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Prove that for $d,n\in \mathbb{Z}$, if for some nonzero integer $a$ we have $ad\mid an$, then $d\mid n$. Conversely, if $d\mid n$, then $ad\mid an$ for all $a \in \mathbb{Z}$.

For the first part, I got: If $ad\mid an$, then $\exists k \in \mathbb{Z}$ such that $ad⋅k=an \Longrightarrow d\cdot k=n\Longrightarrow d\mid n $

However for the second part,I'm not sure how to go about this except for something that seems rather obvious:

$$d⋅k=n \Rightarrow ad⋅k=an \,\, \forall a \in \mathbb{Z} $$

Also, i'm not sure if the rightarrow i'm using actually is being used correctly. I mean is it being used appropriately in this context. Can the right arrow be taken to mean "therefore"? Thanks.

Michael Hardy
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Bucephalus
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    $d,|,n\implies n=md\implies an=amd$. that's really what you wrote, of course...so yes. Your argument is valid. – lulu Jun 17 '17 at 13:46
  • Ahh, so I did it the wrong way around. – Bucephalus Jun 17 '17 at 13:47
  • Well, I'm not sure. It's always a good idea to clearly state assumptions and conclusions. I started with the clear statement $d,|,n$, I ended by writing $an$ as an explicit multiple of $ad$ from which I instantly can conclude that $ad,|,an$. – lulu Jun 17 '17 at 13:48
  • ok, thanks @lulu – Bucephalus Jun 17 '17 at 13:52

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