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        riddles >> hard >> Never an integer
        
(Message started by: meyerjh28 on Jan 19^th, 2005, 9:54pm)

Title: Never an integer
Post by meyerjh28 on Jan 19^th, 2005, 9:54pm

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.

Title: Re: Never an integer
Post by Eigenray on Jan 22^nd, 2005, 7:05pm

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]p-k)^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?

Title: Re: Never an integer
Post by meyerjh28 on Jan 24^th, 2005, 10:00pm

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.