Find the equivalence classes of relation

Question

Let $x,y\in\mathbb{Z}$. Define the relation $$(x,y)\in R\iff x^2+x=y^2+y.$$ Prove $R$ is equivalence relation and find the equivalence classes of relation $R$.

I can prove $R$ is equivalence relation, as below

1. $R$ is reflexive because for any $x\in\mathbb Z$, $$x^2+x=y^2+y$$ so $(x,x)\in R$.

2. $R$ is symmetric because for any $x,y\in\mathbb Z$, if $(x,y)\in R$, $$x^2+x=y^2+y\iff y^2+y=x^2+x,$$ so $(y,x)\in R$.

3. $R$ is transitive because for any $x,y,z\in\mathbb Z$, if $(x,y)\in R$, $$x^2+x=y^2+y$$ and if $(y,z)\in R$, $$y^2+y=z^2+z.$$ So, we have $$x^2+x=z^2+z,$$ so$(x,z)\in R$.

We conclude that $R$ is equivalence relation.

Now I want to find equivalence classes of $R$.

$R$ is relation on $\mathbb Z$, so the equivalence classes of $R$ of $a\in\mathbb Z$ is $$\{x\in \mathbb Z\mid (x,a)\in R\}=\{x\in \mathbb Z\mid x^2+x=a^2+a\}.$$

Is it true answer? Please check. Can we mention one by one equivalence classes of $R$?

Answer 1

You can go further. For a given $a$, the equation $x^{2} + x = a^{2} + a$ has two solutions: $a$ and $-1-a$. Hence the equivalence classes are $\{a, -1-a\}$ for $a\in \mathbb{Z}$.