I know that the answer involves Bezout's theorem in some way. I tried this:
Let $(a, b) = d$ Let $a = dk_1 and b = dk_2$ So, $ax + by = d$ becomes $dk_1x + dk_2x = d$ Dividing by d, we get: $k_1x + k_2x = 1$ So, $(x,y) = 1$ Is this sufficient?
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Yes: this proof is ok. – Crostul Jan 20 '21 at 11:08
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No, it is not OK. There is no $y$ involved. $k_1x+k_2x=(k_1+k_2)x=1$ does not imply $(x,y)=1$. – Dietrich Burde Jan 20 '21 at 11:21
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Sorry, I wrote x instead of y. It will be correct then, right? – Paul Jan 20 '21 at 12:49
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I would prove it in the following way:
Let $(a,b)=d$. By definition, there exists $q_1,q_2$ such that $a=dq_1, b=dq_2$. Then, by hypothesis, $ax + by = dq_1x + dq_2y = d$. As we are in an integral domain, we can simplify the expression to $q_1x + q_2y=1$. Now, notice that you directly concluded that (x,y)=1. This is only true in the case that a linear combination of integers is equal to 1 (or unity, if we were working with polynomials in $\Bbb K[x]$).
By definition of gcd, we say $d'=(x,y)$. Then, $d$ has to divide 1, because $q_1x + q_2y=1$. Hence, the only possibility is $(x,y)=1$.
