## کارگاه نظریه اعداد

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### Problem 2

• نویسنده
نوشته ها
• #13928
Ramin Takloo-Bighash
سرپرست

Show that there is a subset $$A$$ of $$Z^3$$ such that for each point $$(x, y, z)$$ in $$A$$ precisely one of the six neighbouring points $(x, y, z+1), (x, y, z-1), (x, y+1, z), (x, y-1, z), (x+1, y, z), (x-1, y, z)$ is in A

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• #14352
makzeta
سرپرست

This problem doesn’t seem to be correct. Suppose that A consists of only the two elements $$(0,0,1), (0,0,0)$$. I think it should be stated as follows,”…such that for each point $$(x,y,z)$$ not in A precisely one of the six neighbouring points

• #14353
makzeta
سرپرست

…Or maybe, such that for each point $$(x,y,z)$$ in $$Z^3$$ precisely one of the six neighbouring points

• #14360
کیوان م.
مشارکت کننده

I am assuming that in Ramin’s statement, the first instance of $$A$$ is replaced by $$\mathbb{Z}$$, since otherwise the problem is trivial. Now, instead of directly dealing with $$\mathbb{Z}^3$$ , consider the quotient $$(\mathbb{Z}/4 \mathbb{Z})^3$$  as the vertex set of a finite graph with two vertices $$v_1$$ and $$v_2$$ joined by an edge when $$v_1 – v_2$$ is one of the vectors $$\pm e_1, \pm e_2, \pm e_3$$, where, with some abuse of notation, $$e_i$$ also denote the images of the standard basis in the quotient

Now the finite torus” $$(\mathbb{Z}/4 \mathbb{Z})^3$$ is a six-regular bipartite graph, and hence has a perfect matching (one can invoke Philip Hall, but it should be easy to construct one by hand as well). Take one vertex from each edge of this perfect matching and define $$A$$ to be the lift of this set to $$\mathbb{Z}^3$$. This set clearly satisfies the problem

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