Prove that $R$ is reflexive, symmetric, and transitive. Define a relation $R$ on $\Bbb Z$ by declaring that $xRy$ if and only if $x^2\equiv y^2\pmod{4}$. Prove that $R$ is reflexive, symmetric, and transitive.
Suppose $x\in\Bbb Z$. Then $x^2\equiv x^2\pmod {4}$ means that $4\mid (x^2-x^2)$, so $x^2-x^2=4a$ where $a=0\in\Bbb Z$. Therefore $R$ is reflexive. 
Now suppose $x^2\equiv y^2\pmod {4}$. This means $4\mid (x^2-y^2)$ and so $x^2-y^2=4a$, for some $a\in\Bbb Z$. Multiplying by $-1$ we have $-1(x^2-y^2=4a)\\\rightarrow -x^2+y^2=-4a\\\rightarrow y^2-x^2=4(-a)$
so $4\mid(y^2-x^2)$ and $y^2\equiv x^2\pmod{4}$. This shows that $R$ is symmetric. 
Now we assume that $x^2\equiv y^2\pmod{4}$ and $y^2\equiv z^2\pmod{4}$. This means $4\mid(x^2-y^2)$ and $4\mid(y^2-z^2)$. Then we have $x^2-y^2=4a$ and $y^2-z^2=4b$ for some $a,b\in\Bbb Z$. Rearranging we get $x^2=4a+y^2$ and $z^2=y^2-4b$. 
Then
$\begin{align*}x^2-z^2&=(4a+y^2)-(y^2-4b)\\&=4a+4b\\&=4(a+b)\end{align*}$
This shows that $4\mid(x^2-z^2)$ and $x^2\equiv z^2\pmod{4}$, therefore $R$ is transitive.
$\blacksquare$
Please forgive my rushed formatting, just wondering if my arguments here work and if this is a valid proof. Any feedback is appreciated, thanks!

This is straight forward and I think you did ok.  The transitivity follows immediately from transitivity of congruence.   In fact all three follow from the fact that congruence mod $n$ is indeed an equivalence relation. 
