In article <email@example.com>, Robert Israel <firstname.lastname@example.org> wrote: >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.
On the other hand, even a rather weak lower bound on the approximation of log(3)/log(2) by rationals would be useful in the other direction. I believe there are such bounds, due to Gelfond (with later improvements by Baker and others).