Prove that if $n|(mn+k)$ then $n|k$ My attempt at a proof:
Let $x=mn+k$, as $n|x$,by the Distributive Property and the fact that division is the inverse of multiplication $x=n(m+k/n)$ . It follows from the definition of divisibility that $(m+k/n)$ is an integer. As the integers are closed under addition, therefore $k/n$ must be an integer and thus $n|k$.
This last sentence seems as though I'm missing something important, or that I am making a logical error. Any assistance in correcting my proof is greatly appreciated.
 A: Yes, but it's harder to parse than necessary. You don't need to state every axiom you are using at every step (if you did this, you would struggle to get far in maths because you would fail to see the wood from the trees). This is how I would have written it:
If $n=0$ then $k=0$ and the result is immediate. So suppose $n\neq 0 $ and let $x=mn+k=n(m+k/n).$ Since $n|x,$ it follows that $(m+k/n)$ is an integer and since $m$ is an integer, $k/n$ must be an integer also. Therefore, $n|k$.
A: there is an integer $t$  with $$  nt = mn + k$$
$$ nt - nm = k $$
$$n(t-m)  = k $$
As $t-m$  is an integer, this means...
A: $n \mid (mn + k) \Longrightarrow \exists t, \; tn = mn + k \Longrightarrow tn - mn = k \Longrightarrow n(t - m) = k \Longrightarrow n \mid k. \tag{1}$
A: It's nice using rationals to prove things about integers, but in this case there's an awkwardness because $n$ might be zero, and you can't divide by that.
Here's a proof:
Proposition. Suppose $m$, $n$, and $k$ are integers. If $n\mid (mn+k)$ then $n\mid k$.
Proof. Suppose $n\mid (mn+k)$. Let $h\in\mathbb{Z}$ be such that $hn=mn+k$, using the definition of divisibility. Then $(h-m)n=k$, so $n\mid k$ by the definition of divisibility. QED
Another proof makes use of two nice lemmas that you should make sure to know: (1) if $a,b,c$ are integers such that $a\mid b$ then $a\mid bc$, and (2) if $a,b,c$ are integers such that $a\mid b$ and $a\mid c$, then $a\mid (b+c)$. Both are proved in a similar way to the above proposition.
Using these, here's the proof:
Proof. Suppose $n\mid (mn+k)$. Since $n\mid n$, then $n\mid (-mn)$. Then using the assumption, $n\mid ((mn+k)+(-mn))$, so $n\mid k$. QED
