Can $\sqrt{p}^{\sqrt{p}^{\sqrt{p}}}$ be an integer, when $p$ is a non-square positive integer?

Of course, it seems it would never but is there a proof of the fact, or maybe we have some spooky $p$ that makes it valid?

  • $\begingroup$ The usual $a^{(b^c)}$ precedence is understood? $\endgroup$ Commented May 12, 2014 at 12:49
  • $\begingroup$ @enzotib Yes, of course :) $\endgroup$
    – Sawarnik
    Commented May 12, 2014 at 13:03
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    $\begingroup$ The results in here might be of your interest. $\endgroup$ Commented May 12, 2014 at 15:33
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    $\begingroup$ Surely this question is widely open? But if we assume the Schanuel Conjecture, then such towers are transcendental, according to this recent paper by Marques and Sondow, arxiv.org/pdf/1212.6931.pdf $\endgroup$
    – John M
    Commented May 16, 2014 at 2:47
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    $\begingroup$ @JohnM Perhaps you could exposit on the relationship to this conjecture, and the general surrounding situation in the literature. $\endgroup$
    – Alexander Gruber
    Commented May 16, 2014 at 23:11

1 Answer 1


Although such a number almost surely can not be an integer, it seems that this should be an open problem still.

However the claim that it can not be an integer would follow from Schanuel's conjecture:

Given any $n$ complex numbers $z_1,\dots,z_n$ which are linearly independent over $\mathbb{Q}$, the extension field $\mathbb{Q}(z_1,\dots,z_n, e^{z_1},\dots,e^{z_n})$ has transcendence degree at least $n$ over $\mathbb{Q}$, i.e. the set $\{z_1,\dots,z_n, e^{z_1},\dots,e^{z_n}\}$ contains at least $n$ algebraically independent numbers.

This is a strong statement. For example, let $z_1 = 1$ and $z_2 = i\pi$. Then the set $\{1, i\pi, e, -1\}$ must contain two algebraically independent numbers. Then $\pi$ and $e$ would be algebraically independent, showing that, for example, $e+\pi$ and $e\pi$ are transcendental (which otherwise is an open problem).

Now consider an algebraic number $\alpha$ which is irrational (such as $\sqrt{m}$ for $m$ not a square). By the Gelfond-Schneider theorem, $\alpha^\alpha$ is transcendental. Therefore, $\log \alpha$, $\alpha \log \alpha$, and $\alpha^\alpha \log \alpha$ are $\mathbb{Q}$-linearly independent.

By Schanuel's conjecture, the set $$\log \alpha, \alpha \log \alpha, \alpha^\alpha \log \alpha, \alpha, \alpha^\alpha, \alpha^{\alpha^\alpha}$$ contains at least three elements that are algebraically independent. But $\alpha$ is algebraic. Also $\alpha \log \alpha$ and $\alpha^\alpha \log \alpha$ are algebraically dependent on $\log \alpha$ and $\alpha^\alpha$, so $\log \alpha$, $\alpha^\alpha$ and $\alpha^{\alpha^\alpha}$ must be algebraically independent.

In particular, $\alpha^{\alpha^\alpha}$ must be transcendental.

For more on this subject, including more and stronger consequences on Schanuel's conjecture, please see the papers by Marques and Sondow, Schanuel's conjecture and algebraic powers $z^w$ and $w^z$ with $z$ and $w$ transcendental and The Schanuel Subset Conjecture implies Gelfond's Power Tower Conjecture.

I would be interested if anyone thought that, with this kind of problem, that proving something is not rational or not an integer might be substantially more tractable than proving it is transcendental.


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