GCD proof - by contradiction? 
$a, b$ are relatively prime and $a > b$.  Prove that $\text{gcd}(a-b,a+b) = 1 \,\,\text{or}\,\, 2$.

Attempt:
Since $a, b$ are relatively prime, I know that $\text{gcd}(a,b) = 1$.  Also, know that $ax + by = 1 \,\,\text{for some}\,\, x,y \in \mathbb{Z}$.  Not sure where to go from there.  I suppose a proof by contradiction with the assumption that $\gcd(a-b,a+b) \neq 1 \,\,\text{or}\,\, 2$.  Thanks.

Hint $\,\ {\rm mod}\ a\!-\!b\!:\,\ a\equiv b\,\Rightarrow\, \color{#0a0}{a\!+\!b\equiv 2b}\,$ so, by the Euclidean Algorithm $\rm\color{#c00}{(EA)}$
$\qquad (a\!-\!b,\,\color{#0a0}{a\!+\!b})\overset{\rm\color{#c00}{EA}} = (a\!-\!b,\,\color{#0a0}{2b}) = (a\!-\!b,\,2),\ $ by $\ (a\!-\!b,b)\overset{\rm\color{#c00}{EA}}= (a,b)=1,\,$ and Euclid's Lemma.
Here $\rm\color{#c00}{EA}$ means $\ (m,n)\, =\, (m,n')\ $ if $\ n'\!\equiv n\pmod m\,\ $  [= descent step of Euclidean Algorithm].
