# Could a Mersenne prime divide an odd perfect number?

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.

• 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
• @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
• I didn't see any proofs that odd perfect numbers cannot be multiple of $3$. – didgogns May 28 '18 at 12:37
• @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

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.)