If $\gcd (a,b) = 1$, what can be said about $\gcd (a+b,a-b)$? [duplicate]
Solution 1:
Hint If $d= \gcd(a+b,a-b)$ the $d| (a+b)+(a-b)$ and $d \mid (a+b)-(a-b)$.
Solution 2:
You can see that
$$\gcd(a+b,a-b)=\gcd(a+b, a+b+a-b)=\gcd(a+b,2a)$$
but then this divides $2a$, and since any divisor of $a$ does not divide $b$, it must be that it divides $2$, i.e. it is either $1$ or $2$.