Greatest common divisor of two relatively primes Here is the question I am trying to prove:

If $a,b$ are relatively prime and $a>b$ prove that $\gcd(a-b, a+b) \in \{1, 2\}$.

Can I begin with something like $(a-b)k + (a+b)l = d$ where $k,l$ are integers and $d=\gcd(a-b,a+b)$?
 A: One can use a Bézout's lemma argument related to how you started. Since $a$ and $b$ are relatively prime, there exist integers $x$ and $y$ such that $ax+by=1$. 
Note that then $(a-b)(x+y)+(a+b)(x-y)=2ax+2by=2$. It follows that the gcd of $a-b$ and $a+b$ divides $2$. It is easy to come up with examples where the gcd is $1$, and also with examples where the gcd is $2$.
Remark: The above argument is by no means my favourite way to prove the result. 
A: Let $p$ a prime that divides $\gcd(a-b,a+b)$ so $p$ divides $a$ and $b$ and then divides $(a-b)+(a+b)=2a$ and divides $(a+b)-(a-b)=2b$. Now by Euclid's lemma we see that: 


*

*the case $p$ divides $a$ and $b$ isn't possible since $a$ and $b$ are coprime numbers

*if for example $p$ divides $2$ and $b$ then $b=2$
and we conclude that $p=2$ and then $\gcd(a-b,a+b)=1$ or $2$.

A: Hint $\ $ By the Lemma below $\rm\ \gcd(a,b)\mid gcd(a\!-\!b,a\!+\!b)\mid 2\gcd(a,b)$
Lemma  $\ $ If $\rm\,(x,y)\overset{A}\mapsto (X,Y)\,$ is linear then $\: \rm\gcd(x,y)\mid \gcd(X,Y)\mid \Delta \gcd(x,y),\ \ \Delta = \det A$
Proof $\ $ Inverting the linear map $\rm\,A\,$ by Cramer's Rule (multiplying by the adjugate) yields 
$$\rm \begin{eqnarray} a\ x\, +\, b\ y &=&\rm X\\ \\ \rm c\ x\, +\, d\ y &\ =\ &\rm Y\end{eqnarray}
\quad\Rightarrow\quad \begin{array} \rm\Delta\ x\ \ \ =\ \ \ \rm d\ X\, -\, b\ Y \\\\ \rm\Delta\ y\ =\ \rm -c\ X\, +\, a\ Y \end{array}\ ,\quad\ \Delta\ =\ ad-bc\qquad $$
Hence, by RHS system, $\rm\ n\ |\ X,Y\ \Rightarrow\ n\ |\ \Delta\:x,\:\Delta\:y\ \Rightarrow\ n\ |\ gcd(\Delta\:x,\Delta\:y)\ =\ \Delta\ gcd(x,y)$.
In particular $\rm\ n = \gcd(X,Y) \mid \Delta\, \gcd(x,y) $. 
Further, by LHS system $\rm\,n\mid x,y\ \Rightarrow\ n\mid X,Y\ \Rightarrow\ n\mid\gcd(X,Y)$.
In particular $\rm\ n = gcd(x,y)\mid \gcd(X,Y).\quad $  QED
A: Hint May be easier to go the other way. Let's say $\gcd(a-b, a+b) = k$ where $k \neq 1$. Then, it must be that $k|(a-b)$ and $k|(a+b)$. So either


*

*$k|a$ and $k|b$ (this cannot be, why?) or

*both $a,b$ are odd (what happens then?)

