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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 $\underbrace{\pi^{\pi^{\pi^{\pi^\pi}}}}_{5 \text{ times}}$.

Vladimir Reshetnikov
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    If you need to ask whether something can be proved, then it is not _obviously_ true. (Except if your question is "do we need to take this as an axiom or does it follow from simpler principles?"). – hmakholm left over Monica Dec 13 '11 at 23:08
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    Excuse my obvious irony, but I find it ironical that the obviously is obviously ironical. – Myself Dec 13 '11 at 23:27
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    @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
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    @AndresCaicedo: Do you have a strong reason to believe that both questions will be resolved in the same direction or by similar methods? – Vladimir Reshetnikov Dec 13 '11 at 23:30
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    For some reason, WolframAlpha says it is not an integer: http://www.wolframalpha.com/input/?i=Is+pi%5Epi%5Epi%5Epi%5Epi+an+integer%3F – Vladimir Reshetnikov Dec 13 '11 at 23:36
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    If we use [fraction](http://www.wolframalpha.com/input/?i=Is+pi%5Epi%5Epi%5Epi%5Epi+a+fraction%3F) instead integer the problem looks hard^^ – GarouDan Dec 13 '11 at 23:49
  • @GarouDan: the problem is actually hard even if we use 'integer', because we simply don't know pi to enough places to be able to use interval arithmetic to bound it between two adjacent integers. – Steven Stadnicki Dec 14 '11 at 00:15
  • I don't know how the [IntegerQ](http://www.google.com.br/url?sa=t&rct=j&q=integerq&source=web&cd=1&ved=0CCgQFjAA&url=http%3A%2F%2Freference.wolfram.com%2Fmathematica%2Fref%2FIntegerQ.html&ei=9uznTpWQNYrFtgfx1ZH0CQ&usg=AFQjCNF4tzs8WA9yCYV8bDgHScLVMOmg3Q) Mathematica algorithm works, but if we know this, I think is done. – GarouDan Dec 14 '11 at 00:25
  • The words "by the way" seem inappropriate here. – Michael Hardy Dec 14 '11 at 00:27
  • Four upvotes seems inappropriate! – The Chaz 2.0 Dec 14 '11 at 00:49
  • @VladimirReshetnikov, by "converse" I take it you mean "negation". – Gerry Myerson Dec 14 '11 at 00:56
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    What is the motivation for asking about this particular number? – Jonas Meyer Dec 14 '11 at 00:59
  • @GerryMyerson: Yes, I meant "negation". Sorry for confusion. – Vladimir Reshetnikov Dec 14 '11 at 01:04
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    @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 01:15
  • @TheChaz What do you mean? – Vladimir Reshetnikov Dec 14 '11 at 01:17
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    I mean that people use votes to indicate something other than "this shows research effort". That's not to say that you won't get an upvote from me (eventually). I DO appreciate your continued interaction with those who would help you. – The Chaz 2.0 Dec 14 '11 at 01:24
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    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: http://www.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 01:38
  • @MichaelLugo lol. `IntegerQ[(((1 + Sqrt[5])/2)^5 - ((1 - Sqrt[5])/2)^5)/Sqrt[5]]` doesn't works too. Bad implemented this function. – GarouDan Dec 14 '11 at 02:05
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    @GarouDan "IntegerQ[expr] returns False unless expr is manifestly an integer (i.e. has head Integer)." http://reference.wolfram.com/mathematica/ref/IntegerQ.html – Vladimir Reshetnikov Dec 14 '11 at 03:30

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