prove that if gcd(a,b) = 1, then gcd(a,a+b) = 1

Question

So I need to prove that if gcd(a,b) = 1, then gcd(a,a+b) = 1. How do I go about it? I tried going with the ax + by = 1 method but it didn't work out so well.

Answer 1

I think it should have worked out well, as $$ax+by = 1$$ can be rewritten as $$a(x-y) + (a+b)y = 1$$ And since $x-y$ is an integer, we can just call it $x'$, and then we have $$ax'+(a+b)y = 1$$