If $a$ and $b$ are odd, prove $\gcd(a,b) = \gcd(\frac {\left| {a-b} \right |} {2}, b)$ Honestly I don't have a strong idea. I don't know where to even begin, I have considered that the $\gcd(a,b)$ is somehow less than $a-b$, but I'm not even sure why that would be the case.. Any help would be great!

I am pretty sure you know Euclidean algorithm, which, basically, says $gcd(a,b)=gcd(|a-b|,b)$. Since $b$ is odd, and $|a-b|$ is even, $2$ divides the latter, but not the former, so that $gcd(|a-b|,b)=gcd(\frac{|a-b|}{2},b)$.
