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 $a \notin \Bbb Z$?
 A: 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.
A: We need to prove

*

*$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} \geqslant n $$

*$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} < n+1 $$


*

*$$ 1+\left\lfloor \frac{1+na^2}{a}\right\rfloor  > \frac{1+na^2}{a} > an$$
so dividing by $a$ we obtain what we wanted to.


*$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} \leqslant \frac{1+\frac{1+na^2}{a}}{a}$$
and
$$ 1+na^2 \leqslant na^2+a^2-a $$
since $0 \leqslant a^2 - a-1 = \left(a - \frac{1+\sqrt{5}}{2}\right)\left(a - \frac{1-\sqrt{5}}{2}\right)$, so after some transformations we arrive at
$$ \frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} \leqslant n+1 $$
Now, the equality may hold only if both inequalities we used are actually equalities, so $a = \frac{1+\sqrt{5}}{2}$ and $\frac{1+na^2}{a}$ is an integer. But since $a^2=a+1$,
$$ \frac{1+na^2}{a} = \frac{(n+1) + na}{a} = \frac{n+1}{a} + n, $$
which is irrational, in particular it can't be integer.
A: We wish to show that $\exists\epsilon\in[0,1)$ such that $$1+\left\lfloor\frac1a+na\right\rfloor=a(n+\epsilon)=na+\epsilon a\implies\left\lfloor\frac1a+\{na\}\right\rfloor=\{na\}+b$$ where $b=\epsilon a-1\in[-1,a-1)$. As the LHS is either $0$ or $1$, either $b=-\{na\}$ or $1-\{na\}$ so we will always find an $\epsilon\in(0,1)$ except the case where $$b=1-\{na\}>a-1\implies\{na\}<2-\phi$$ as $a\ge\phi$. This means that $$\frac1a+\{na\}<\frac1a+2-\phi\le\frac1\phi+2-\phi=1\implies\left\lfloor\frac1a+\{na\}\right\rfloor=0$$ which is a contradiction as $b=1-\{na\}$.
