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I have proven the forward direction. By BeZout's identity: $1=ar+(bc)s$ for some integers $r,s$. A corollary states that $(a,b) =1$ if and only if $1=ar+bs$ for some integers $r, s$. So we can conclude that $(a,b)=1$ since $1 = ar +b(cs)$ and also that $(a,c)=1$ since $1=ar + c(bs)$.

But I do not know how to prove the reverse direction.

Paul
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mmm
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2 Answers2

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Suppose that $(a,b)=1$ and $(a,c)=1$. Then we can write $$ ra+sb=xa+yc=1. $$ since this property is equivalent to finding a linear combination of $a,b$ or $a,c$ so that we sum up to $1$. Multiplying this out, $$ (ra+sb)(xa+yc)=1\cdot 1=1. $$ When we expand, we'll have a $(sy)(bc)$ term and some other terms depending on $a$. Therefore we have found the required linear combination adding to $1$.

In particular, $$ (rxa+ryc+sbx)a+(sy)bc=1. $$

Ian Coley
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  • So I found that there is a proposition that states that for all integers a, b, c (a,bc)|(a,b)(a,c). So I think that since (a,b)(a,c)=1 then (a,bc)|1 so (a,bc)=1. Does that suffice to prove the given statement? – mmm Feb 20 '14 at 02:35
  • Certainly. But this way uses what you know already. – Ian Coley Feb 20 '14 at 02:36
  • Also, your suggestion does make sense for me. Thank you. – mmm Feb 20 '14 at 02:37
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Suppose that $(a,b)=m$. then $a=mk$ and $b=mj$. Then $bc=mjc$. So $m$ divides $bc$ too. meaning $(a,bc)$ is at least m.

You can do the same form $(a,c)=n$

Asinomás
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