If $q^k n^2$ is an odd perfect number with Euler factor $q^k$, can $q = 73$ hold?

Question

Call a number $N$ perfect if $\sigma(N)=2N$ where $\sigma$ is the classical sum-of-divisors function.

If $N = q^k n^2$ is an odd perfect number, can $q = 73$ hold?

Here is my attempt:

Since $37=(q+1)/2 \mid \sigma(q^k)/2 \mid n$, it follows that $${37}^2 \mid n^2 \mid q^k n^2 = N$$ where we have used the fact that an odd perfect number only has one Euler prime.

Using factor/sigma chains, I get:

$$\sigma({37}^2) = 1407 = {3}\cdot{7}\cdot{67} \mid N$$ Again, we use the fact that none of the resulting factors from the chain are the same as the Euler prime $q = 73$: $$3^2 \mid n^2$$ $$7^2 \mid n^2$$ $${67}^2 \mid n^2$$

Again, via chaining we obtain:

$$\sigma(3^2) = 13 \Longrightarrow {13}^2 \mid n^2$$ $$\sigma(7^2) = 57 = {3}\cdot{19} \Longrightarrow {19}^2 \mid n^2$$ $$\sigma({67}^2) = 4557 = {3}\cdot{7^2}\cdot{31} \Longrightarrow {31}^2 \mid n^2$$

Estimating the abundancy of $q^k n^2$, we get:

$$2 = I(q^k)I(n^2) > I(n^2) \geq I({37}^2)\cdot{I(3^2)}\cdot{I(7^2)}\cdot{I({67}^2)}\cdot{I({13}^2)}\cdot{I({19}^2)}\cdot{I({31}^2)}$$ $$ = \frac{328939969561158501}{158894526512072169} = \frac{1453933719}{702323011} > 2.07$$

Does this conclusively prove that $q \neq 73$? If not, what else needs to be done? And what was/(were) the obstruction(s), if any?

Thanks @Joffan! This is my first time to implement sigma / factor chains (albeit manually), so please bear with me. I guess this chain only terminates the node for ${37}^2$, correct? — Jose Arnaldo Bebita Dris, May 25 '15 at 19:26
It will work again at $\sigma(37^8)$. If you have a factor for $\sigma(37^{n})$, I think it will generally also divide $\sigma(37^{2n+1})$ and $\sigma(37^{3n+2})$. — Joffan, May 25 '15 at 19:33
Ohh okay. So just to confirm, I guess no computer checking on $q = 73$ has finished running yet? — Jose Arnaldo Bebita Dris, May 25 '15 at 19:35
There might well be another argument to show that $q=73$ is infeasible; but it's not this one. — Joffan, May 25 '15 at 19:37

score 2 · Accepted Answer · answered May 25 '15 at 19:42

Unfortunately for your argument, $1407 \nmid \sigma(37^4)$ and $1407 \nmid \sigma(37^6)$, so we can't identify the components of $n$ quite so readily, and the same sort of issue will give problems for your other factor inferences.

$1407$ will work again at $\sigma(37^8)$. If you have a factor for $\sigma(37^{n})$, I think it will generally also divide $\sigma(37^{2n+1})$ and $\sigma(37^{3n+2})$. Unfortunately this pattern would not allow us to construct a finite set of alternative divisors of $n^2$.

If $q^k n^2$ is an odd perfect number with Euler factor $q^k$, can $q = 73$ hold?

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