If $n\mid m$ and $m\mid r$, then $n\mid(r+r^2)$

Question

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.

Answer 1

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 $n\mid r+r^2$.

Answer 2

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!