Find the quotient set of this relation

Question

Find the quotient set of $R$ defined over $\mathbb Z \times \mathbb Z$ as: $$(a,b) \sim (c,d) \iff (3|(a^2-c^2) \land bd \ne \sqrt 2)$$

When I look at this relation I think that the secon coordinate does not matter at all - it is simply impossible that a product of two integers equals $\sqrt 2$ thus, I think this is only a 'trick' to make this relation look more "scarry".

And so, regardless of whatever valid first coordinate we pick, we may pick any integer as the second coordinate and this pair will still be valid. Thus, let's examine the meaning of the first coordinate.

The first coordinate divides the set of integers into three groups, based on the remainder they give in division by $3$. Now,
$$(3n)^2 = 3 \cdot 3n^2 +0 \\ (3n+1)^2 = 9n^2 + 6n + 1 = 3 \cdot(3n^2 +2n) +1 \\ (3n+2)^2 = 3n^2 + 12n + 4 = 3(n^2+4n+1) +1$$ And so there are only two possibilities - either zero or one. So now, we can define the quotient set - the set of all abstract equivalence classes as $$\{[(3,0)], [(1,0)] \}$$ Do you think that my solution is correct?

Answer 1

I think that the secon coordinate does not matter at all - it is simply impossible that a product of two integers equals $\sqrt 2$

Correct.

And so there are only two possibilities - either zero or one.

Correct again. You could maybe simplify the derivation a tiny bit by noting that any integer is either of the form $n=3k$ and then $n^2 = 3 \cdot 3k^2$, or $n=3k \pm 1$ and then $n^2 = 3 \cdot(3k^2 \pm 2k)+1\,$.

we can define the quotient set - the set of all abstract equivalence classes as $\{[(3,0)], [(1,0)] \}$

Or just $\,\{[(0,0)], [(1,0)] \}\,$. I think this could use some explanation for the step from calculating the remainders $\,\operatorname{mod}{3}\,$ to determining the equivalence classes: $\,3 \mid a^2 - c^2\,$ if both $a,c \in 3\mathbb{Z}\,$, or both $\,a,c \in \mathbb{Z} \setminus 3\mathbb{Z}\,$. Other than that, I'd say well done.