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**?