amparvardi wrote: Prove that for every natural number $k$ ($k \geq 2$) there exists an irrational number $r$ such that for every natural number $m$, \[[r^m] \equiv -1 \pmod k .\] Remark. An easier variant: Find $r$ as a root of a polynomial of second degree with integer coefficients. Proposed by Yugoslavia. Choose $r=k+\sqrt{k^2-k}$ (since $(k-1)^2<k^2-k<k^2$, this an irrational number) and look at the sequence : $a_1=2k$ $a_2=4k^2-2k$ $a_{n+2}=2ka_{n+1}-ka_n$ You get $a_n=r^n+(\frac kr)^n$ and so $r^n=a_n-(\frac kr)^n$ and, since $0<k<r$ : $[r^n]=a_n-1$ And since $a_n\equiv 0\pmod k$, we get the result.

This problem is a particular case of a theorem more general that was proved by Pascual Restrepo in his article "Irrational numbers whose powers have a nice property". I attached the file: http://reflections.awesomemath.org/2008_4/Irrationals.pdf