Show that $8 \mid (a^2-b^2)$ for $a$ and $b$ both odd 
If $a,b \in \mathbb{Z}$ and odd, show $8 \mid (a^2-b^2)$.

Let $a=2k+1$ and $b=2j+1$. I tried to get $8\mid (a^2-b^2)$ in to some equivalent form involving congruences and I started with
$$a^2\equiv b^2 \mod{8} \Rightarrow 4k^2+4k \equiv 4j^2+4j \mod{8}$$
$$\Rightarrow k^2+k-j^2-j=2m$$
for some $m \in \mathbb{Z}$ but I am not sure this is heading anywhere that I can tell.
Second attempt: Use Euler's Theorem and as $\gcd(a,8)=\gcd(b,8)=1$ and $\phi(8)=4$, $a^4 \equiv b^4 \equiv 1 \mod 8$ so $a^4-b^4\equiv 0 \mod{8}$.
I haven't gotten too much further are there any hints?

HINT:
$k^2-j^2+k-j=(k-j)(k+j+1)$
As $(k+j+1)-(k-j)=2j+1$ which is odd, they must be of opposite parity, exactly one of them must be divisible by $2$ 
Method 2:
If $a,b$ are odd, observe that one of $(a-b),(a+b)$ is divisible by $4,$ the other by $2$
Method 3:
$(2a+1)^2=4a^2+4a+1=8\frac{a(a+1)}2+1\equiv 1\pmod 8$
