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So obviously because of the if and only if we must first prove that

If there exist integers $x$ and $y$ such that $x+y=S$ and $(x,y)=G$ then $G\mid S$.

And then if $G\mid S$, then there exist integers $x$ and $y$ such that $x+y=S$ and $(x,y)=G$.

Please help

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Cindy, please do not delete the question, leaving only "prove if and only if": Someone took the time to answer your question, which makes little sense without the question! –  amWhy Dec 9 '12 at 22:43
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I have reverted the question to the original question. Please do not again overwrite it with a new question (which invalidates the answer below). Instead, please post the new question separately. –  Bill Dubuque Dec 10 '12 at 0:53
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1 Answer

$G|x$ and $G|y$, hence $Gp=x$ and $Gq=y$ for some integers $p,q$. Thus, $G(p+q)=x+y$. therefore $G|x+y$

Converse: If $S=kG$, let $x=G$ and $y=(k-1)G$

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I forgot a part of the question I will edit my answer –  Amr Dec 9 '12 at 22:03
    
could you elaborate on the converse? –  Cindy Dec 9 '12 at 22:12
    
Check that $x,y$ given in the converse satisfy $(x,y)=G$ and $x+y=S$ –  Amr Dec 9 '12 at 22:14
    
@Cindy: Just do the algebra. With Amr’s suggested values of $x$ and $y$, what are $x+y$ and $(x,y)$? –  Brian M. Scott Dec 9 '12 at 22:14
    
So with the converse y=(k-1)G => y=kG-G => y=S-G => y=S-x => x+y=S but how do you show (x,y)=G? –  Cindy Dec 9 '12 at 22:17
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