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I know that if $a \mid b$ and $a \mid c$ then $a \mid b+c$. Which is easy to see since

$a \mid b \Leftrightarrow \exists k \in \mathbb{N} : a \cdot k=b$ and

$a \mid c \Leftrightarrow \exists n \in \mathbb{N}: a \cdot n=c$.

Thus $b+c=ak+an=a(n+k)$, which implies $a \mid c+b$.

But does $a \mid b$ and $a \nmid c$ imply $a \nmid (b+c)$?

I think that this is true, but I do not know how to proof this.

Thomas Andrews
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  • Yes. In short: $$a\mid b\implies (a\mid b+c\iff a\mid c)$$ – Thomas Andrews Feb 15 '24 at 16:28
  • Suppose $a | b+c$ and try to reach a contradiction. – jjagmath Feb 15 '24 at 16:29
  • Perhaps explicitly considering the remainder of $(b + c)$ when divided by $a$ will be a good approach. – davidlowryduda Feb 15 '24 at 16:30
  • Easier: Start with $a\mid b\iff a\mid -b.$ So $a\mid b+c$ and $a\mid b$ means $a\mid (b+c)+(-b)=c.$ But that assumes you are working in the integers. If you are working in the natural numbers, you'll have to prove it seperately. – Thomas Andrews Feb 15 '24 at 16:31
  • $!\bmod a!:\ b\equiv 0,, \color{#c00}{c\not\equiv 0},\Rightarrow, b+c\equiv \color{#c00}{c\not\equiv 0},$ by basic congruence laws, cf. linked dupe. $\ \ $ – Bill Dubuque Feb 15 '24 at 16:36
  • This is a much more general principle: If $G$ is a group, $H$ is a subgroup, $h\in H$, and $g\in G\setminus H$, then $gh\in G\setminus H$. From this we see lots of examples: Your result with $G=\mathbb{Z}$ and $H={am:m\in\mathbb{Z}}$. But also, the sum of a rational and an irrational is irrational, the product of an invertible $n\times n$ matrix and a non-invertible $n\times n$ matrix is non-invertible, if $V$ is a vector space and $W$ is a subspace, then for $w\in W$ and $v\in V\setminus W$, $v+w\in V\setminus W$. –  Feb 15 '24 at 16:41
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    @jwhite That too is a dupe, see the complementary subgroup test. – Bill Dubuque Feb 15 '24 at 16:42
  • My comment was not to oppose vote to close as a duplication. Just to provide a potentially beneficial viewpoint, which I see you've already expounded nicely upon. –  Feb 15 '24 at 16:50

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This is using the more general (and very important, this will come up all the time!) fact that if $a \mid x$ and $a \mid y$, then $a \mid mx + ny$ for all $m, n \in \mathbb Z$, where something of the form $mx + ny$ is called an integer combination of $x$ and $y$ for future note. I'd suggest trying to prove this on your own using the definitions you gave, ask for help if you need though.

From there, if you assume for a contradiction that $a \mid b$ and $a \nmid c$, but $a \mid b + c$, then using the above, $a \mid b$ and $a \mid b + c$ imply $a \mid (-1)b + (b + c)$, that is, $a \mid c$, which is a contradiction, so you must have $a \nmid b + c$.

Alex Pawelko
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  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Feb 15 '24 at 16:38
  • Apologies, I did not see the duplicate linked as I answered (I believe it was not linked yet, which is my fault for jumping the gun). – Alex Pawelko Feb 15 '24 at 16:39
  • Of course basic questions like this have been answered many times over the past $13$ years. Please delay answering them so that they can be processed appropriately. – Bill Dubuque Feb 15 '24 at 16:41