# Could a Fermat prime divide an odd perfect number?

This question is a follow-up to this earlier post.

A positive integer $$N$$ is said to be perfect if $$\sigma(N)=2N$$, where $$\sigma(x)$$ is the sum of divisors of $$x \in \mathbb{N}$$. If $$M$$ is odd and $$\sigma(M)=2M$$, then $$M$$ is said to be an odd perfect number. Euler proved that an odd perfect number, if it exists, must have the form $$p^k m^2$$, where $$p$$ is the special/Euler prime satisfying $$p \equiv k \equiv 1 \pmod 4$$ and $$\gcd(p,m)=1$$.

A number of the form $$F_n = 2^{2^n} + 1$$ is said to be a Fermat number. If $$F_n$$ is prime, then it is called a Fermat prime.

Here is my question in this post:

Could a Fermat prime divide an odd perfect number?

MY ATTEMPT

(1) It is currently unknown if $$5 = 2^{2^1} + 1$$ divides an odd perfect number.

(2) Here is a related question for the case $$p = 17 = 2^{2^2} + 1$$. In fact, in a recent preprint (Dris, Tejada [2018]), it is shown that the Descartes-Frenicle-Sorli Conjecture for odd perfect numbers (that is, $$k=1$$) is incompatible with the set of assumptions $$\bigg(m^2 - p^k \text{ is a power of two}\bigg) \land \bigg(p \text{ is a Fermat prime}\bigg).$$

Updated December 12, 2018 Said preprint has now been published in NNTDM.

• It is not known if Fermat primes $F_0$, $F_1$ etc. may divide an odd perfect number. It is known that $3\cdot 5\cdot 7=105$ cannot divide a perfect number. It is known that if $3$ does not divide an odd perfect number, then that odd perfect number will have at least 12 distinct prime factors. Commented May 26, 2021 at 7:10