On the rational Beatty sequence Let $S(p/q, b) = \{[pn/q + b]|n\in\mathbb{Z}\}$, where $p, q$ are coprime positive integers and $b$ is any integer, be a rational Beatty sequence. I can't see why the following conclusion is true: $S(p/q, b_1) = S(p/q, b_2)$ iff $b_1\equiv b_2\mod p$. Could anybody help me? Thanks.
 A: Sorry about not noticing the other direction. It isn't difficult, but not as trivial as the other. I'll treat both directions here.
If $p\le q$, then we have 
$S(p/q,b)=\mathbb{Z}$ irrespective of the value of $b$, so we can (hopefully?) assume that $p>q$ and that $\gcd(p,q)=1$. As $p(n+q)/q=(pn/q)+p$, the Beatty sequences are stable under addition of $p$. Thus they are unions of cosets of $p\mathbb{Z}$. To be more precise, we have
$$
S(p/q,b)=\bigcup_{i=0}^{q-1}(\left[\frac{pi}q\right]+b+p\mathbb{Z}).
$$
Consider the set
$$
R(p,q)=\{\left[\frac{pi}q\right]\mid i\in\mathbb{Z},0\le i<q-1\}.
$$
It is a subset of $\{0,1,\ldots,p-1\}$ of $q$ elements, and we can view it as a subset of the additive group $G=\mathbb{Z}_p$.
I denote by $R+a$ the subset 
$$R+a=\{a+r\in G\mid r\in R\}\subset G$$
for all $a\in G$, $R\subset G$. 
I claim that the relation $R(p,q)=R(p,q)+a$ holds only, if $a=0_G$.
Consider the set $H(p,q)=\{a\in G\mid R(p,q)+a=R(p,q)\}$.
If $a,b\in H(p,q)$, then
$$
R(p,q)+(a+b)=(R(p,q)+a)+b=R(p,q)+b=R(p,q),
$$
so we see that $a+b\in H(p,q)$, and thus $H(p,q)$ is a subgroup of $G$. But we easily see that $R(p,q)$ is a union of cosets of $H(p,q)$. Let $k$ be the number of such cosets. Then 
$q=|R(p,q)|=k|H(p,q)|$. But from basic group theory (Lagrange's theorem) we also get that $p=|G|$ is divisible by $|H(p,q)|$. We assumed that $\gcd(p,q)=1$, and thus $H(p,q)$ has only a single element, zero.
When can we have $S(p/q,b_1)=S(p/q,b_2)$? Our earlier considerations imply that this happens, iff $R(p,q)+b_1=R(p,q)+b_2$. This means that $b_2-b_1\in H(p,q)$. We just saw that this happens, iff $b_2-b_1=0_G$, in other words $b_2\equiv b_1\pmod p$.
