<s>[INST] Prove that $\left \lfloor \frac{1+\lfloor na+1/a\rfloor}{a} \right \rfloor=n$ If $a \geq \frac{1+\sqrt{5}}{2}$ and $n \in \Bbb W$, prove that
$$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=n.$$
I could prove only when $a$ is an integer, that is $a \geq 2$. If $a \in \Bbb Z$ we have: $$\left \lfloor \frac{1+na^2}{a} \right \rfloor=na+\left\lfloor \frac{1}{a} \right\rfloor=na$$ so we get: $$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=\left \lfloor \frac{1+na}{a} \right \rfloor=n+\left\lfloor \frac{1}{a} \right\rfloor=n.$$ But what if $a \notin \Bbb Z$?
 [/INST] From the well known
$$x-1<\lfloor x\rfloor  \leq x \tag{1}$$
Thus
$$\frac{1+na^2}{a}-1<
\left\lfloor \frac{1+na^2}{a}\right\rfloor\leq 
\frac{1+na^2}{a}$$
and
$$n<n+\frac{1}{a^2}<
\frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a}\leq
n+\frac{1}{a^2}+\frac{1}{a}\tag{2}$$
Finally, for $\color{red}{a>\phi}$
$$\frac{1}{a}+\frac{1}{a^2} <\frac{2}{1+\sqrt{5}}+\frac{4}{(1+\sqrt{5})^2}=
\frac{2+2\sqrt{5}+4}{(1+\sqrt{5})^2}\\
=\frac{6+2\sqrt{5}}{6+2\sqrt{5}}=1$$
and from $(2)$
$$n<
\frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a}<
n+1\tag{3}$$
the result follows from $(3)$.

The corner case for $\color{red}{a=\phi}$ leads to
$$\left\lfloor \frac{1+n\phi^2}{\phi}\right\rfloor=
\left\lfloor \frac{\sqrt{5}-1}{2}+n\cdot\frac{\sqrt{5}+1}{2}\right\rfloor=\\
\left\lfloor (n+1)\cdot\frac{\sqrt{5}}{2}+\frac{n+1}{2}-1\right\rfloor=
\left\lfloor (n+1)\cdot\frac{\sqrt{5}+1}{2}-1\right\rfloor=\\
\left\lfloor (n+1)\cdot\phi-1\right\rfloor < ...$$
because $\phi$ is irrational, thus $(n+1)\cdot\phi-1$ can never be an integer
$$...< (n+1)\cdot\phi-1$$
Then $(2)$ becomes
$$n<
\frac{1+\left\lfloor \frac{1+n\phi^2}{\phi}\right\rfloor}{\phi}<
n+1$$

I wanted to make it a short answer, but the corner case and the explanatory notes spoiled the effort.
</s>