<s>[INST] Let $a=\dfrac{3+\sqrt{5}}{2}.$ Show that $\lfloor a \lfloor an \rfloor \rfloor+n$ is divisible by $3$. 
Let $a=\dfrac{3+\sqrt{5}}{2}.$
Show that for all $n\in\mathbb N$, $\lfloor a \lfloor an \rfloor \rfloor+n$ is divisible by $3$.

My teacher solve this problem with induction, I am just curious if we can do this exercise without it ?
 [/INST] Let $\alpha=\dfrac{1+\sqrt{5}}{2}, \beta=\dfrac{3+\sqrt{5}}{2}$, then
$$\beta= \alpha^2= \alpha+1$$
Since 
 $$\alpha\lfloor\beta n\rfloor-\lfloor\beta n\rfloor=\alpha n-(\alpha-1)\{\alpha n\}=\lfloor\alpha n\rfloor+(2-\alpha)\{\alpha n\}$$ 
so $\lfloor\alpha n\rfloor\lt \alpha\lfloor\beta n\rfloor-\lfloor\beta n\rfloor\lt \lfloor\alpha n\rfloor+1$. we get that
$$\lfloor\alpha\lfloor\beta n\rfloor\rfloor=\lfloor\alpha n\rfloor+\lfloor\beta n\rfloor$$
then, we have
\begin{equation}
\begin{split}\lfloor\beta\lfloor\beta n\rfloor\rfloor+n &=\lfloor\alpha\lfloor\beta n\rfloor+\lfloor\beta n\rfloor \rfloor+n\\&=\lfloor\alpha\lfloor\beta n\rfloor\rfloor+\lfloor\beta n\rfloor +n\\
&=\lfloor\alpha n\rfloor+\lfloor\beta n\rfloor+\lfloor\beta n\rfloor +n\\
&=\lfloor\alpha n\rfloor+2\lfloor\beta n\rfloor +n\\
&=\lfloor\alpha n\rfloor+2\lfloor\alpha n +n\rfloor +n\\
&=3\lfloor\alpha n\rfloor +3n
\end{split}
\end{equation}
</s>