cdot\left(\frac{\sigma(m^2)}{m^2}\right) = 1. \tag{*}$$ Since $m$ is odd, we know $m^2 \equiv 1 \pmod 8$. Thus, looking at the given equality (Equation $(*)$) modulo $8$, we know that the class of $\sigma(m^2)$ modulo $8$ will be the inverse of the class of $\sigma(p^k)/{2p^k}$ modulo $8$. The latter class is exactly half of the class of $\sigma(p^k)/p^k$ modulo $16$. These considerations make us expect a connection between $\sigma(m^2)$ modulo $8$, and information about $p$ and $k$ modulo $16$." $\ldots$