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Topic: Number theory question continued
Replies: 9   Last Post: Mar 27, 2006 5:13 PM

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Robert Israel

Posts: 11,902
Registered: 12/6/04
Re: Number theory question continued
Posted: Mar 27, 2006 4:31 PM
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In article <44283c39$1@news.ysu.edu>,
Eric J. Wingler <wingler@math.ysu.edu> wrote:
>
>"ManOfLight" <mladensavov@yahoo.com> wrote in message
>news:1143276915.440583.51260@z34g2000cwc.googlegroups.com...

>> Hello everybody,
>> I post it again to rapair one mistake
>> Could you give a clue how I can start the following problem or
>> propose a solution.
>>
>>
>> "Is it true that for every sufficiently large interval there will be a
>> integer in it of the form
>> 2^n-3^m where m,n are integers?"
>>
>>
>> As far as I understand it we are supposed either to prove that there
>> exists number H : every interval with length H contains such a number
>> or disprove it.

>
>If x = 2^n - 3^m, then log(1 - x/2^n) = m*log(3) - n*log(2), so it's
>possible that you may be able to make use of the density of the set of
>numbers of the form m*log(3) - n*log(2).


I doubt it.

This implies there are infinitely many integers of the form x = 2^n - 3^m
with c < x/2^n < d for any c and d with 0 < c < d < 1. But unfortunately
the interval for x grows as n -> infty. You can do slightly better
using results about Diophantine approximation, but still you won't get
an interval of constant length.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada







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