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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$>,
Eric J. Wingler <> wrote:
>"ManOfLight" <> wrote in message

>> 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
Department of Mathematics
University of British Columbia Vancouver, BC, Canada

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