If $a \equiv b$ mod n, then $ac \equiv b(c+n)$ mod n Show that $a\equiv b$ mod n implies that $ac \equiv b(c+n)$ mod n.
My proof attempt: 
If $a \equiv b$ mod n, then $n|(b-a)$ which implies that $(b-a) = nx$ for some $x \in \mathbb{Z}$. 
Which implies that $ac \equiv bc$ mod n. Because $n|(bc-ac) \Rightarrow n|(c(b-a)) \Rightarrow n|(n(cx))$.
So now, $ac = b(c+n)$ mod n can be written as $n|(bc+bn-ac)$ which can be reduced to $n|(c(b-a) + bn) \Rightarrow n| (n(cx)+nb) \Rightarrow n| n(cx + b)$. 
Anyways, I got marked off for this proof and I'm not positive why. So any input would be valuable.
 A: I don't think you've written anything incorrect but I also don't see how you proved the result. As a side note, statements like $n|(n(cx))$ are meaningless since $n$ divides any multiple of itself. It isn't wrong, but you will only waste space.
Try starting with writing or what you are trying to prove. You want to show $n|(ac-bc-bn)$. Note that $n|(a-b)$ so $n|(ac-bc)$. Thus $ac-bc-bn=mn+bn=(m+b)n$ for some integer $m$. This gives the result.
A: I think it is because your written implication flow is reversed. Your flow should be from given information (like $a\equiv b\pmod n$) or from true statement (like $n\mid ncx$). So I might write that in this way:
$$\begin{align*}
&&a&\equiv b\pmod n\\
\implies&& n&\mid (b-a)\\
\implies&& n&\mid c(b-a)\\
\implies&& n&\mid c(b-a) + nb \\
\implies&& n&\mid b(c+n) - ac\\
\implies&& ac&\equiv b(c+n) \pmod n
\end{align*}$$
or
Let $a = b + kn$, $k\in\mathbb Z$, then
$$\begin{align*}
ac &= (b+kn)c\\
&= bc + knc\\
&= bc + bn + knc - bn\\
&= b(c+n) + n(kc-b)\\
\implies ac &\equiv b(c+n) \pmod n
\end{align*}$$
