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?
If $s$ and $g>0$ are integers, how can I prove that $x$ and $y$ exist satisfying $x+y=s$ and $(x,y)=g$ if and only if $g \mid s$?
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elementary-number-theory
divisibility