To be proved: for all integers $a,b,x,y$, if $ax + by = \gcd(a, b)$ then $\gcd(x,y)=1$.
Suppose that $ax+by = \gcd(a,b)$. I first want to prove that every common divisor of $x$ and $y$ divides $\gcd(a,b)$.
Let $n$ be such a common divisor. We have: $n$ divides $x$ and $n$ divides $y$, implying that $x = kn$ and $y = k'n$. Substituting these expressions for $x$ and $y$ in the equation: $ax+by = \gcd(a,b)$, we get: $a(nk)+ b( k'n)= \gcd(a,b)$, which is equivalent to: $n(ak + bk') = \gcd(a,b)$. This shows that $n$ divides $\gcd(a,b)$.
My question is : can I continue in the following way? That is by saying that the problem amounts to finding the greatest value of number $n$, if $n$ divides $\gcd(a,b)$ for all values of $a$ and of $b$? In that case, we would have a value of $\gcd(a,b)$ ranging from $1$ to infinity, thus having a possible minimum value of $1$, and only the value $n=1$ would guarantee that $n$ is a divisor of $\gcd(a,b)$ in all possible cases.
I suspect a fallacy in my attempt due to an incorrect use of quantifiers. Is it the case?
Am I on a good track at least in the first part of the above attempt?
