<s>[INST] If a and b are odd integers, then $8\mid (a^2-b^2)$ 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!
 [/INST] $$4n^2+4n+1-4m^2-4m-1=4n^2+4n-4m^2-4m=4(n^2-m^2+n-m) \\
4[(n-m)(n+m)+(n-m)]=4(n-m)(n+m+1)$$
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 $8|b^2-1$.
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