## 6.2.17

### On a conjectured biconditional for odd perfect numbers - Part 2

Can this heuristic on odd perfect numbers be made rigorous?

Preliminaries

Euler showed that an odd perfect number, if one exists, must take the form $N=q^k n^2$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  (It follows that $q \geq 5$.)

Let $\sigma(x)$ be the sum of the divisors of $x \in \mathbb{N}$.  Then by definition,
$$\sigma(q^k)\sigma(n^2) = \sigma(q^k n^2) = \sigma(N) = 2N = 2 q^k n^2$$
from which it follows that
$$I(q^k)I(n^2) = \frac{\sigma(q^k)}{q^k}\cdot\frac{\sigma(n^2)}{n^2} = 2$$
where $I(x)=\sigma(x)/x$ is the *abundancy index* of $x$.

Numerics

By a well-known formula for the sum of divisors of a prime power, we have
$$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{{q^k}(q - 1)}.$$

This is bounded by
$$\frac{q+1}{q} = I(q) \leq I(q^k) < \frac{q^{k+1}}{{q^k}(q - 1)} = \frac{q}{q - 1}.$$

Now, note that, when $k=1$, we have the bounds
$$1 < I(q^k) = \frac{q+1}{q} = 1 + \frac{1}{q} \leq 1 + \frac{1}{5} = \frac{6}{5},$$
from which it follows that
$$\frac{5}{3} \leq I(n^2) = \frac{2}{I(q^k)} = \frac{2}{I(q)} < 2.$$

Additionally, when $k>1$, we have the bounds
$$1 < I(q^k) < \frac{q}{q - 1}.$$
Since
$$\frac{q}{q - 1} = \frac{1}{1 - (1/q)}$$
we obtain
$$q \geq 5 \iff 1/q \leq 1/5 \iff 1 - (1/q) \geq 4/5 \iff \frac{1}{1 - (1/q)} \leq 5/4$$
so that
$$1 < I(q^k) < \frac{q}{q - 1} \leq \frac{5}{4}.$$

This then implies that
$$\frac{8}{5} < I(n^2) = \frac{2}{I(q^k)} < 2.$$

Implications

Note that we have proved the following implications:

(1) If $N=q^k n^2$ is an odd perfect number with $k=1$, then $I(n^2) \geq 5/3$.

(2) If $N=q^k n^2$ is an odd perfect number with $k>1$, then $I(n^2) > 8/5$.

Contrapositives

(3) Contrapositive of (1) If $N=q^k n^2$ is an odd perfect number with $I(n^2) < 5/3$, then $k>1$.

(4) Contrapositive of (2) If $N=q^k n^2$ is an odd perfect number with $I(n^2) \leq 8/5$, then $k=1$.

Analysis

By (3) and (2), we have
$$I(n^2) < 5/3 \Longrightarrow k>1 \Longrightarrow I(n^2) > 8/5 \Longrightarrow 8/5 < I(n^2) < 5/3$$
resulting in no contradiction, since ${8}\cdot{3} = 24 < 25 = {5}\cdot{5}$.

By (4) and (1), we get
$$I(n^2) \leq 8/5 \Longrightarrow k=1 \Longrightarrow I(n^2) \geq 5/3 \Longrightarrow 5/3 \leq I(n^2) \leq 8/5$$
resulting in a contradiction, since ${5}\cdot{5}= 25 \leq 24 = {8}\cdot{3}$ is false.

This implies that $I(n^2) > 8/5$ is unconditionally true.

Now here is the heuristic:

Heuristic

If we identify the lower bound $I(n^2) \geq 5/3$ with $k=1$ and similarly identify the lower bound $I(n^2) > 8/5$ with $k>1$, and since assuming the negation of the latter lower bound contradicts (4) and (1) above, does this mean that we can predict $k>1$ to be true?

That is, can we "assume" or prove that the following biconditionals hold?
$$k=1 \iff I(n^2) \geq 5/3$$
$$k>1 \iff I(n^2) > 8/5$$

Response by MSE user Noah Schweber
(Link to MSE question)

You can do nothing of the sort, and the axiom of choice plays no role here.

Let's ignore the specific math leading up to the "Analysis" part, and just look at the logic of what you're trying to do. (In particular, I make no claim about whether said math is correct.)

You have two numbers, "$k$" and "$I=I(n^2)$," that you care about; and you know $$k\ge 1,\quad k=1\Longrightarrow I\ge{5/3},\quad\mbox{ and }\quad k>1\Longrightarrow I>{8/5}.$$ All this lets you conclude is that $I<{5/3}$ implies $k>1$. If $I\ge{5/3}$, you can't conclude anything - both "$k=17$, $I=17$" and "$k=1, I=17$" satisfy the three facts above. In particular, from what you know so far there is no way to identify $k=1$ with $I\ge{5/3}$. (It's always a good idea to think through a few specific examples before trying to jump to a conclusion.)

You are trying to invoke the axiom of choice to get around a logical gap - that you can't conclude the converse of a statement from the statement. But this isn't a thing the axiom of choice does.

****

EDIT: Note that by Shoenfield absoluteness (actually that's massive overkill but oh well), the truth of Sorli's conjecture can't possibly depend on the axiom of choice: if there is a proof using AC, there is also a proof not using AC. (Incidentally, this doesn't mean that the proof via AC might be easier to find, just that AC won't ultimately be necessary.) I strongly suspect that the same goes for each of the Millennium problems - indeed $P=NP$, Birch Swinnerton-Dyer, and Navier-Stokes are all not hard to cast in arithmetic terms - but I am worried about the Hodge Conjecture on the general principle that it looks scary to me (but it also does look $\Pi^1_2$).