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

  • If you tried, what did you try? It seems unlikely (but certainly possible) that no $^n\pi$ is ever even rational, much less prime. – Thomas Andrews Apr 2 at 18:14
  • 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). – Kevin Meier Apr 2 at 18:22
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    I think to check whether $^n \pi$ is an integer is hard enough. – Sisyphus Apr 2 at 18:54

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.

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