$[n \sqrt{2}] = [m (2+\sqrt{2})]$ for $m,n$ natural.

Question

Does $[n \sqrt{2}] = [m (2+\sqrt{2})]$ for $m,n$ natural have no solution where $[x]$ is the floor function of $x$? I tried calculating some examples ($1,000,000$ examples on Python) and it seems as though the left and right hand side never equal the same number- no matter the choice of $m$ and $n$. This is from a Calculus $1$ worksheet.

Answer 1

Here's a partial answer (let me know if partial answers are unwelcome on math.SE). The equation has no solutions in natural numbers $\mathbb{N}$. Suppose for contradiction that there exists $m,n\in\mathbb{N}$ with $[n\sqrt{2}]=[m(2+\sqrt{2})]$. Then there are three cases:

Case 1: $m=n$. Impossible because then $$[n\sqrt{2}]=[m(2+\sqrt{2})]=[n(2+\sqrt{2})]=[n\sqrt{2}]+2n$$ which implies that $2n=0$ and $n=0\notin\mathbb{N}$.

Case 2: $m>n$. Then if we write $m=n+l$ for some $l\in\mathbb{N}$, $$\begin{align*}[n\sqrt{2}]&=[(n+l)(2+\sqrt{2})] \\ & = [(n+l)\sqrt{2}]+2(n+l) \\ & >[n\sqrt{2}] \end{align*}$$ which is a contradiction.

Case 3: $m<n$. Then if $m=n-l$ for some $l\in\mathbb{N}$, we have $$\begin{align*}[n\sqrt{2}]&=[(n-l)(2+\sqrt{2})] \\ & = [(n-l)\sqrt{2}]+2(n-l). \end{align*}$$ But this won't get us anywhere (unless if someone can continue from here and prove me wrong).

And of course the result follows easily from the link Qiaochu provided: https://en.m.wikipedia.org/wiki/Beatty_sequence. In layman's terms, it states that $$[nr]\neq [m\frac{r}{r-1}]$$ for all positive integers $n,m$ and positive irrationals $r$. In particular, if $r=\sqrt{2}$, then $\frac{r}{r-1}=\frac{\sqrt{2}}{\sqrt{2}-1}=2+\sqrt{2}$, so $$[n\sqrt{2}]\neq [m(2+\sqrt{2})]$$ for all $n,m$ positive integers.

Answer 2

The less detailed treatment, but more intuitve, is that the two real numbers are extremely close only when their ratio is a continued fraction convergent for $1+ \sqrt 2.$ The numerators and denominators are consecutive elements in the sequence $$ 1, 2, 5, 12, 29, 70, 169, 408, $$

https://oeis.org/A000129

https://en.wikipedia.org/wiki/Pell_number#Pell_numbers

while the relevant pair of real numbers, rounded to the nearest integer, are in the sequence $$ 1, 3, 7, 17, 41, 99, 239, 577, 1393, $$

https://oeis.org/A001333

These are one half the Pell-Lucas numbers,

https://en.wikipedia.org/wiki/Pell_number#Pell%E2%80%93Lucas_numbers

Both integer sequences obey the linear recursion $$ w_{n+2} = 2 w_{n+1} + w_n $$

The characteristic roots are $1+ \sqrt 2$ and $1- \sqrt 2.$ Both sequences can be written in the form $$ A (1 + \sqrt 2)^n + B (1- \sqrt 2)^n $$ The exact coefficients should be given in the OEIS entries for each sequence.