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supposition: $A_1=\{k \in \mathbb{Z}\ \colon\ k | (bc +a) \text{ and }k |b\}$, $A_2=\{k \in \mathbb{Z}\ \colon\ k|a \text{ and }k|b\}$
claim: $A_1=A_2 $

(my) proof: Let's show that $A_1 \implies A_2 $. Let $k \in A_1$, so $k \mid (bc +a)$ and $k \mid b$. Now if $k \mid (bc+a)$, then $k \mid 1 \cdot (bc+a) -bc$ $\implies$ $ k \mid a $. So $A_1 \implies A_2 $.

I'm not sure how to show $k \mid 1 \cdot (bc+a) -bc$?

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    One does not usually write that a set implies another set. Rather, one write "$A_1\subseteq A_2$", and "$A_2\subseteq A_1$".2011-10-14
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    Since $k|b$, then $k|bc$. Since $k|(bc+a)$ and $k|bc$, then $k|(bc+a)-bc$. Will you be accepting any more of the 45 questions you've asked so far, or just the 7 you accepted so far (including today's acceptance of a question that was last modified on November 2, 2010)?2011-10-14
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    If $k \mid x$ and $k\mid y$ then $k\mid (x-y)$. Let $x=bc+a$, $y=bc$. We are told that $k \mid x$. We are also told that $k\mid b$, which implies $k\mid bc$, that is, $k \mid y$.2011-10-14
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    Explicitly, what is needed is the following: If $k|u$ and $k|v$, then $k |(au + bv)$, for all integers $a,b$.2011-10-14
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    @Arturo: I will be accepting as soon as I have time to accept. Just one question: How do you get from k|b to k|bc? I have hard time to figure that out. Oh now I see. Is it as following? k|b <=> b=kl, $l \in N$ <=> cb =ckl = k(lc) => k|cb2011-10-15
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    @alvoutila: Are you really having a hard time figuring out how $k|b$ implies $k|bc$? $k|b$, and $b|bc$, so by transitivity $k|bc$. Yes, you *could* go by the definition, but you are just repeating the proof that "divides" is transitive.2011-10-15

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