Prove: If $a$ and $b$ are odd integers, then $8\mid (a^2-b^2)$
So far I have: $$8\mid \left((2n+1)^2-(2m+1)^2\right)\Longleftrightarrow 8\mid \left(4n^2+4n+1-4m^2-4m-1\right) $$ Is this right so far? Where do I go from here? Step by step explanation please!
$$4n^2+4n+1-4m^2-4m-1=4n^2+4n-4m^2-4m=4(n^2-m^2+n-m) \\ 4[(n-m)(n+m)+(n-m)]=4(n-m)(n+m+1)$$
Now
$$(n-m)+(n+m+1)=2n+1$$ is odd, thus one of $(n-m)$ or $(n+m+1)$ must be even.
P.S. A simpler solution is the following
$$a^2-1=(a-1)(a+1)$$ is the product of two consecutive even numbers. Thus one is divisible by $4$ and the other is even. This shows that $8 |a^2-1$. Same way $8|b^2-1$.
Using the difference of squares formula $a^2-b^2=(a+b)(a-b)$ shows us that $$\begin{align}(2n+1)^2-(2m+1)^2 &= \bigl((2n+1)+(2m+1)\bigr) \bigl((2n+1)-(2m+1)\bigr)\\ &= (2n+2m+2)(2n-2m)\\ &= 4(n+m+1)(n-m).\end{align}$$ Now, use the fact that, for integers $n,m,$ we have that $n-m$ is even if and only if $n+m$ is even. (If you haven't seen this fact before, it's a nice exercise to prove.) Hence, exactly one of $n+m+1,n-m$ is even, and so....
${\rm mod}\ 8\!:\,\ {\rm odd}^2\equiv \overbrace{\{1,3,5,7\}^2}^{\Large \{\pm1,\,\pm3\}^2} \equiv 1,\ $ so $\rm\,\ odd^2\!-odd^2 \equiv 1-1\equiv 0$
Alternatively $\ (2n\!+\!1)^2 = 4\color{#c00}{n(n\!+\!1)}+1\equiv 1\pmod 8\ $ by $\ 2\mid \color{#c00}{n(n\!+\!1)}$.
