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 $a \notin \Bbb Z$?

  • 1
    $\begingroup$ What does $W$ stand for? $\endgroup$
    – Gary
    Commented Jan 31, 2021 at 10:02
  • $\begingroup$ The set of whole numbers $\endgroup$ Commented Jan 31, 2021 at 10:04
  • $\begingroup$ If $1<a\in \Bbb R$ then $\lfloor 1/a \rfloor=0.$ $\endgroup$ Commented Jan 31, 2021 at 10:38
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    $\begingroup$ 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. $\endgroup$ Commented Jan 31, 2021 at 10:46

3 Answers 3


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.

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


$$ 1+na^2 \leqslant na^2+a^2-a $$

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


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