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We have to prove $b|a$ and $b|c \Rightarrow b|ka+lc$ for all $k,l \in \mathbb{Z}$.

I thought it would be enough to say that $b$ can be expressed both as $b=ka$ and $b=lc$. Now we can reason that since $ka+lc=2b$ and $b|2b$, it directly follows that $b|ka+lc$?

But I'm not sure if that works for any value of $k$ and $l$ (namely $k$ and $l$ are defined through quotient between $a$ and $c$, respectively).

What am I missing?

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Ahhh, I think I figured it on my own... I have to show not that b divides the sum of a and c, but the sum of any multiples of a and b... – smihael Dec 11 '12 at 11:39
Please see the answers bellow: $b|a \not\Rightarrow b = ka$ but $a=kb$!!! – smihael Dec 11 '12 at 11:56
up vote 4 down vote accepted

But $b\mid a$ does not mean that $b$ can be expressed in the form $ka$: it means the exact opposite, that $a=mb$ for some integer $m$. For instance, $3\mid 12$ because there is an integer $m$ such that $12=3m$, specifically, $m=4$. Similarly, $b\mid c$ means that $c=nb$ for some integer $n$. Thus, $$ka+\ell c=k(mb)+\ell(nb)\;.$$ Can you see how to proceed from here to conclude that $b\mid ka+\ell c$?

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Yes, thank you. Such a stupid mistake. – smihael Dec 11 '12 at 11:45
@smihael: You’re welcome. (And we’ve all made ’em at one time or another.) – Brian M. Scott Dec 11 '12 at 11:48

You have things backwards. $b|a$ means $a = mb$, where $m \in \mathbb{Z}$. You should be able to take it from there.

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Stupid me, I should ge more sleep... Thanks! – smihael Dec 11 '12 at 11:43

An alternative presentation of the solution (perhaps slightly less elementary than the already proposed answers) is to work in the quotient ring.

You can write that in $\mathbb{Z}/b\mathbb{Z}$

\begin{align*} \overline{ka+lc}&=\bar{k}\bar{a}+\bar{l}\bar{c}\\ &=\bar{0}+\bar{0}\\ &=\bar{0} \end{align*}

where $\bar{a}=\bar{c}=\bar{0}$ because $b | a$ and $b | c$.

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