suppose gcd (a,b) = k.
k|a and k|b => k|(a+b)
So since k is a common divisor.
So if then
Taking the contrapositive gives us gcd (a, a+b) = 1 => gcd (a,b) = 1
suppose gcd (a, a+b) = p. Then the proof for this direction is very very similar to the previous direction.