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A positive integer $n$ is called perfect if $\sigma(n)=2n$, where $$\sigma(n)=\sum_{d \mid n}{d}$$ is the sum of divisors of $n$.

If $n$ is odd and $\sigma(n)=2n$, then $n$ is called an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the form $n=p^k m^2$ where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

It is not known if any odd integers $< 100$ divide an odd perfect number.

It is known, however, that:

An odd perfect number cannot be divisible by 105

An odd perfect number cannot be divisible by 825

An odd perfect number cannot be divisible by 5313

We also have the following MSE question:

Can an odd perfect number be divisible by 165?

where the accepted answer is that: "An odd perfect number, if one exists, could in principle be divisible by 165."

Note that $$105 = 3\cdot 5\cdot 7$$ $$825 = 3\cdot {5}^2 \cdot 11$$ $$5313 = 3\cdot 7\cdot 11\cdot 23$$ and that $$165 = 3\cdot 5\cdot 11.$$

Now, I am intrigued by a possible relationship of these non-divisors of odd perfect numbers to coefficients of cyclotomic polynomials. Indeed, from the Wikipedia page on Cyclotomic polynomial, we have the following penultimate observation:

The case of the 105th cyclotomic polynomial is interesting because 105 is the lowest integer that is the product of three distinct odd prime numbers (3*5*7) and this polynomial is the first one that has a coefficient other than 1, 0, or −1:

WolframAlpha computation of the 105th cyclotomic polynomial

A similar case holds for the 825th and 5313th cyclotomic polynomials:

WolframAlpha computation of the 825th cyclotomic polynomial

WolframAlpha computation of the 5313th cyclotomic polynomial

Lastly, we have the following expression for the 165th cyclotomic polynomial:

$$x^{80} + x^{79} + x^{78} - x^{75} - x^{74} - x^{73} - x^{69} - x^{68} - x^{67}$$ $$+ x^{65} + 2x^{64} + 2x^{63} + x^{62} - x^{60} - x^{59} - x^{58} - x^{54}$$ $$- x^{53} - x^{52} + x^{50} + 2x^{49} + 2x^{48} + 2x^{47} + x^{46} - x^{44}$$ $$- x^{43} - x^{42} - x^{41} - x^{40} - x^{39} - x^{38} - x^{37} - x^{36}$$ $$+ x^{34} + 2x^{33} + 2x^{32} + 2x^{31} + x^{30} - x^{28} - x^{27} - x^{26}$$ $$- x^{22} - x^{21} - x^{20} + x^{18} + 2x^{17} + 2x^{16} + x^{15} - x^{13}$$ $$- x^{12} - x^{11} - x^7 - x^6 - x^5 + x^2 + x + 1$$

WolframAlpha computation of the 165th cyclotomic polynomial

I therefore dare to conjecture that:

CONJECTURE 1 An odd perfect number $n = p^k m^2$ is not divisible by $165$.

CONJECTURE 2 An odd perfect number $n = p^k m^2$ is not divisible by an odd number $l$ if the $l$-th cyclotomic polynomial has a coefficient other than $1, 0$, or $-1$.

Alas, I have no proof. So here is my question:

Can anybody here (yes, somebody with more computing power) come up with a counterexample?

1 Answers1

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$$\frac{\sigma(5.3^2.11^2)}{5.3^2.11^2}< 2$$ thus there is no barrier for an odd perfect number divisible by $3,5,11$.

On the other hand $$\frac{\sigma(5.3^2.7^2)}{5.3^2.7^2}> 2$$

reuns
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  • Thank you for your answer, @reuns. (Upvoted as well!) Would you care to comment more on why "there is no barrier for an odd perfect number divisible by $165$"? I am sure you are aware of the factor chain method of producing consequent factors from an assumed (initial) set of factors for an odd perfect number. – Jose Arnaldo Bebita Dris Feb 29 '20 at 07:42