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Given the tetration as

\begin{align} {^{n}a} = \underbrace{a^{a^{\cdot^{\cdot^{a}}}}}_n \end{align}

and the set of prime numbers as $\mathbb{P}$.

Can you prove or to disprove the following statement?

\begin{align} \forall p\in\mathbb{P}:(^{p^2+1}\pi)\in\mathbb{P} \end{align}

I tried to disprove it, but I failed to do it.

Thank you

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  • $\begingroup$ If you tried, what did you try? It seems unlikely (but certainly possible) that no $^n\pi$ is ever even rational, much less prime. $\endgroup$ – Thomas Andrews Apr 2 '18 at 18:14
  • $\begingroup$ Initially, I played around by searching some rules about $^n\pi^$ for small $n$, but I did not find anything. Computing some values also failed (if the statement is wrong this would probably the easiest way to disprove it). $\endgroup$ – Kevin Meier Apr 2 '18 at 18:22
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    $\begingroup$ I think to check whether $^n \pi$ is an integer is hard enough. $\endgroup$ – I was suspended for talking Apr 2 '18 at 18:54
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I believe that nothing close to this is known currently; see e.g. this answer and the ensuing discussion. There are lots of "basic" problems about integrality/nonintegrality, rationality/irrationality, algebraicity/transcendence, etc. which are wide open.

There are many number-theoretic conjectures with good heuristic evidence which are known to resolve many such questions, but I don't know of any which would resolve this particular one. I suspect that all heuristics currently known suggest that $^n\pi$ cannot even be an integer for $n\in\mathbb{N}$ (number theorists, correct me if I'm wrong about this), but I don't think we're anywhere close to proving that or any related claim. Even numerical evidence is hard to come by: $^4\pi$ is already so huge that we don't obviously have a method to check experimentally whether it "looks like" an integer.


EDIT: Adding to address questions along the lines of xakepp35's comments below, the order of exponentiation matters. For example, $$2^{(2^{(2^2)})}=2^{16}=65536$$ but $$(2^2)^{(2^2)}=4^4=256.$$ Of course the latter is much easier to calculate, and it's not hard to show that $(\pi^\pi)^{(\pi^\pi)}$ is not an integer.

Given that it's so intractable, why to people care about tetration? Well, many don't - it certainly doesn't have any real-world applications I know of, and it's a very niche topic in pure mathematics. It is however a simple example example of a very fast-growing function, and the general study of fast-growing functions (indeed, ones much faster than tetration) is quite interesting; they arise in Ramsey theory and proof theory, for example. (That first link is to Graham's number, which is unspeakably gigantic; it's definitely worth a read.)

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  • $\begingroup$ Why not check that 4^pi on a computer? Would not the value fit in a.. say it, 256GiB of RAM on a typical (older) server hardware? $\endgroup$ – xakepp35 Jun 25 '19 at 20:31
  • $\begingroup$ 1) pi^pi^pi^pi = (pi^pi)^(pi^pi), so we have to 2) calculate pp=pi^pi, which is 36,462159607 and put it to large enough memory cell. How large? well, lets estimate bits for integer part: ln(2^bits)=ln(pp^pp)) so bits = pp/log_base_pp(2). So kinda 190 bits will be enough to store integer part of tetration(pi,4). Then, quadruple this amount and take 1 kilobyte to avoid any losses.. Take same amount for mantissa. Take bigint library.. go! what's the problem? $\endgroup$ – xakepp35 Jun 25 '19 at 20:37
  • $\begingroup$ @xakepp35 Your first equation is wrong: it's $$\pi^{(\pi^{(\pi^\pi)})},$$ which is much larger. E.g. for contrast, consider the case of $2$: we have $$(2^2)^{(2^2)}=2^{2\cdot 2^2}=2^8=256$$ but $${}^42=2^{(2^{(2^2)})}=2^{(2^4)}=2^{16}=65536.$$ At a glance, I think ${}^4\pi$ is indeed computationally infeasible (although I'm not an expert of course). $\endgroup$ – Noah Schweber Jun 25 '19 at 21:18
  • $\begingroup$ What is an operation, where you raise base to a power first? Eg we can have 2 constructions - of 3,3. First should give you 27^3, second is 3^27. In computer science first is more useful, second is nowhere encountered. So I think you have wrong order and it thus overflows. $\endgroup$ – xakepp35 Jun 26 '19 at 6:34
  • $\begingroup$ Imagine Mandelbrot fractal process. You first raise Z to power 2 then add some C. And you got this very complex shape. Sane here . you may raise pi to pi first. Got new base, raise it to pi again, and then again. Resulting number is not so big, and it is irrational. And you may call it a power tower off, because it is a repeating process of raising to same power $\endgroup$ – xakepp35 Jun 26 '19 at 6:37

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