I am already able to prove that $g \mid s$ assuming $x+y=s$ and $(x,y)=g$, but I am having some trouble showing that assuming $g \mid s$, there exists an $x$ and $y$ such that $x+y=s$ and $(x,y)=g$. So far, I've started with saying that $g \mid 0$ necessarily. Therefore we also know that $g \mid (0x+sy)$. I'd like to be able to set the values of $x$ and $y$ to something to show that there exist an $x$ and a $y$ that satisfy $x+y=s$ and $(x,y)=g$, but I'm not really sure where to go from here. Can anyone offer any help?
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HINT. $(gn,g) = g(n,1)=g$.
The converse follows because $a|b$ and $a|c$ implies $a|b+c$.
$\rm\ \ g = (x,s-x) = (x,s)\ \iff\ 1 = (x/g,\:s/g)\:,\ $ so choose $\rm x/g\ $ coprime to $\rm s/g\:,\ $ e.g. $\rm\ s/g + 1$