Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Let $a_n = an - \lfloor an \rfloor$, $n = 1, 2, 3, \dots$. Since $a^2 = a + (k-1)/4$, it follows that $\lfloor a^2 n \rfloor = \lfloor an \rfloor + n(k-1)/4$, and $(a-1)\lfloor an \rfloor = (a-1)(an - a_n) = n(k-1)/4 - (a-1)a_n$, so, adding $\lfloor an \rfloor$ to each side, $a\lfloor an \rfloor = \lfloor an \rfloor + n(k-1)/4 - (a-1)a_n = \lfloor a^2 n \rfloor - (a-1)a_n$. Since $a$ is irrational, the $a_n$ form a dense subset of the open unit interval $(0, 1)$, so, by the preceding, the differences $\lfloor a^2 n \rfloor - a\lfloor an \rfloor = (a-1)a_n$ form a dense subset of the open interval $(0, a-1)$. Finally, since $\lfloor a^2 n \rfloor - \lfloor a \lfloor an \rfloor \rfloor = \lfloor a^2 n \rfloor - a\lfloor an \rfloor \rfloor = \lfloor (a-1)a_n \rfloor$, the conclusion follows.