# $\lfloor a n\rfloor \lfloor b n\rfloor \lfloor c n\rfloor = \lfloor d n\rfloor \lfloor e n\rfloor \lfloor f n\rfloor$ for all $n$

Simple task, but I know only extremely overkill solution :)

Let $a,b,c,d,e,f$ be positive irrational numbers.

Suppose that for any positive integer number $n$,

$$\lfloor a n\rfloor \cdot \lfloor b n\rfloor \cdot \lfloor c n\rfloor = \lfloor d n\rfloor \cdot \lfloor e n\rfloor \cdot \lfloor f n\rfloor.$$

Prove that the sets $\{a,b,c\}$ and $\{d,e,f\}$ are equal.

• Your question is incomplete. What fo you mean by {a,b,c} and what are you doing with the product of the floors? – Scott Burns Nov 19 '16 at 13:01
• {a,b,c} is an unordered set – kotomord Nov 19 '16 at 13:06
• Interesting question +1. It follows quickly that $abc = def$ from $\lim x / \lfloor x \rfloor = 1$. Would you have a simple (counter)example why the irrational condition is necessary. – dxiv Nov 21 '16 at 4:38
• @dxiv - I have no simple (counter)example, but my solution uses irrationality of (at least one) number – kotomord Nov 21 '16 at 8:06
• @AlexYoucis, I'll write it today (Sorry, I'm working now) It uses en.wikipedia.org/wiki/Equidistribution_theorem – kotomord Nov 23 '16 at 14:58

Let $E=(\mathbb{R}/\mathbb{Z})^6$. Let $x=(\overline{a},\overline{b},\overline{c},\overline{d},\overline{e},\overline{f})\in E$. Let $F=\mathbb{N}x$.

$\overline{F}$ is a subgroup of $E$.

$a,b,c,d,e,f$ are irrationnal, so it exists $(u_n)\in\mathbb{N}^{\mathbb{N}}$ and $(\delta_a,\delta_b,\delta_c,\delta_d,\delta_e,\delta_f)\in \{0,1\}^6$ such that $u_nx \in E \mapsto 0_E$ with $u_nx=(\overline{a_n},\overline{b_n},\overline{c_n},\overline{d_n},\overline{e_n},\overline{f_n})$ and $0<a_n<1,0<b_n<1,0<c_n<1,0<d_n<1 ,0<e_n<1 ,0<f_n<1$.

and $(a_n,b_n,c_n,d_n,e_n,f_n)\in[0,1]^6 \mapsto (\delta_a,\delta_b,\delta_c,\delta_d,\delta_e,\delta_f)$.

So it exists $(v_n)\in\mathbb{N}^{\mathbb{N}}$ such that $v_nx \in E \mapsto 0_E$ with $v_nx=(\overline{a'_n},\overline{b'_n},\overline{c'_n},\overline{d'_n},\overline{e'_n},\overline{f'_n})$ and $0<a'_n<1,0<b'_n<1,0<c'_n<1,0<d'_n<1 ,0<e'_n<1 ,0<f'_n<1$.

and $(a'_n,b'_n,c'_n,d'_n,e'_n,f'_n)\in[0,1]^6 \mapsto (1-\delta_a,1-\delta_b,1-\delta_c,1-\delta_d,1-\delta_e,1-\delta_f)$

So $\lfloor au_n\rfloor \lfloor bu_n\rfloor \lfloor cu_n\rfloor=(au_n-a_n) (bu_n-b_n)(cu_n-c_n)=abc(u_n)^3-abc_n(u_n)^2-acb_n(u_n)^2-bca_n(u_n)^2+\dots$

So, as already shown by dxiv, $$\lim\frac{\lfloor au_n\rfloor \lfloor bu_n\rfloor \lfloor cu_n\rfloor}{u_n^3}=abc$$

So $abc=def$ because $\lfloor au_n\rfloor \lfloor bu_n\rfloor \lfloor cu_n\rfloor=\lfloor du_n\rfloor \lfloor eu_n\rfloor \lfloor fu_n\rfloor$

And $$\lim \frac{\lfloor au_n\rfloor \lfloor bu_n\rfloor \lfloor cu_n\rfloor-abc(u_n)^3}{u_n^2}=-ab\delta_c-ac\delta_b -bc\delta_a$$

We have also $$\lim \frac{\lfloor av_n\rfloor \lfloor bv_n\rfloor \lfloor cv_n\rfloor-abc(v_n)^3}{v_n^2}=-ab(1-\delta_c)-ac(1-\delta_b) -bc(1-\delta_a)$$

So $$ab+ac+bc=-\lim \frac{\lfloor au_n\rfloor \lfloor bu_n\rfloor \lfloor cu_n\rfloor-abc(u_n)^3}{u_n^2}-\lim \frac{\lfloor av_n\rfloor \lfloor bv_n\rfloor \lfloor cv_n\rfloor-abc(v_n)^3}{v_n^2}$$

So $ab+ac+bc=de+df+ef$.

