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.
 A: 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].
A: Let $c$ be any common divisor of $a+b$ and $a-b$. Then $c$ divides $(a+b)+(a-b)$ and $(a+b)-(a-b)$. So $c$ divides $2a$ and $2b$.
Note that there are integers $x$ and $y$ such that $ax+by=1$ and therefore $(2a)x+(2b)y=2$. It follows that $c$ divides $2$. Thus no number greater than $2$ can divide both $a+b$ and $a-b$. 
For completeness, you should show that each of the possibilities $\gcd(a+b,a-b)=1$ and $\gcd(a+b,a-b)=2$ can happen. 
A: $\begin{eqnarray}{\bf Hint}\ \ \text{determinant of}\,\  \color{#c00}{\left[ \begin{array}{cr} 1 &\!\! -1\\ 1 & 1\end{array}\right]} \left[ \begin{array}{cc} a & x\\ b & y\end{array}\right] &=& \left[ \begin{array}{cc} a\!-\!b &\! x\!-\!y\\ a\!+\!b &\! x\!+\!y\end{array}\right] \\[.6em]
\Longrightarrow_{\phantom{|_{|_|}}}\   \color{#c00}2\ {{(\color{#0a0}{ay\! -\! bx})}} &=&\, (\color{#90f}{a\!-\!b})(x\!+\!y)-(\color{#90f}{a\!+b})(x\!-\!y) 
\end{eqnarray}$
By Bezout there are $\,\color{#0A0}{x,y}\in\Bbb Z\,$ with $ \color{#0a0}{ay\!-\!bx\!=\!1}\,$ so $\,d\mid\color{#90f}{a\!-\!b,a\!+\!b}\Rightarrow d\,|\, {\rm\small RHS}\Rightarrow d\,|\,\color{#c00}2(\color{#0a0}1)\! =\! \rm\small LHS$
Remark $\ $ The method generalizes to any linear map $\ (m,n)\mapsto (m,n)A = (am\!+\!bn,cm\!+\!dn)$
$\begin{eqnarray}\phantom{\bf Hint}\ \ \text{determinant of}\,\  \left[ \begin{array}{cr} a&\!\! b\\ c & d\end{array}\right] \left[ \begin{array}{cc} m & x\\ n & y\end{array}\right] &=& \left[ \begin{array}{cc} am\!+\!bn &\! ax\!+b\!y\\ cm\!+\!dn &\! cx\!+\!dy\end{array}\right] \\[.3em]
\Rightarrow\ \   D\!\!\!\!\!\!\!\!\! \smash[b]{\underbrace{(my - nx)}_{\quad\ \ \large =\,(m,n)\ {\rm by\ Bezout}}}\!\!\!\!\!\!\!\!\!\!\! &=& (am\!+\!bn)(cx\!+dy)-(cm\!+\!dn)(ax\!+\!by)\\ \\
\end{eqnarray}$
Therefore we deduce that  $\ (am\!+\!bn,\,cm\!+\!dn)\mid D\, (m,n)\ $ since it divides  $\rm\small RHS$ so also  $\rm\small LHS$. Alternatively, this can also be deduced using Cramer's Rule, e.g. see this answer. Notice that the question is simply the special case when the determinant $\,D = 2,\,$ and the gcd $\,(m,n) = 1.\,$
