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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: First Sage Math Cell 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: Second Sage Math Cell Output

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.

Here is our:

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

Jose Arnaldo Bebita Dris
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    Nice project, but I do not understand how the examples (both the squarefree ones and the non-squarefree ones) help us to prove the conjecture. And what is meant with "the conjecture is plausible" ? How does the data support the conjecture ? – Peter Jan 14 '23 at 08:18
  • Thank you for your time and attention, @Peter. If you (barely) noticed in the outputs, it does appear that the implication $p = k$ implies that $\sigma(p^k)/2$ is not squarefree holds, in the sense that I did not get an instance of when "the hypothesis $p = k$ is true but the conclusion $\sigma(p^k)/2$ is not squarefree does not hold". (By the contrapositive, $\sigma(p^k)/2$ is squarefree would conjecturally imply that $p \neq k$, which is what the data indeed shows.) – Jose Arnaldo Bebita Dris Jan 14 '23 at 08:35
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    One can click on the image to get a better view, @Peter. And yes, quite indeed, we do not need a double loop over $k$ and $p$, since it would simply be $\sigma(p^p)/2$ which has to be checked. – Jose Arnaldo Bebita Dris Jan 14 '23 at 08:51
  • Though we differentiated between $p$ and $k$ in order to double-check that, indeed, $\sigma(p^k)/2$ is squarefree would imply that $p \neq k$, @Peter. (I mean, they are two sides of the same coin.) – Jose Arnaldo Bebita Dris Jan 14 '23 at 08:52
  • @Peter: The updated Pari-GP scripts

    for(x=1, 10000, if((isprime(x)) && (Mod(x,4) == 1) && !(issquarefree(sigma(x^x)/2)),print(x)))

    and

    for(x=1, 10000, if((isprime(x)) && (Mod(x,4) == 1) && (issquarefree(sigma(x^x)/2)),print(x)))

    are working flawlessly and as designed, with the second script returning $0$ rows. This gives further computational evidence in support of the Conjecture.

     – Jose Arnaldo Bebita Dris Jan 14 '23 at 09:39
  • @Peter: The case of $p$ being a Mersenne prime is ruled out by the requirement that the special prime ought to satisfy $p \equiv 1 \pmod 4$, since if $2^n - 1$ is prime, then $2 \leq n$ is prime. (Note that $2^n - 1 \equiv 3 \pmod 4$ for $n \geq 2$.) – Jose Arnaldo Bebita Dris Jan 14 '23 at 11:29
  • Let us continue this discussion in chat. – Peter Jan 14 '23 at 11:30
  • @Peter: The fact that $q \mid (p + 1)$ is guaranteed by the fact that $(p + 1)/2$ is odd. Can you check if putting $q = (p + 1)/2$ proves the Conjecture? – Jose Arnaldo Bebita Dris Jan 14 '23 at 11:33

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I prove the slight stronger conjecture :

Let $p$ be an odd prime , not being a Mersenne prime and $q$ be the smallest odd prime factor of $p+1$. Then, we have $q^2\mid p^{p+1}-1$ from which we can conclude that $\frac{\sigma(p^p)}{2}$ is not squarefree.

Proof : With $s:=p+1$ we have $$p^{p+1}-1=(s-1)^{p+1}-1\equiv (-1)^p\cdot s(p+1)+(-1)^{p+1}-1=(-1)^p\cdot (s(p+1)-1)-1=-s(p+1)=-s^2\equiv 0 \mod s^2$$ So , we have $(p+1)^2\mid p^{p+1}-1$.Because of $q\mid p+1$ , this finishes the proof.

Peter
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    If $p$ is of the form $4k+3$ (covering the Mersenne prime case) , then $p-1$ is not divisible by $4$ , but $p^{p+1}-1$ is divisible by $16$. Hence $\frac{\sigma(p^p)}{2}$ is divisible by $4$ in this case, hence again not squarefree. – Peter Jan 14 '23 at 13:42