# Proving a relation is an equivalence relation specifically proving transitivity

I'm currently studying for an exam and I've come across this question:

• Define a relation R on Z by

xRy ⇔ 6|($x^{2} − y^{2}$)

for x, y ∈ Z.

Prove that R is an equivalence relation and describe the equivalence classes of R

I understand how to prove it's reflexive, and I've tried to prove it's symmetric but I used the fact that -6|($y^{2}-x^{2}$) which doesn't seem like the correct way to answer this question, and I have no idea how to prove it's transitive, any help would be greatly appreciated. Thanks in advance!

Your proof that it's symmetric is almost certainly correct (but if you want to post the details, I could critique it).

To prove transitivity, assume $x \sim y$ and $y\sim z$.

By definition, $6| (x^2-y^2)$ and $6|(y^2-z^2)$. Another way of putting this would be to say there are integers $m,n$ such that

$x^2-y^2 = 6m$

and

$y^2-z^2 = 6n$.

If we just add those two equations together, we get

$x^2 - z^2 = 6(m+n)$.

That is, $6|(x^2 -z^2)$, or $x\sim z$.

• Thank you so much, I almost feel a bit silly I spent so long on this question and didn't spot that! – Lucy Jan 13 '17 at 17:40
• You're welcome! Math intuition comes with time. – Jared N Jan 13 '17 at 17:53