Author |
Topic: Never an integer (Read 790 times) |
|
meyerjh28
Newbie
Gender: 
Posts: 2
|
 |
Never an integer
« on: Jan 19th, 2005, 9:54pm » |
 Quote  Modify
|
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.
|
|
 IP Logged |
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
 |
Re: Never an integer
« Reply #1 on: Jan 22nd, 2005, 7:05pm » |
Quote Modify
|
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?
|
| « Last Edit: Jan 24th, 2005, 3:04pm by Eigenray » |
IP Logged |
|
|
|
meyerjh28
Newbie
Gender:
Posts: 2
|
|
Re: Never an integer
« Reply #2 on: Jan 24th, 2005, 10:00pm » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print