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.

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.)

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