Divisibility Proof $8\mid (x^2 - y^2)$ for $x$ and $y$ odd $x,y \in\Bbb Z$. Prove that if $x$ and $y$ are both odd, then $8\mid (x^2 - y^2)$.
My Proof Starts:
Assume $x$ and $y$ are both odd. So, $x = 2k + 1$ and $y = 2l +1$ for some integers $k$ and $l$. Thus,
\begin{align}
x^2 - y^2 &= (2k + 1)^2 - (2l + 1)^2 \\
&= 4k^2 + 4k + 1 - (4l^2 + 4l + 1) \\
&= 4k^2 + 4k - 4l^2 - 4l
\end{align}

My two concerns:
1) Is this correct so far?
2) How would I deal with the “$8\;\mid$” part?

$8 \mid x$ if and only if $x\equiv 0\mod 8$. Then you have $4k^2 + 4k - 4l^2 - 4l$ If $k$ is odd, $k^2$ is too. Then $4k^2 + 4k \equiv 0 \mod 8$ if $k$ is even then so is $k^2$ then $4k^2 + 4k \equiv 0\mod 8$. A similar argument for $l$ will finish your proof.
