“I am sure there are infinitely many perfect numbers”

The question Are there infinitely many perfect numbers? is a classic old unsolved problem. However, we keep finding perfect numbers (via Mersenne primes) and produce a lot of knowledge on perfect numbers, e.g. distribution and more. And when I am reading about this topic perfect numbers are often treated (maybe it is just my perception) as if they were infinitely many. And to be honest: After learning about perfect numbers etc. my intuition now tells me that there have to be infinitely many of them, no?.

I just realized this personal bias and want ask those who are experts in this field: Isn't it more likely that there are infinitely many perfect number? Are there some mathematical statements that make the existence of infinitely many perfect numbers more likely? And: Does it even make sense to ask this question?

$~$

Remark 1: In the book The Riemann Hypothesis: A Resource for the Afficionado and Virtuoso Alike the authors point out various aspects from different mathematical fields that all point to the validity of the Riemann hypothesis. They concluded that it is quite unlikely (whatever this means) for the RH to be false (they emphasized: not because of a lot of empirical data, but because of some other mathematical connections that make the validity more likely). This motivated my questions: So are there some mathematical facts that make the existence of infinitely many perfect numbers more likely?

Remark 2: Sorry for such a provocative title :)

• I have adjusted a few uses of "infinite" to "infinitely many", since I believe you want to know how many perfect numbers there are, not whether there is some sensical way to talk about infinitely large individual perfect numbers. If I have changed the intent of your question, please feel free to revert my changes. – Eric Towers Sep 22 '17 at 21:02
• @EricTowers Thanks for the edit! – Fabian Schn. Sep 22 '17 at 21:04
• @Arthur Thanks, but I mentioned Mersenne primes already and was aware of the search for odd perfect numbers. You said "but there might be"... thats what I am asking for, why can you say this? Is it really more likely that there are some out there? (However, for me is the question for the number of perfect numbers more important) – Fabian Schn. Sep 22 '17 at 21:22
• The point about "I am sure" in mathematics is the response "well can you prove it?" There have been many eminent mathematicians who have been sure but wrong (perhaps a "big list" question on the topic is indicated). Without a proof you only have a conjecture. There are not many conjectures dignified with community recognition. – Mark Bennet Sep 22 '17 at 21:22
• "I am sure that the twin prime conjecture is true," or "I am sure that the Reimann hypothesis is correct" run through mathematics. But there is a big gap between you being sure that they are true and there being a proof that says so. Which gives two paths. One is to assume that it is true and investigate the implications. The other is to find what it is about perfect numbers that we don't already understand in order to move toward being able to write the proof. – Doug M Sep 22 '17 at 21:35

This is not a complete answer. Just my views:

Assuming that you know that a perfect number is the one which is of the form $2^{p-1}\times (2^p - 1)$ where p and $2^p-1$ are primes (the latter being well known Mersenne primes).

Euler established this well known one to one correspond hence between Mersenne primes and perfect numbers. So if you have a Mersenne primes you get an even perfect number and vice versa (it is still unknown whether there are any odd perfect numbers or not).

Sadly, this is the only way of finding perfect numbers. So in order to discuss the infinitude of perfect numbers, you need to ultimately discuss infinitude of Mersenne primes which is not a piece of cake. Many great mathematical minds think that answer for the question are there infinitely many Mersenne primes? is yes!! but again there is no proof yet.

See this

• A quibble: this is the correspondence between even perfect numbers and Mersenne primes. I don't think it is known whether there are any odd perfect numbers (but I've not been paying attention to this issue for some years, so maybe there is progress...) – paul garrett Sep 22 '17 at 22:19

(Note: What follows was previously posted as an answer to this related MSE question, but has now been deleted.)

This is not a direct answer to your question, but it is certainly related.

We do know that $$I(q^k) < \frac{5}{4} < \frac{3}{2} \leq I(2^{p-1})$$ where $N=q^k n^2$ is an odd perfect number in Eulerian form and $M=(2^p - 1){2^{p-1}}$ is an even perfect number in Euclidean form.

What follows are incorrect, as rightly pointed out by reuns. I leave the rest of the original answer untouched, mainly for my own benefit.

Because of a property satisfied by the abundancy index $I(x)=\sigma(x)/x$ at prime powers, it follows that $$2^{p-1} < q^k.$$

It follows that, since $q^k < n^2$ [Dris, 2012], if $N$ is bounded, then $M$ is also bounded.

Thus, if there exists an odd perfect number, we would infer that there are only finitely many even perfect numbers.

This is in response to user243301's other answer to the OP.

These images are from WolframAlpha, and were a result of two (similar yet different) inequalities. The significance of these images is that the inequalities which produced them have roots in the theory of odd perfect numbers:  • @reuns, $q^k < n^2$ implies $q^{2k} < q^k n^2 = N$. Thus, $q^k$ is bounded from above by $\sqrt{N}$. Therefore, if there exists at least one odd perfect number $N$, then all the $p$'s corresponding to $2^p - 1$ (and therefore, to $2^{p-1}$) are bounded above by $\sqrt{N}$. (If you prefer to be picky, if we set $M_p = 2^p - 1$, then ${M_p}(M_p + 1)/2 < 2\sqrt{N}(\sqrt{N} - 1)$. – Jose Arnaldo Bebita-Dris Oct 8 '17 at 5:05
• @reuns, consequently, we conclude that if there exists an odd perfect number, then there are only finitely many even perfect numbers. Note that the same conclusion follows if we assume that there are finitely many (i.e. at least two) odd perfect numbers. – Jose Arnaldo Bebita-Dris Oct 8 '17 at 5:08
• What $p$ ? Why $2^p-1$ and $2^{p-1} (2^p-1)$ is related to $N$ ? – reuns Oct 8 '17 at 5:09
• Did you read my answer thoroughly, @reuns? $I(q^k) < 5/4 < 3/2 \leq I(2^{p-1})$ implies $2^{p-1} < q^k$ (since both $2$ and $q$ are primes), right? That is, an upper bound for $N = q^k n^2$ implies an upper bound for $M = (2^p - 1){2^{p-1}}$. – Jose Arnaldo Bebita-Dris Oct 8 '17 at 5:25
• Thank you for the downvote. =) – Jose Arnaldo Bebita-Dris Oct 8 '17 at 5:27