# Is ${^5\pi}$ an integer? [duplicate]

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

Is ${^5\pi}$ an integer? It is "obviously" not, right? But can we prove it?

Here ${^5\pi}$ means the result of tetration $\underbrace{\pi^{\pi^{\pi^{\pi^\pi}}}}_{5 \text{ times}}$.

## marked as duplicate by Andrés E. Caicedo, Gerry Myerson, Asaf Karagila♦, J. M. is a poor mathematician, t.b.Dec 14 '11 at 5:05

• @HenningMakholm: Of course, that is why I put "obviously" in quotes. I just mean that a possible proof that ${^5\pi}$ is not an integer would hardly be a surprising result for anyone, but the converse would really surprise many people. – Vladimir Reshetnikov Dec 13 '11 at 23:27
• @JonasMeyer: This is the smallest number of the form ${^n\pi}$ for which I do not know the answer. Tetration is the first hyperoperator (after addition, multiplication and exponentiation) for which the question is not trivial. And $\pi$ is just a natural example of a transcendental number. – Vladimir Reshetnikov Dec 14 '11 at 1:15
• Wolfram Alpha says that $(\phi^5-\tau^5)/\sqrt{5}$, where $\phi = (1+\sqrt{5})/2$ and $\tau = (1-\sqrt{5})/2$, is not an integer: wolframalpha.com/input/?i=Is+%28%28%281%2Bsqrt%285%29%29%2F2%29^5+-+%28%281-sqrt%285%29%29%2F2%29^5%29%2Fsqrt%285%29+an+integer%3F -- but it clearly is, because it's the fifth Fibonacci number, namely 5. (WA then gives a "decimal approximation" which is 5 followed by a couple thousand zeroes.) – Michael Lugo Dec 14 '11 at 1:38