If there exist x and y for which ax+by=GCD(a,b) then GCD(x,y)=1.
Let g=GCD(a,b) and x and y solutions to ax+by=g.
Since g|a and g|b, there exist integers n and m such that a=gn and b=gm.
The GCD(n,m)=1 since g is the GCD(a,b).
Let d be a divisor of x. If d also divides my, then d must divide 1, which isn’t possible. So, GCD(x,y)=1.