Lemma: We show $\overline{F}=\overline{\mathbb{N}x}$ is a subgroup. For all $k \in \mathbb{N}$, let the compact $A_k=kx+\overline{\mathbb{N}x}\subset \overline{\mathbb{N}x}$. And for all $k$, $0\in A_k$. So $0 \in G=\cap_k A_k$. $x+G\subset G$. So $\mathbb{N}x+G\subset G$.

$G$ is compact so $\overline{\mathbb{N}x}\subset G$.

As $G \subset A_0=\overline{\mathbb{N}x}$, we have $G=\overline{\mathbb{N}x}$.

As $0 \in\overline{\mathbb{N}x}=G \subset A_1=x+\overline{\mathbb{N}x}$, we have $-x\in \overline{\mathbb{N}x}$. So $\overline{\mathbb{N}x}= \overline{\mathbb{Z}x}$

Remark: for all $m\in\mathbb{N}$, we can choose $(\delta_a,\delta_b,\dots,\delta_f)=(\{ma\},\{mb\},\dots,\{mf\})\in \overline{F}$, and $(u_n)$ such that $u_n x\mapsto \delta$. So we have $ab\{mc\}+ac\{mb\}+bc\{ma\}=de\{mf\}+df\{me\}+ef\{md\}$ for all $m\in\mathbb{N}$.

• But you have no proof what a+b+c = d+e+f? – kotomord Nov 23 '16 at 13:40
• No, but my comment was too long. – marco2013 Nov 23 '16 at 13:41
• I use the property: $\overline{F}$ is a subgroup. So if $(\overline{y_1},\overline{y_2},\overline{y_3},\overline{y_4},\overline{y_5},\overline{y_6})\in \overline{F}$, we have $(\overline{1-y_1},\overline{1-y_2},\overline{1-y_3},\overline{1-y_4},\overline{1-y_5},\overline{1-y_6})\in \overline{F}$ too. – marco2013 Nov 23 '16 at 13:57
• In your lemma, why do we have $0\in A_k$ ? – Ewan Delanoy Nov 23 '16 at 14:58
• Because $E$ is compact. So from the sequence $nx$, we can find a subsequence $w_nx\mapsto \ell\in E$ (with $w_n$ strictly increasing). Let$\epsilon>0$,$\exists N\in\mathbb{N}$ such that $n>N\implies \|w_nx-\ell\|<\epsilon$. So $\forall n,m>N$, $\|(w_n-w_m)x\|<2\epsilon$. Take $w_n-w_m>k$. So $0\in \overline{A_k}$. But $A_k$ is compact. – marco2013 Nov 23 '16 at 15:08

So, the overkill proof;

One of the equidistribution theorems: https://en.wikipedia.org/wiki/Equidistribution_theorem

a is irrational, then sequence $a*p_n$ mod 1 is uniformly distributed.

Lemma: If a > 0 is irrational, then sequence { $\lfloor a*n\rfloor$ } contains infinitely many prime numbers.

Proof of lemma:

Case (a>1):

$\lfloor a*n\rfloor = p_m$ <=> $p_m < a*n < p_m + 1$ <=> $\frac{p_m}{a} < n < \frac{p_m+1}{a}$ <=> n = $\lfloor \frac{p_m+1}{a}\rfloor$, {$\frac{p_m}{a}$} $\in (1-\frac{1}{a}, 1)$ (where are infinitely many such m by thm.)

Case (a<1) is trivial.

Without loss of generality $a \ge b \ge c, d \ge e \ge f, a\ge d$
Let a > d. Fix n what $n*a > n*d + 2, n*c > 1, \lfloor n*a \rfloor =p_m$
$\lfloor n*d \rfloor \lfloor n*e \rfloor \lfloor n*f \rfloor > 0$ and not divisible to $p_m$ - it is the absurd.
So, a = d and we need to prove what if $\lfloor n*e \rfloor \lfloor n*f \rfloor = \lfloor n*b \rfloor \lfloor n*c \rfloor$, then sets {b, c} and {e,f} are equals.