## 14.1.23

### If $p^k m^2$ is an odd perfect number with special prime $p$ and $p = k$, then $\sigma(p^k)/2$ is not squarefree.

While researching the topic of odd perfect numbers, we came across the following implication, which we currently do not know how to prove:

> CONJECTURE: If $p^k m^2$ is an odd perfect number with special prime $p$ and $p = k$, then $\sigma(p^k)/2$ is not squarefree.

Here, $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of the positive integer $x$.  (Note that both $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$ hold.)

OUR ATTEMPT

Here are the details of our search in the range $5 \leq p < 50$, $1 \leq k < 50$ using the following Sage Math Cell - Pari-GP scripts:

> (1) Searching for examples where $\sigma(p^k)/2$ is not squarefree

for(x=1, 50, for(y=1, 50, if((isprime(x)) && (Mod(x,4) == 1) && (Mod(y,4) == 1) && !(issquarefree(sigma(x^y)/2)),print(x,"   ",y,"   ",factor(sigma(x^y))))))

> Output:

> (2) Searching for examples where $\sigma(p^k)/2$ is squarefree

for(x=1, 50, for(y=1, 50, if((isprime(x)) && (Mod(x,4) == 1) && (Mod(y,4) == 1) && (issquarefree(sigma(x^y)/2)),print(x,"   ",y,"   ",factor(sigma(x^y))))))

> Output:

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As you can see, the Conjecture does appear plausible.  However, computational searches are very far from a complete proof, though they certainly add to the evidence supporting the Conjecture.

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Here is our:

QUESTION:
Does anybody here have any ideas on how to prove the Conjecture?