I was asked the following question. Where $[x]$ denotes the equivalence class of an integer $x$ with regards to the modular arithmetic relation.
Question: Prove that if $[x]=[y]$ then $[x^2]=[y^2]$ without using the fact that multiplication behaves well with respect to equivalence classes.
My Proof:Given that $[x]=[y]$. We see that $x\equiv y$ (mod d) as the equality implies $x\in [y]$ and $y\in [x]$. Using a property of congruences that if $a\equiv b$ (mod d) then $a^k\equiv b^k$ (mod d) we see that $x\equiv y$ (mod d) $\implies x^2 \equiv y^2$ (mod d)
Because congruences form an equivalence relation, the above result implies that if $a\in [y^2]$ then $a\in [x^2]$ and that if $b\in [x^2]$ then $b\in [y^2]$. Thus, $[x^2]\subset [y^2]$ and $[y^2]\subset [x^2]$. Hence, $[x^2]=[y^2]$.
I was told that this is wrong as I use that multiplication of equivalence classes. Now I get why that is being said, but I never multiply an equivalence class with another as that would make the proof trivial.
Proof Given that $[x]=[y]$. Multiply both sides by $[x]$ to get $[x^2]=[yx]$. Again with the equality $[x]=[y]$, multiply both sides by $[y]$ to get $[yx]=[y^2]$. Thus, $[x^2]=[yx]=[y^2]$.
I never use that fact in that manner. Is my proof wrong with regards to the question?
Hint:
If $d\mid(x-y)$ then $x-y=kd$ for some integer $k$.
So, $x=y+kd$ and $x^2=y^2+2kd+k^2d^2$.
Then $x^2-y^2=(2k+d)d$ that is $d\mid(x^2-y^2)$.
Addendum:
With this same technique you could prove that if $[a]=[b]$ and $[c]=[d]$ then $[ac]=[bd]$.
This is equivalent to suppose $a\equiv b\!\!\!\!\mod d$ and $c\equiv d\!\!\!\!\mod d$ and then imply $ac\equiv bd\!\!\!\!\mod d$.