Could a Mersenne prime divide an odd perfect number?

by Jeppe Stig Nielsen   Last Updated May 28, 2018 10:20 AM

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.

Answers 1

For starters: Suppose that a Mersenne prime $q = 2^r - 1$ divides an odd perfect number $N$, and suppose further that $q^2 || N$. Then we have

$$2^{2r} - 2^r + 1 = 2^{2r} - 2^{r+1} + 1 + 2^r - 1 + 1= q^2 + q + 1 = \sigma(q^2) \mid 2N$$

so that $(2^{2r} - 2^r + 1) \mid N$ (since $2^{2r} - 2^r + 1$ is odd).

So, basically you now have a priori knowledge of (at least) two (proper) factors of $N$, namely $2^r - 1$ and $2^{2r} - 2^r + 1$. (It might then be possible to derive a contradiction from assuming these, though I still do not possess any proof of this fact.)

And then, after disproving $q^2 || N$, you assume $q^4 || N$, etc.

(Essentially, what I am thinking of is that $q^{2i}$ is a divisor of $N$ (for a certain $i \in \mathbb{N}$), and then $\sigma(q^{2i})$ is odd, which then cannot divide $\sigma(p^{\alpha})$ where $p$ is the Euler prime of the odd perfect number $N$. Looks like we will be left with one unmatched prime, in the spirit of Simon Davis's proof attempt at the Odd Perfect Number conjecture.)

Jose Arnaldo Bebita Dris
Jose Arnaldo Bebita Dris
May 28, 2018 10:14 AM

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