Finding an equivalence classes for the relation $(x_1,y_1)\sim(x_2,y_2)\ \iff\ x_1^2+y_1^2=x_2^2+y_2^2$. Let R be the relation defined on the set of integer pairs by $(x_1,y_1)R(x_2,y_2)$ when $x_{1}^{2}$ + $y_{1}^{2}$ = $x_{2}^{2}$ + $y_{2}^{2}$.
Prove that R is an equivalence relation and determine the equivalence
classes.
So to prove it's an equivalence relation
I said
Symmetric - Yes, because  $x_{2}^{2}$ + $y_{2}^{2}$ = $x_{1}^{2}$ + $y_{1}^{2}$.
Reflexive - Yes, because  $x_{1}^{2}$ + $y_{1}^{2}$ = $x_{1}^{2}$ + $y_{1}^{2}$.
Transitive = Yes, because 
$x_{1}^{2}$ + $y_{1}^{2}$ = $x_{2}^{2}$ + $y_{2}^{2}$.
$x_{2}^{2}$ + $y_{2}^{2}$ = $x_{3}^{2}$ + $y_{3}^{2}$.
$x_{1}^{2}$ + $y_{1}^{2}$ = $x_{3}^{2}$ + $y_{3}^{2}$.
I think this is enough to prove it's an equivalence relation.
Now I am new to this, and my understanding of equivalence classes is basically list all the things that satisfy this relation.
I'm not sure if that's correct, but here is what I got from that 
$$[x,y] =\{(x,y),(x,-y),(-x,y),(x,y)\}$$
is this correct, are there alternate ways?
 A: Your proof is rather sloppy. To prove, for example, that the relation is symmetric, you should prove that
$$\forall (x_1,y_1),(x_2,y_2)\in\Bbb{Z}^2:\qquad
(x_1,y_1)R(x_2,y_2)\quad\implies\quad(x_2,y_2)R(x_1,y_1).$$
Similarly, to prove reflexivity, you should prove that
$$\forall (x_1,y_1)\in\Bbb{Z}^2:\qquad (x_1,y_1)R(x_1,y_1),$$
and something similar for transitivity.
Your characterization of the equivalence classes is incorrect. As noted in the comments $(4,3)R(5,0)$, for example. In stead, consider the fact that for every $(x_i,y_i)\in\Bbb{Z}^2$ we can set $c_i:=x_i^2+y_i^2$, and then for all $(x,y)\in[(x_i,y_i)]$ we have
$$x^2+y^2=x_i^2+y_i^2=c_i.$$
Can you tell what these sets (these equivalence classes) look like geometrically?
A: More generally, if $f\colon X\to Y$ is a map, then $$a\operatorname{R}b:\Leftrightarrow f(a)=f(b) $$
is always an equivalence relation. This is readily proved and after that comes in handy in mayn situations of life.
Moreover, we can loosely identify the set of equivalence classes with the image of $f$.
Here $f\colon \Bbb Z^2\to \Bbb N_0$ is given by $(x_1,x_2)\mapsto x_1^2+x_2^2$, and with some non-trivial results from number theory, $\operatorname{im}f$ contains $0$ and all naturals $n$ such that for all primes $p$ with $p\equiv 3\pmod4$, the highest power of $p$ dividing $n$ is of even exponent.
