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Topic: Never an integer (Read 799 times) |
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meyerjh28
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Never an integer
« on: Jan 19th, 2005, 9:54pm » |
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Show, without using the Gelfond-Schneider 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 22nd, 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 {ar} and {br} such that
([sqrt]p-k)r = ar + br[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=nsqrt(p), then
nb < ma < nb+1/a,
where the first two terms are integers.
Does such a proof actually exist, or are you just asking?
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| « Last Edit: Jan 24th, 2005, 3:04pm by Eigenray » |
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meyerjh28
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Re: Never an integer
« Reply #2 on: Jan 24th, 2005, 10:00pm » |
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I am just asking. As you know, the result follows immediately from the G-S 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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