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Apparently it is unknown whether $\pi \uparrow\uparrow 4$ is an integer (I learned this from this tweet). I'm curious about which real numbers have some power tower which is an integer. That is, facts about the set:

$S = \left\{x \in \mathbb{R} \vert \exists k\in\mathbb{N} \text{ s.t. }(x\uparrow\uparrow k) \in \mathbb{Z}\right\}$

What is known about this set? For instance, is it measure 0?

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    $\begingroup$ Ummm $x \in \mathbb{Z}^+$ as a start? $\endgroup$
    – David G. Stork
    Jan 4, 2021 at 22:35
  • $\begingroup$ @DavidG.Stork all integers not just the positive ones, but of course I'm more curious about irrational numbers. $\endgroup$
    – Davis Yoshida
    Jan 4, 2021 at 22:37
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    $\begingroup$ Do you want to limit it to $x > 0$? $x^x = e^{x \log x}$ becomes complex for negative $x$, and you have to consider branch cuts and such. If you consider $x >0$, then $x\uparrow\uparrow k$ is increasing, and so injective, so there is at most one $x$ with $x\uparrow\uparrow k = n$ for any $n$, so your set is countable (and thus measure $0$). $\endgroup$
    – Jair Taylor
    Jan 4, 2021 at 22:41
  • $\begingroup$ @JairTaylor Positive $x$ is fine. $\endgroup$
    – Davis Yoshida
    Jan 4, 2021 at 22:44
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    $\begingroup$ I'm pretty sure $S$ is dense in $[1,\infty)$, though it looks kinda tedious to show - but it's for not very interesting reasons involving words like "increasing function" and "continuous" and "goes to infinity" rather than anything involving the specifics of the problem. $\endgroup$
    – Milo Brandt
    Jan 4, 2021 at 23:01

1 Answer 1

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The Gelfond-Schneider theorem implies that

  • if $x \upuparrows 2 = x^x \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ then $x$ is transcendental, and
  • if $x \upuparrows 3 = x^{x^x} \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ then $x$ is irrational.

But I don't think it's known whether or not $x \upuparrows 3 \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ implies that $x$ is transcendental, or whether or not $x \upuparrows 4 \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ implies that $x$ is irrational.

It is at least easy to see the following: for fixed $k$ the function $f(x) = x \upuparrows k$ satisfies $f(1) = 1$, is strictly increasing on $[1, \infty)$, continuous, and goes to infinity, so takes on every real value in $[1, \infty)$ exactly once. So for every positive integer $n \in \mathbb{N}$ there is a unique positive real $x \in [1, \infty)$ such that $f(x) = n$, which we might call the $k^{th}$ power-tower-root of $n$. But I think very little is known about these reals. I would not expect them to be expressible in terms of other familiar constants but I expect this to be quite out of reach. Here's another example of "this is obviously true and nobody has the slightest clue how to prove it": nobody knows whether or not $\pi + e$ or $\pi e$ are irrational (although at least one of them must be transcendental)!

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  • $\begingroup$ A small detail: $f$ is not strictly increasing on $(0,\infty)$. For $k = 2$, $f$ is strictly decreasing on $(0,1/e)$ and strictly increasing on $(1/e, \infty)$ and has a local minimum at $1/e$. I think you mean that $f$ is strictly increasing on $[1,\infty)$, right? $\endgroup$
    – Marktmeister
    Jan 4, 2021 at 23:06
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    $\begingroup$ @Marktmeister: yes, I meant to say it's strictly increasing on $[1, \infty)$, thanks for the correction. $\endgroup$
    – Qiaochu Yuan
    Jan 4, 2021 at 23:07

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