Indija TS 2008

First observe that $A$ and $B$ are separated by $a$ numbers and as such $A\cup B$ is not a set of consecutive integers.

a. ($\Rightarrow$) If $r=t$ and $a=kr$ for some $k\in \mathbb{N}$, then it is clear that $Z$ is covered by the mutually disjoint sets $(A\cup B)+s$, where $s$ ranges over the set $$\bigcup _{m\in \mathbb{Z}}\{2(k+1)rm,2(k+1)rm+r,\dots ,2(k+1)rm+kr\}=2(k+1)rZ\cup \{2(k+1)rZ+r\}\cup \cdots \cup \{2(k+1)rZ+kr\}.$$ (Here $xP$ denotes the set $\{x:p\in P\}$.)

b. ($\Rightarrow$) If $r<t$ and $a=k(r+t)$ for some $k\in \mathbb{N}$, then $Z$ is covered by the mutually disjoint sets $(A\cup B)+s^{\prime }$, where $s^{\prime }$ ranges the set $\{(r+t)s:s\in Z\}=(r+t)Z$.

a. ($\Leftarrow$) If $r=t$, and $Z$ is covered by mutually disjoint sets of the form $(A\cup B)+s$, for $s$ in a fixed subset of $Z$, then any translate of $A\cup B$ has a gap of size $a+r+1-(r+1)=a$, which should be covered by mutually disjoint sets of size $r$. This proves that $a=kr$, for some $k\in \mathbb{N}$.

b. ($\Leftarrow$) If $r<t$, and $Z$ is covered by a certain class of mutually disjoint translates of $A\cup B$, we will prove that the translates of $A$ and $B$ must alternate in any covering of $Z$ by translates of $A\cup B$. This would imply that $a=k(r+t)$ for some $k\in \mathbb{N}$. If a covering contained two consecutive translates $A+x$ and $A+x+r$ of $A$, it would contain the corresponding matching translates $B+x$ and $B+x+r$ of $B$. But

$B+x=\{a+r+1+x,a+r+2+x,...,a+r+t+x\}$ and $B+x+r=\{a+2r+1+x,a+2r+2+x,...,a+2r+t+x\}$. As $a+2r+1+x\le a+r+t+x$, we see that $B+x$ and $B+x+r$ overlap. Hence this is impossible.

Suppose now that a covering of $Z$ contained two consecutive translates $B+x$ and $B+x+t$ of $B$. Then the covering must contain the corresponding matching translates $A+x$ and $A+x+t$ of $A$.

Now $$A+x=\{1+x,2+x,...,r+x\}\text{ and }A+x+t=\{1+x+t,2+x+t,...,r+x+t\}.$$ The gap between $A+x$ and $A+x+t$ corresponds to the set $\{r+x+1,r+x+2,...,t+x\}$ which is of size $t-r$, which is less than $t$. So $A+x$ must be followed immediately by another translate of $A$ itself. But this is exactly what we proved to be impossible just above. Thus two consecutive translates of neither $A$ nor $B$ can occur in any covering of $Z$. This completes the proof.