if GCD of $(a, b) = 1$, prove that GCD $(a+b, a-b) = 1$ or $2 .$ The proof goes like: Let GCD $( a+b, a-b ) = d$ and let there exist integers m and n such that $ a+b =md$ and $ a-b = nd.$ By adding and subtracting these two equations we get: $2a = (m+n)d$ and $2b = (m-n)d$ , because $a, b$ are coprime then $2$ GCD $(a,b)$ = GCD $(2a, 2b),$ and so on.
My question is, why do we have to add and subtract above equations? I need to understand the concept of this prove in some more details. Thanks!
Letting $(x,y)$ denote the gcd of $x$ and $y$, you have $(a,b)=1$ and want to show that $(a+b,a-b)=1 \text{ or }2$. As such, you begin by letting $d=(a+b,a-b)$ and observing that there exist $m,n \in \mathbb{Z}$ such that $$a+b=md \text{ and } a-b = nd.$$ If you add the above equations, you get $2a = (m+n)d$. If you subtract, you get $2b = (m-n)d$. This tells you that $d|2a$ and $d|2b$. As you note, $(2a,2b)=2(a,b)$. This implies that $(2a,2b)=2$. This means that $d=1$ or $d=2$ (since $d$ is a divisor of $2a$ and $2b$, we must have $d\leq(2a,2b)$).
To answer the question about why you add/subtract them, the answer is to isolate $a$ and $b$ on the LHS. This is what allows you to conclude that $d|2a$ and $d|2b$. As mentioned in one of the comments, you don't HAVE TO add/subtract (there are other ways to proceed), but that is why it was done in this case.