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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.

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