$a\mid b,c\Rightarrow a\mid bx+cy\,$ for any integers $x,y$

Question

Prove that if $$a \mid b$$ and $$a \mid c$$ then $$a \mid bx+cy$$ for any integers $x$ and $y$.

Here's my proof: $$b = ak$$ $$c = am$$

$$bx+cy = akx+amy = a(kx+my)$$

Notice that $kx+my$ is an integer.

Therefore by the fundamental Theorem of Arithmetic/Theorem of divisibility,

$$ a \mid bx+cy $$
Is there another way to prove this? How would you do this another way?

I think your way is nice enough. I especially like how you wrote Amy's name there. — Paul Hurst, May 15 '14 at 07:42
But you don't need to invoke the fundamental theorem of arithmetic — Jean-Claude Arbaut, May 15 '14 at 07:47
@drhab Consider this: $,S = a\Bbb Z,$ is a subgroup of $,\Bbb Z,$ by the Subgroup Test (i.e. $,S\ne \emptyset,$ is closed under subtraction). Therefore, since $,S,$ is a group, $,b,c\in S,\Rightarrow, nb+mc\in S,$ for all $,n,m\in \Bbb Z.\ $ Do you consider that essentialy different? — Bill Dubuque, May 15 '14 at 14:24
@BillDubuque No. The OP notes in fact that $b$ and $c$ both belong to $a\mathbb Z$ and concludes almost directly that this leads to $nb+mc\in a\mathbb Z$. This without mentioning the concept of group, but nevertheless applying (or discovering if you wish) characteristics of a group. — drhab, May 15 '14 at 20:28

score 2 · Answer 1 · answered May 15 '14 at 07:43

I wouldn't do it any other way. However I would improve the write-up of the proof.

You say initially that $c=aj$ but after this you clearly mean that $c=am$. You need to be consistent.
Your third equality sign is missing its left hand side. It should be $$bx+cy=akx+amy=[\hbox{etc}]\ .$$
The reason for your final conclusion is nothing to do with the fundamental theorem of arithmetic, it is simply the definition of divisibility.

1 Answers1