# Odd perfect numbers having as prime factors exclusively Mersenne primes and Fermat primes

I don't know if the following question is in the literature, please add a commment if it is in the literature. I add my thoughts and motivation below in last paragraph, it is discursive and speculative, if this post, that is crossposted on Mathematics Stack Exchange (I've asked it as MSE 3636345 with a similar title a month ago) isn't suitable for this MathOverflow please add your feedback in comments that I can to delete it.

An odd perfect number is an odd integer $$N\geq 1$$ such $$\sigma(N)=\sum_{1\leq d\mid N}d=2N.$$ I add the Wikipedia article for Perfect number.

Question. Is it possible to rule out/discard that the only prime factors of an odd perfect number are (a suitable choice of) Mersenne primes and/or Fermat primes? I'm asking if we can to disprove the existence of odd perfect numbers having as prime divisors exclusively Mersenne primes and Fermat primes (it is unknown if there exist infinitely many Mersenne primes and it is unkonwn if there exists finitely many Fermat primes). Many thanks.

I'm asking it as a reference request to know if this question is in the literature, then refer it or add a comment with the bibliography and I try to search and read it from the literature. In other case I'm asking about what work can be done for my Question, and after some feedback in answers I should to choose an answer.

My only idea to do some work about the veracity of the question is try to compare to Euler's thorem for odd perfect numbers, and the theory of odd perfect numbers.

I don't know Florian Luca, The anti-social Fermat number, American Mathematical Monthly, 107 (2): pp. 171–173 (2000), I know about it from an informative point of view: what refers the Wikipedia section Other interesting facts from the Wikipedia article Fermat number.

It seems reasonable to think that the problem of the existence of odd perfect numbers is unrelated to the problem concerning even perfect numbers. Then we can to think in my question as a question of miscellany in mathematics. I've persuaded myself about certain things about some unsolved problems in mathematics. I know that this isn't scientific. Thus to avoid these ideas we propose this exercise just as a miscellaneous problem. What I evoke is that a different option that odd perfect numbers are unrelated to certain constellations of primes, is the speculative option that there is a close relationship.

## References:

[1] Could a Mersenne prime divide an odd perfect number?, MSE 2798459 from Mathematics Stack Exchange (May 27 '18).

[2] Could a Fermat prime divide an odd perfect number?, MSE 2960850 from Mathematics Stack Exchange (Oct 18 '18).

• I hope that this post is interesting for professors and users here, please feel free to add your feedback in comments. Jun 2, 2020 at 20:39

This should be provable by standard although laborious methods. What follows is a proof sketch (I have not checked all the computational details but this method should work).

We recall a few basic facts whose proofs will be omited:

Let $$h(n) = \sigma(n)/n$$, and let $$H(n) = = \prod_{p|n} \frac{p}{p-1}$$.

1. For all $$n$$, $$h(n) \leq H(n)$$ with equality iff $$n=1$$.

2. For all $$a$$ and $$b$$, $$h(ab) > h(a)$$ whenever $$b>1$$.

3. If $$N$$ is an odd perfect number then we may write $$N = q^e m^2$$ where $$q$$ is a prime, $$q \equiv e \equiv 1$$ (mod 4), and $$(q,m)=1$$. (This result is due to Euler and in some sense is very weak: it actually applies to any $$N$$ where $$N$$ is odd and $$\sigma(N) \equiv 2$$ (mod 4).

4. No perfect number can have as an abundant or perfect proper divisor.

5. No odd perfect number can be simultaneously divisible by 3, 5 and 7.

Now let us consider an odd perfect number $$N$$ only divisible by Fermat and Mersenne primes.

Let us first consider the case where $$N$$ is not divisible by 3. Then $$H(N)$$ is bounded above by $$\prod_{p}\frac{p}{p-1}$$ where $$p$$ is any Fermat prime or Mersenne prime other than 3. Call this product $$S$$. Then $$S \leq \prod_{i=2}^\infty \frac{2^{2^i}+1} {2^{2^i}} \prod_{i=5}^{\infty} \frac{2^i-1}{2^i-2}.$$ And it is not too hard to see that $$S<2$$, which is impossible if $$N$$ is perfect.

So we may assume that $$3|N$$. Thus, we can only have one of 5 and 7 dividing $$N$$. We can using the same approach get a contradiction if $$7|N$$, and get a contradiction if $$(35,N)=1$$. Thus we must have $$5|N$$.

Continuing in this way we get that $$(3)(5)(17)(257)|N$$ and that no primes smaller than 257 divide $$N$$. With a little work, one should also be able to show that one needs all of these Fermat primes raised to large powers.

The smallest Mersenne prime after 257 is 8191. Assume $$8191|N$$. But this will force $$N$$ to be divisible by $$(3)(5)(17)(257)(8191)$$ and with larger powers of all those primes, which would force $$N$$ to be divisible by an abundant number. So $$N$$ is not divisible by 8191. Then using that the next three Mersenne primes are 131071, 524287, and 2147483647, one should be able to get that the relevant product must be smaller than 2, and thus have a contradiction.

Note that the vast majority of the work is needed to handle the situation where we have all the known Fermat primes dividing $$N$$. This is a common difficulty in proving things with OPNs because the relevant product when taken over of all Fermat primes and Fermat pseudoprimes is exactly 2. So $$(3/2)(5/4)(17/16)(257/256)$$ is just a tiny bit below 2.

• What excellent! Many thanks for share such results. I'm going to study it, and as was said after there are feedback in answers in next days/weeks I should accept an available answer. Many thanks again for your attention on questions posted on MathOverflow. Jun 2, 2020 at 21:26
• Thank you very much again, I am understanding your argument in a first reading. It seems to me a miracle that from a question of mine a professional mathematician of this MathOverflow can to deduce such results. Jun 2, 2020 at 21:33