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The relationship between Mersenne primes $2^r-1$ and even perfect numbers $2^{r-1}(2^r-1)$ is well-known (Euclid, Euler).

In a video on the web I heard the statement that it is known that a Mersenne prime cannot divide an odd perfect number (quote: We do know, if we find an odd perfect number, it is not going to have a Mersenne prime as a factor). Is that true? Does anyone have a reference or a proof?

We know odd perfect numbers are of the form $$p^\alpha Q^2$$ where $p$ is a prime and $p\equiv\alpha\equiv 1 \pmod 4$ and $p\nmid Q$ (Euler). Clearly the special prime $p$ cannot be a Mersenne prime (Mersennes are $3\pmod 4$), so my question is if $Q$ (which is known to be composite of course) could contain a Mersenne prime factor.

Jeppe Stig Nielsen
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  • The video should have mentioned at least the main idea of the proof. – Peter May 27 '18 at 20:48
  • @Peter Yes. Although the video had another scope (the even perfect numbers), and the comment about the odd ones was kind of parenthetic. – Jeppe Stig Nielsen May 27 '18 at 20:52
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    @JeppeStigNielsen, +1 for sparking my interest and curiosity! Care to link to the video in question? – Jose Arnaldo Bebita Dris May 28 '18 at 10:16
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    I didn't see any proofs that odd perfect numbers cannot be multiple of $3$. – didgogns May 28 '18 at 12:37
  • @JoseArnaldoBebitaDris It is youtube.com/watch?v=q8n15q1v4Xo, see 1:27 into the video. – Jeppe Stig Nielsen May 28 '18 at 14:02
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    @didgogns That is actually an excellent remark. Since research articles such as Odd perfect numbers have at least nine distinct prime factors (Pace P. Nielsen) treat the case where $3$ is a factor of the odd perfect number, specifically, this seems to indicate that no-one has ruled out that Mersenne prime $3=M_2$ divides an odd perfect number. – Jeppe Stig Nielsen May 28 '18 at 15:41
  • Your comment qualifies as an actual answer to this question, @didgogns. – Jose Arnaldo Bebita Dris May 28 '18 at 23:50
  • @JeppeStigNielsen, I agree. It is currently an open problem to determine whether or not $3$ divides an odd perfect number. See this MO question for some discussions surrounding this inquiry. – Jose Arnaldo Bebita Dris May 29 '18 at 09:21
  • (1/2) I think that the users are right. On the other hand I wondered if the following question is in the literature Is it possible that the only prime factors of an odd perfect number are (a suitable choice of) Mersenne primes and/or Fermat primes? I don't know Florian Luca, The anti-social Fermat number, American Mathematical Monthly, 107 (2): pp. 171–173 (2000), I know about it from an informative point of view: what refers the Wikipedia section Other interesting facts from the Wikipedia article Fermat number. The behaviour of Euler's totient function $\varphi(m)$ and Dedekind psi – user759001 Apr 08 '20 at 12:57
  • (2/2) function $\psi(m)$ seems to me interesting (and their nested compositions as $\varphi(\psi(m)),\psi(\varphi(\psi(m)))\ldots$ or on the other hand $\psi(\varphi(m)),\varphi(\psi(\varphi(m)))\ldots$) for certain combinations of (I mean products of a suitable choice of the following factors), in particular, powers of Mersenne primes or powers of Fermat primes, provided values that are the sides related to some regular polygons that can be constructed with compass and straightedge (as reference I refer the related section from Wikipedia Fermat number). – user759001 Apr 08 '20 at 13:03
  • Are you talking about Numberphile's video? But, it is not proved that Mersenne Primes cannot divide odd perfect numbers. They are of form 2^n - 1 and all powers of 2, except 1 and 2, are multiples of 4. So, they are of form 4n - 1 or 4n + 3. They cannot have an odd power on it as Euler proved the prime factor which has odd power, both the prime factor and the odd power are of form 4n + 1. They can only have even powers on them. – Syed Owais Ali May 19 '24 at 12:57

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Comment by didgogns converted to an answer: It is currently an open problem to determine whether or not $M_2 = 2^2 - 1 = 3$ does divide an odd perfect number. In fact, it is likewise currently an open problem to determine an odd number less than $105$ that does not divide an odd perfect number. (See this MO question for some discussions surrounding this latter problem.)

Jose Arnaldo Bebita Dris
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