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I need to find equivalence classes of relation $\sim$ \begin{array}{l} x\sim y \Leftrightarrow 4 \mid\left(x^{3}-y^{3}\right) \\ \text { for } x, y \in \mathbb{Z} \end{array}

I have shown that $\sim$ is equivalence relation. I blind guessed that all even numbers will be one equivalence class, so my question is is there some systematic approach to this problem? Any help is appreciated :)

ArsenBerk
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Algebrah
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  • Essentially, you want to use $x^3-z^3=x^3-y^2+y^3-z^3$. But you can also use the fact that $x\sim y$ if and only if either both numbers are odd or both numbers are even. –  Nov 10 '20 at 13:27

1 Answers1

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We have $x\sim y \iff $ $x^{3}$ and $y^{3}$ leave the same remainder when divided by $4$.

Write $x=4q+r$ with $r\in\{0,1,2,3\}$. Then $x \sim r$ (check!) and so it suffices to consider the equivalence class of $r$. In particular, there are at most four equivalence classes. In fact, it turns out that there are three equivalence classes: $[0]$, $[1]$, $[3]$, because $[2]=[0]$.

In other words, all even numbers are in the same class, but the odd numbers fall into two classes.

lhf
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