This is an equivalence relations exercise. It has two parts. The first is about proving that the relation is of equivalence, which seems to be fine to me, but I'll put it there anyway. With the second part, however, is where I'm having trouble. It is about getting the equivalence classes of certain pairs. I wrote some answers, probably wrong, but whatever the case, I remain ignorant on how to approach that kind of problem anyway. Can you check that out?
Let $E = \mathbb{Z} \times \mathbb{Z}$. Over $E$ is defined the relation $R$ such as
$$(a,b)R(c,d) \iff a^2 + 7b^2 = c^2 +7d^2$$
Prove that $R$ is of equivalence.
Reflexive
Have some $(a,b) \in E$.
$$(a,b)R(a,b)$$
$$a^2 + 7b^2 = a^2 +7b^2$$
$$7 = 7$$
Symmetric
Have some $(a,b),(c,d) \in E$.
$$(a,b)R(c,d)$$
$$a^2 + 7b^2 = c^2 +7d^2$$
$$c^2 +7d^2 = a^2 + 7b^2$$
$$(c,d)R(a,b)$$
Transitive
Have some $(a,b),(c,d),(e,f) \in E$.
$$(a,b)R(c,d) \land (c,d)R(e,f)$$
$$(a^2 + 7b^2 = c^2 +7d^2) \land (c^2 + 7d^2 = e^2 +7f^2)$$
$$(a^2 + 7b^2 + c^2 + 7d^2) = ( c^2 +7d^2 + e^2 +7f^2)$$
$$(a^2 + 7b^2) = (e^2 +7f^2)$$
$$(a,b)R(e,f)$$
Determine the equivalence classes for $(0,0)$ and $(1,1)$.
Uh, as far as I'm concerned, for $(0,0)$ I need to find pairs $(x,y)$ where $x^2 +7y^2 = 0$. Well, $(0,0)$ is the only one in my mind (which is already obvious since the relation is reflexive right?). Similar logic goes for $(1,1)$.
This is pretty fishy. Is this correct?
$$[(0,0)] = \{(0,0)\}$$
$$[(1,1)] = \{(1,1)\}$$
Even if it was correct, I fear I don't have a good strategy for this kind of question. I literally just try to think up some random pairs that might work. And that's it. Do you have any advice regarding this?
Suppose that $(c, d) R (0, 0)$. Then by definition of $R$, we have
$$c^2 + 7d^2 = 0^2 + 7 \cdot 0^2 = 0$$
This implies that $c = d = 0$, since the only way you can add two non-negative numbers and get $0$ is if they were both $0$. So $[(0, 0)] \subseteq \{(0, 0)\}$, and the reverse inclusion is obvious.
Now suppose that $(c, d) R (1, 1)$. Then by definition, we have
$$c^2 + 7d^2 = 1^2 + 7 \cdot 1^2 = 8$$
It's easiest here to start by analyzing some cases:
If $d > 1$ or $d < -1$, can this ever be possible? Consider how big $7d^2$ would have to be in this case.
If $d = \pm 1$, then we've got $c^2 + 7 = 8 \implies c^2 = 1$.
If $d = 0$, we have $c^2 = 8$.