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?
 A: All is correct; now the last expression can be written
$$
4\bigl(k(k+1)-l(l+1)\bigr)
$$
and you just have to prove that $k(k+1)-l(l+1)$ is even.
Hint: Can you tell whether $m(m+1)$ is even, for an integer $m$?
A: $8 \mid x$ 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.
A: You're correct so far. What you need to finish is this
Hint: $4u^2+4u = 8v$
Solution:

 $ 4u^2+4u=4(u+1)u=8\binom{u+1}{2}$

A: if $K$and $l$ are even, then $K=2K_1$ and $l=2K_2$
then $4K^2+4K-4l^2-4l=16K_1^2+8K_1-16K_2^2-8K_2$ which is clearly divisble by 8
now if $K$and $l$ are odd, then $K=2K_1+1$ and $l=2K_2+1$
then $4K^2+4K-4l^2-4l=4(4K_1^2+4K_1+1)+8K_1+4 -4(4K_2^2+4K_2+1)-8K_2+4= 16K_1^2+16K_1+8K_1+8-16K_2^2-16K_2-8K_2-8$ which is clearly divisble by 8
if $K$ is even and $l$ is odd or $K$ is odd and $l$ is even, it is the same calculation, try it! 
