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Harvard-MIT Mathematics Tournament

Number theory Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $S=\{s_{0}, \ldots, s_{n}\}$ be a finite set of integers, and define $S+k=\{s_{0}+k, \ldots, s_{n}+k\}$ . We say that $S$ and $T$ are equivalent, written $S \sim T$ , if $T=S+k$ for some $k$ . Given a (possibly infinite) set of integers $A$ , we say that $S$ tiles $A$ if $A$ can be partitioned into subsets equivalent to $S$ . Such a partition is called a tiling of $A$ by $S$ .

Suppose that $S$ tiles the set of all integer cubes. Prove that $S$ has only one element.

Solution

Solution:

Let the difference between the smallest and largest element of $S$ be $a$ . Then the set equivalent to $S$ that contains $b^{3}$ can only contain integers between $b^{3}-a$ and $b^{3}+a$ , inclusive. But for sufficiently large $b$ , $b^{3}$ is the only cube in this range, so $S$ can only have one element.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.