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Topic: Never an integer (Read 766 times) 

meyerjh28
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Never an integer
« on: Jan 19^{th}, 2005, 9:54pm » 
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Show, without using the GelfondSchneider Theorem, that n^sqrt(p) is never an integer, where n and p are positive integers > 1, and p is not a square.


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Eigenray
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Re: Never an integer
« Reply #1 on: Jan 22^{nd}, 2005, 7:05pm » 
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It's been a few days, so I thought I'd officially say I've gotten nowhere. A proof by descent seems possible but unlikely. The only thing that comes to mind is that there's an integer k such that 0 < [sqrt]p  k < 1, and sequences of integers {a_{r}} and {b_{r}} such that ([sqrt]pk)^{r} = a_{r} + b_{r}[sqrt]p decays exponentially to 0. Or there are infinitely many integers a,b, such that 0 < a[sqrt]p  b < 1/a, so if m=n^{sqrt(p)}, then n^{b} < m^{a} < n^{b+1/a}, where the first two terms are integers. Does such a proof actually exist, or are you just asking?

« Last Edit: Jan 24^{th}, 2005, 3:04pm by Eigenray » 
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meyerjh28
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Re: Never an integer
« Reply #2 on: Jan 24^{th}, 2005, 10:00pm » 
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I am just asking. As you know, the result follows immediately from the GS Theorem, but I thought that there must be a simpler way to show this, as sqrt(p) is only a degree 2 irrational. Maybe there is a simpler way to take care of this specific case. I don't know.


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