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It is currently not known whether odd perfect numbers (numbers with the property $\sigma(n)=2n$, where $\sigma(n)$ is the sum of divisors of $n$, including $1$ and $n$) exist.

But suppose, a perfect odd number exists.

Do we then know whether there are infinite many perfect odd numbers ?

This could be possible because various necessary conditions are known for an odd number to be perfect. Perhaps they allow to construct arbitary many perfect odd numbers, supposing that an odd perfect number exists.

Peter
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  • Are you sure this problem not open ? – Emax Jan 27 '17 at 10:59
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    @SSepehr No, therefore I ask – Peter Jan 27 '17 at 10:59
  • Well, we don't even know whether there are infinite perfect numbers (either even or odd), so I'd say this is very close to be an open question... – DonAntonio Jan 27 '17 at 11:31
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    @JoseArnaldoBebitaDris What did you complete ? The proof that there are no perfect numbers, or whether there are infinite many if there is one ? And if it is the latter, are there infinite many or not (assuming that there is one) ? – Peter Sep 08 '17 at 10:50
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    @JoseArnaldoBebitaDris Thank you! The EVEN perfect numbers are closely related to the Mersenne prime numbers and it is conjectured (but not proven) that there are infinite many Mersenne prime numbers which would imply that there are infinite many EVEN perfect numbers. – Peter Sep 08 '17 at 11:35
  • @Peter: I have undeleted my (erroneous) answer from two years ago, mainly for my own benefit. I have indicated where the reasoning is unsound. – Jose Arnaldo Bebita Dris Jan 23 '20 at 09:45
  • @Peter: I am posting a follow-up answer in a bit. – Jose Arnaldo Bebita Dris Feb 03 '20 at 08:55

2 Answers2

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This is not a direct answer to your question, but it is certainly related.

We do know that $$I(q^k) < \frac{5}{4} < \frac{3}{2} \leq I(2^{p-1})$$ where $N=q^k n^2$ is an odd perfect number in Eulerian form and $M=(2^p - 1){2^{p-1}}$ is an even perfect number in Euclidean form.

Because of a property satisfied by the abundancy index $I(x)=\sigma(x)/x$ at prime powers, it follows that $$2^{p-1} < q^k.$$

(NOTE THAT THE REASONING IS UNSOUND IN THE LAST PARAGRAPH, AS IT ONLY FOLLOWS FROM $I(q^k) < I(2^{p-1})$ THAT $2 < q$. WHAT FOLLOWS IS THEREFORE INCONCLUSIVE, AT THIS POINT.)

It follows that, since $q^k < n^2$ [Dris, 2012], if $N$ is bounded, then $M$ is also bounded.

Thus, if there exists an odd perfect number, we would infer that there are only finitely many even perfect numbers.

Jose Arnaldo Bebita Dris
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This answer attempts to prove your conjecture that if there is an odd perfect number, then there are infinitely many of them.

Briefly, the argument makes use of the following lemma:

Lemma If $q^k n^2$ is an odd perfect number with special/Euler prime $q$, then the biconditional $$\frac{q^2}{n}+\frac{n}{q^2} \text{ is bounded from above } \iff \frac{\sigma(q^2)}{n}+\frac{\sigma(n)}{q^2} \text{ is bounded from above}$$ holds.

Details are in the following question: If $q^k n^2$ is an odd perfect number with Euler prime $q$, which of the following relationships between $q^2$ and $n$ hold?

Jose Arnaldo Bebita Dris
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