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Let $P$ be a finite subset of ${\mathbb Z}$ containing at least three elements. I say that $P$ tiles $\mathbb Z$ if $\mathbb Z$ can be written as a disjoint union of translates of $P$.

Trivially if $P$ is an arithmetic progression of the form $\lbrace t+sk \ | \ 1\leq k \leq M \rbrace$ where $s$ and $M$ are coprime, then $P$ tiles $\mathbb Z$ (because of Euclidean division and the Chinese remainder theorem).

Is the converse true ?

My thoughts : It is easy to see that $P=\lbrace 1,2,4\rbrace$ can never tile $\mathbb Z$. For suppose it did, and suppose $\bigcup_{t\in T}t+P$ is such a tiling. We call the $t+P$ (for $t\in T$) distinguished translates of $P$. Choose any $t_1\in T$. Then $t_1+3$ must be in some distinguished translate of $P$, which can only be $(t_1+2)+P$ or $(t_1+1)+P$ or $(t_1-1)+P$. But all those intersect $t_1+P$, contradiction.

Ewan Delanoy
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    $${ 1,2,5,6}$$ tiles $\Bbb Z$ and it's not an arithmetic progression. – Crostul Mar 12 '21 at 10:09
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    There's a lot of literature on this, often relating to cyclotomic polynomials. You're asking for $P(x) \times T(x) = 1/(1-x)$, where $P(x) \neq (1-x^{kn})/(1-x^n)$. There are solutions like $P(x) = (1+x)(1+x^4)$ in Crostul's example, giving $T(x) = 1/[ (1-x)(1+x)(1+x^4)]$ whose coefficients are 1 or 0. – Calvin Lin Mar 12 '21 at 10:12
  • @Crostul I apparently misinterpret "tiling $\mathbb Z$". How do we get the numbers of the form $6k+3$ and $6k+4$ this way ? – Peter Mar 12 '21 at 10:32
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    @Peter $\ldots, {1,2,5,6}, {3,4,7,8},{9,10,13,14},{11,12,15,16},\ldots$. – Milten Mar 12 '21 at 10:41
  • @Milten Thanks for the clarification. – Peter Mar 12 '21 at 10:43
  • @CalvinLin That viewpoint corresponds at first glance to tiling $\mathbb N$. Does it work in general for tiling $\mathbb Z$ too? – Milten Mar 12 '21 at 10:47
  • @CalvinLin You might also give some google keywords so that others can find the literature on this. – Milten Mar 12 '21 at 10:49
  • @CalvinLin Thanks for your feedback. Could you please elaborate on "a lot of literature" and indicate a few mainstream references for ignoramuses like me. – Ewan Delanoy Mar 12 '21 at 10:50
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    @EwanDelanoy Search for "Tiling the Integers" (which was the term that you used). There are several papers (example ) that use that term. – Calvin Lin Mar 12 '21 at 10:59
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    @Milten Essentially yes. In the linked paper above (which deals with $\mathbb{Z}$), see Lemma 1.2. Every tiling by translates of a finite set is periodic. Thus, dealing with $ \mathbb{Z}$ is the same as dealing with some $\mathbb{Z_n}$ which is the same as dealing with $ \mathbb{N}$. – Calvin Lin Mar 12 '21 at 11:04

1 Answers1

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$$\{1;2;5;6 \}$$ tiles $\Bbb Z$ and is not an arithmetic progression.

Crostul
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