Question: Let $S\subseteq \mathbb{Z}\times\mathbb{Z}$ be the equivalnce relation $$ x^{2}\equiv y^{2} (\operatorname{mod}4) $$ Find all the equivalence classes of $S$ on $\mathbb{Z}$.
My Answer: Let $z\in\mathbb{Z}$. The euivalence class of $S$ on $z$ will be the set of all $x$ that satisfy $$ z^{2} - x^{2} = 4m\Leftrightarrow x^{2} = z^{2} - 4m. $$ If $z$ is even, in other words $z = 2c$, then $x^{2} = 4c^{2} - 4m = 0 - 4(c^{2} + m)$ which is the equivalence class of 0. Therefore, the equivalence class of every even integer is the same as the equivalence class of zero.
Respectively, if $z$ is odd, in other words $z = 2c + 1$, then $x^{2} = 4c^{2} + 4c + 1 + 4m = 1 + 4(c^{2} + c + m)$, which is the equivalence class of 1. Therefore, the equivalence class of every odd integer is the same as the equivalence class of 1.
So all the equivalence classes of $S$ on $\mathbb{Z}$ are given by the following \begin{align*} [2n]_{S} = [0]_{S} &= \{x\in\mathbb{Z}\mid 0 - x^{2} = 4m, m\in\mathbb{Z}\} = \{\ldots, -8, -4, -2, 0, 2, 4, 8, \ldots\}\\ [2n + 1]_{S} = [1]_{S} &= \{x\in\mathbb{Z}\mid 1 - x^{2} = 4m, m\in\mathbb{Z}\} = \{\ldots, -9, -5, -3, -1, 1, 3, 5, 9, \ldots\} \end{align*}
Is my thought process correct? Does my answer satisfy the question? Thank you in advance and sorry for any spelling error, I'm translating from Greek:)
You are correct in the two equivalence classes, but you can simplify the presentation a bit: $$[0]_S=\{2x:x\in\mathbb Z\}$$ $$[1]_S=\{2x+1:x\in\mathbb Z\}$$ One class is the even integers, one is the odd integers, both are disjoint and both unify to the integers.