Prove that $\left \lfloor \frac{1+\lfloor na+1/a\rfloor}{a} \right \rfloor=n$
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.