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I am struggling with a direct proof involving divisibility. I have been struggling with proofs all semester.

What I'm trying to prove: $n, m, r \in \mathbb{Z}$. If $n\mid m$ and $m\mid r$, then $n\mid(r+r^2)$

So far for this proof I only have a very basic setup, but not sure where to go from here

I have $$ n\mid m, i \in \mathbb{Z}, m=ni$$

$$ m\mid r, j \in \mathbb{Z}, r=mj$$

and trying to prove that $$\exists k \in \mathbb{Z}, (r+r^2)=nk$$

I then have $$r+r^2 = (mj+(mj)^2)$$ $$=(mj+(mj)(mj))$$ $$=mj(1+mj)$$

Since $r=mj$ I substituted $mj$ in for $r$. I'm not completely sure I'm heading in the right direction, maybe I should substitute $m$ as $ni$ so that it would look more like this:

$$r+r^2=(nij+(nij)^2)$$ $$=(nij+(nij)(nij))$$ $$=nij(1+(nij))$$

As I said, I am really struggling with proofs this semester, so any help would be greatly appreciated to help make sense of this. It's hard to prove anything when you second guess every step you take.

Michael Hardy
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Michael Rector
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    Compensatory upvote. I wish most OPs included this much of an attempt. – Brian Tung Oct 10 '16 at 19:39
  • @MichaelHardy, do you consider changing | into \mid such an important issue, worth editing posts to perform this change, hunting for every occurence of |? – Alex M. Oct 10 '16 at 19:45
  • Thanks, Brian. I wouldn't feel right asking for help if I didn't at least show what I have and where I am struggling. – Michael Rector Oct 10 '16 at 19:45
  • Michael, the transitivity "$a\mid b,b\mid c$ implies $a\mid c$" makes it much easier. I can understand that you want to finish your proof first, but this way the proof becomes really short and clear. Also, it always works, e.g., we also have, ays, $n\mid r+r^2+r^3$, without using more $nij$ and $nj$. – Dietrich Burde Oct 10 '16 at 20:38
  • @AlexM. : $\ldots,$and other copy-editing$,\ldots$ Editing a posting doesn't take much. $\qquad$ – Michael Hardy Oct 10 '16 at 21:24

2 Answers2

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You were almost there! You got as far as

$$r+r^2=nij(1+nij)$$

But this means that $n$ divides $r+r^2$, because $n$ divides $n \times ij(1+nij)$. That's all you need!

TonyK
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  • Thank you. So my second attempt was more accurate than my first. Now I just need to mention that $nij(1+nij)$ is an integer which will divide $r+r^2$? – Michael Rector Oct 10 '16 at 19:47
  • No no no no no....$nij(1+nij)$ is equal to $r+r^2$. As you yourself showed. – TonyK Oct 10 '16 at 20:29
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If we have $n\mid m$ and $m\mid r$ we also have $n\mid r$, this has been shown here on MSE, and is easy to see: write $m=an$ and $r=bm$, then $r=abn$. But then again, apply transitivity. Because $n\mid r$ and $r\mid r^2$ it follows that $n\mid r^2$, and then of course also $n\mid r+r^2$.

Community
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Dietrich Burde
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