Prove that $\{(x,y)\in \mathbb{Z}\times\mathbb{Z}: x^2\equiv y^2 \!\!\!\mod \!\!4\}$ is an equivalence relation

Question

I'm trying to prove that the relation R on $\mathbb{Z}$ is an equivalence relation, where R is defined by:

$$\{(x,y)\in \mathbb{Z}\times\mathbb{Z}: x^2\equiv y^2 \!\!\!\mod \!\!4\}$$

I know I need to show that it is reflexive, symmetric, and transitive, but I'm not sure how to do that when working with congruence. Thank you.

I answered my own question below but I'm not sure if I did it correctly. Any feedback would be really helpful. Thanks.

Answer 1

Hint : Try that $\equiv$ is a relation on $\mathbb{Z}$ and $R$ will be a straightfoward consequence.

Answer 2

Reflexive: Let $x\in\mathbb{Z}$. Then $x^2\equiv x^2\, mod\, 4$ for all $x\in R$. Thus, $(a,a)\in R$, so R is reflexive.

Symmetric: Let $(a,b)\in R$ and let $a^2\equiv b^2\, mod\, 4$. This means that $4\mid a^2-b^2$. Then, it is obvious that $4\mid -(a^2-b^2)$, so $4\mid b^2-a^2$. Therefore, $b^2\equiv a^2\, mod\, 4$. Thus, $(b,a)\in R$, so R is symmetric.

Transitive: Let $(a,b)\in R$ and $(b,c)\in R$. This means that $4\mid a^2-b^2$ and $4\mid b^2-c^2$. Then, there exists integers $l,k$ such that $4l=a^2-b^2$ and $4k=b^2-c^2$. The second expression can be written as $b^2=4k+c^2$. Then, by substitution, we have $4l=a^2-(4k+c^2)=a^2-4k-c^2$. Adding $4k$ to both sides, we get $4l+4k=a^2-c^2$. Factoring out $4$, we get $4(l+k)=a^2-c^2$. Since $l+k$ is an integer, we know that $4\mid a^2-c^2$. Therefore, $a^2\equiv b^2\, mod\, 4$. Thus, $(a,c)\in R$, so R is transitive.

Thus, R is an equivalence relation.

This is what I have for my proof so far. What do you guys think? Any feedback is appreciated. Thank you