Re: Number theory question continued

In article <44283c39$1@xxxxxxxxxxxx>,
Eric J. Wingler <wingler@xxxxxxxxxxxx> wrote:

"ManOfLight" <mladensavov@xxxxxxxxx> wrote in message
news:1143276915.440583.51260@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada




.

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