How to show $e^{e^{e^{79}}}$ is not an integer

In this question, I needed to assume in my answer that $e^{e^{e^{79}}}$ is not an integer. Is there some standard result in number theory that applies to situations like this?

After several years, it appears this is an open problem. As a non-number theorist, I had assumed there would be known results that would answer the question. I was aware of the difficulty in proving various constants to be transcendental — such as $e + \pi$, which is not known to be transcendental at present.

However, I was looking at a question that seems simpler, naively: whether a number is an integer, rather than whether it is transcendental. It seems that what appeared to be possibly simpler is actually not, with current techniques.

The main motivation for asking about this particular number is that it is very large. It is certainly possible to find a pair of very large numbers, at least one of which is transcendental. But the current lack of knowledge about this particular number is even an integer shows just how much progress remains to be made, in my opinion. Any answers that describe techniques that would suffice to solve the problem (perhaps with other, unproven assumptions) would be very welcome.

Answer

The paper Chuangxun Cheng, Brian Dietel, Mathilde Herblot, Jingjing Huang, Holly Krieger, Diego Marques, Jonathan Mason, Martin Mereb, and S. Robert Wilson, Some consequences of Schanuel’s conjecture, Journal of Number Theory 129 (2009) 1464–1467, shows that $e,e^e,e^{e^e},\dots$ is an algebraically independent set, on the assumption of Schanuel’s Conjecture. Maybe a close reading of that paper will suggest a way of applying the result to the 79-question.

Attribution
Source : Link , Question Author : Carl Mummert , Answer Author : Gerry Myerson

Leave a Comment