Fix two positive integers $k,m>1$. Now take some $n$ for which $\{n\sqrt{2003}\}<\frac 1{k^m}$. Now, for each $i\in\overline{1,m-1}$, we have $\lfloor k^{i-1}n\sqrt{2003}\rfloor\le\frac{k^in\sqrt{2003}}k<\lfloor k^{i-1}n\sqrt{2003}\rfloor+\frac 1k$, which, after multiplication with $k$, gives $k\lfloor k^{i-1}n\sqrt{2003}\rfloor\le k^in\sqrt{2003}<k\lfloor k^{i-1}n\sqrt{2003}\rfloor+1\Rightarrow\lfloor k^in\sqrt{2003}\rfloor=k\lfloor k^{i-1}n\sqrt{2003}\rfloor$. This means that the numbers $\lfloor k^in\sqrt{2003}\rfloor,\ i\in\overline{0,m-1}$ are $m$ numbers in geometric progression with ratio $k$.