Prove that if $a$ and $b$ are relatively prime, then $\gcd(a+b, a-b) = 1$ or $2$ Prove that if $a$ and $b$ are relatively prime, then $\gcd(a+b, a-b) = 1$ or $2$.
I started off by putting $\gcd(a+b, a-b) = d$.
This implies that there are two relatively prime integers $x_1, x_2$, such that 
$dx_1 = a+b$ 
$dx_2 = a -b$
Adding the first equation to the second gives us: $d(x_1 + x_2) = 2a$, and subtracting the second from the first gives us $d(x_1 - x_2) = 2b$. This implies that $d\mid2a, d\mid2b \implies \gcd(2a, 2b) = d \implies d = 2\cdot \gcd(a,b) \implies d = 2$. 
This is based on the fact that $\gcd(x_1 + x_2, x_1 - x_2) = 1$, which is in fact the statement of the problem. How do I prove this? Can I somehow use the Euclidean Algorithm for this.  

If $ax+by=1$ then $$\begin{align}2&=2ax+2by \\&=(a-b)x + (a+b)x + (b-a)y+(b+a)y \\&= (a-b)(x-y)+(a+b)(x+y)\end{align}$$
