In article <email@example.com>, Eric J. Wingler <firstname.lastname@example.org> wrote: > >"ManOfLight" <email@example.com> wrote in message >news:firstname.lastname@example.org... >> 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.