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 ?
 A: 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}
A: It is in fact the case that,
for $a = \frac{3 + \sqrt{5}}{2}$,
$$
\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
\boldsymbol{\floor{a \floor{ a n}} + n = 3 \floor{a n}.}
$$
Proof. Let $r = a n - \floor{a n}$, $0 < r < 1$.
Then $\floor{a n} = (a n - r)$, so
we need to show that
\begin{align*}
3(a n - r) - n &\le a (a n - r) < 3(a n - r) - n + 1 \\
3 a n - 3 r - n &\le a^2 n - r < 3 a n - 3 r - n + 1 \\
\end{align*}
Notice that $a$ satisfies $a^2 = 3 a - 1$.
This reduces the above to
$$
3 a n - 3 r - n \le 3 a n - a r  - n < 3 a n - 3 r - n + 1
$$
Adding $3r + n - 3a n$,
$$
0 \le (3 - a) r < 1
$$
which is true since $0 < r < 1$ and
$$
3 - a = \frac{3 - \sqrt{5}}{2} < \frac{3 - 2}{2} <  1.
$$
