<s>[INST] Prove $\gcd(a+b, a-b) = 1$ or $2\,$ if $\,\gcd(a,b) = 1$ 
I want to show that for $\gcd(a,b) = 1$ $a,b \in Z$   $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ holds.
I think the first step should look something like this:
$d = \gcd(a+b, a-b) = \gcd(2a, a-b)$

From here I tried to proceed with two cases.
1: $a-b$ is even, which leads to $\gcd(a+b, a-b) = 2$
2: $a-b$ is odd, which leads to $\gcd(a+b, a-b) = 1$
My main problem I think is, that I do not know how I should include $\gcd(a,b) = 1$ in the proof.
Any help is appreciated. Thx in advance.
Cherio Woltan
 [/INST] If $\gcd (a,b)=1$, then both $a$ and $b$ can't be even simultaneously. Now two cases arise,
Case $1$. Both $a$ and $b$ are odd
        In this case, both $a+b$ and $a-b$ will be even, so their gcd must be an even number,
        and so $\gcd (a+b, a-b) = \gcd (2a, a-b) = 2$, as $a-b$ is even.
Case $2$. One of the $a$ and $b$ is even and the other is odd
        In this case, both $a+b$ and $a-b$ will be odd, so their $\gcd$ must be an odd number,
        and in this case $\gcd (a+b, a-b) = \gcd (2a, a-b) = 1$, as $a-b$ is odd.
Note: $\gcd (a, a-b)$ can't exceed $1$ as $a$ and $b$ are relatively prime.
</s>