# 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 $$a \notin \Bbb Z$$?

• What does $W$ stand for?
– Gary
Commented Jan 31, 2021 at 10:02
• The set of whole numbers Commented Jan 31, 2021 at 10:04
• If $1<a\in \Bbb R$ then $\lfloor 1/a \rfloor=0.$ Commented Jan 31, 2021 at 10:38
• On this site the custom is that \Bbb Z is the integers and \Bbb N is the positive integers and \Bbb N_0 is the non-negative integers. I prefer \Bbb Z^+ for the positive integers and \Bbb N for the non-negative integers but I adhere to the custom of this site when I'm here. Commented Jan 31, 2021 at 10:46

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

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.

• Why can't the equality holds for $n+1$, because when $a=\phi$, we get $\frac{1}{a}+\frac{1}{a^2}=1$ Commented Jan 31, 2021 at 11:20
• Updating the answer ... Commented Jan 31, 2021 at 11:24

We need to prove

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

1. $$\frac{1+\lfloor \frac{1+na^2}{a}\rfloor}{a} \leqslant \frac{1+\frac{1+na^2}{a}}{a}$$

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.

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