# For what $x$ is there some $k$ such that $x\uparrow\uparrow k$ is an integer?

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:

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

• Ummm $x \in \mathbb{Z}^+$ as a start? Commented Jan 4, 2021 at 22:35
• @DavidG.Stork all integers not just the positive ones, but of course I'm more curious about irrational numbers. Commented Jan 4, 2021 at 22:37
• 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 $0$). Commented Jan 4, 2021 at 22:41
• 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. Commented Jan 4, 2021 at 23:01
• I asked for which $x$ there is any $k$ such that $x \uparrow \uparrow k$ is an integer. $k=1$ satisfies this requirement for any integer. Commented Jan 4, 2021 at 23:33

• 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)!
• 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? Commented Jan 4, 2021 at 23:06
• @Marktmeister: yes, I meant to say it's strictly increasing on $[1, \infty)$, thanks for the correction. Commented Jan 4, 2021 at 23:07