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 $a \notin \Bbb Z$?

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 $b=1-\{na\}$.
