Let $k$ be a positive integer congruent to $1$ modulo $4$ which is not a perfect square and let $a=\frac{1+\sqrt{k}}{2}$.
Show that $\{\left \lfloor{a^2n}\right \rfloor-\left \lfloor{a\left \lfloor{an}\right \rfloor}\right \rfloor : n \in \mathbb{N}_{>0}\}=\{1 , 2 , \ldots ,\left \lfloor{a}\right \rfloor\}$.
Note that $a^2 -a $ is an integer. Hence, $a^2 n - a n \in \mathbb{Z}$.
Let us denote $\varepsilon := an-\lfloor an \rfloor$. Then $a^2 n - \lfloor a^2 n \rfloor = \varepsilon$ and $\lfloor a^2 n \rfloor - a\lfloor an\rfloor = (a-1)\epsilon $.
It means that $\lfloor a^2 n \rfloor - \lfloor a\lfloor an\rfloor \rfloor = \lceil (a-1)\varepsilon \rceil $. Since $\varepsilon$ is dense in (0,1) when $n$ runs through integers, the result follows.