## 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
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.)
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 $\Pi^1_2$).