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Problem: Prove that if $(a,b)=1$ and $c|a+b$ , then $(a,c)=(b,c)=1$.

My Attempt: Let $(a,c)=e$ and $(b,c)=f$. Then $e|a$ and $e|c$. This implies that $e|a+b$, which further implies that $e|b$. Thus we deduce that $e\leq f$ (from the fact that $e|b$ and $e|c$) and that $e\leq 1$ (from the fact that $e|a$ and $e|b$).Similarily since $f|b$ and $f|c$, we deduce that $f\leq e.$ Thus we can write that $f\leq e\leq f\Rightarrow e=f$ and since $e\leq 1\Rightarrow e=1$, we have $f=e=1.$

Since I don't have a teacher, I would like to know whether this proof is correct or not. Thanks in advance!

Student
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    Looks correct to me, although the last few lines seem a litte redundant. After concluding that $e$ divides $b$ you're done.. Becaue then $e = 1$ follows from $e | b$ combined with $(a, b) = 1$, and, completely analogously (and also because the statement is symmetric in $a$ and $b$) $f = 1$ follows from $f | a$ and $(a, b) = 1$ . That being said, if anything is 'wrong', you could explain a little bit more why $e$ has to divide $b$. – Woett Nov 02 '16 at 10:58
  • Thank you for your reply. If I write $e|b$ because $e|a+b-a$, will that be sufficient? – Student Nov 02 '16 at 12:40

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