TITLE: Suppose $[a],[b],[c]\in\Bbb Z_n$, and are residue classes which satisfy $[a][b]=[1]$ and $[a][c]=[1]$. Prove $[b]=[c]$.
QUESTION [5 upvotes]: I have a partial solution to this problem, but I am now stuck and unsure where to go to. I don't want a solution to the problem, just guidance on where to go, or if what I'm doing is wrong.
Proof: As $[a][b]=[1]$, we have $[ab]=[1]$. Thus $ab\equiv 1\pmod n$, and similarly as  $[a][c]=[1]$, we have $[ac]=[1]$. Hence we obtain $ac\equiv 1 \pmod n$. From this we see that $n\mid ab-1$ and $n\mid ac-1$. By definition of division, we have that $ab-1 =nt$, and $ac-1 =np$ where $t,p\in\Bbb Z$. Thus, $ac-np=1=ab-nt$. Thus, by factoring, we obtain $a(c-b)=n(p-t)$ where $p-t\in\Bbb Z$. Thus we have that $n\mid a(c-b)$.
At this point I'm stuck, and I feel as if the problem is close to being done/nearly done.
Any guidance is appreciated, thank you.

REPLY [2 votes]: Note that
$[a][b] = 1, \; [a][c] = 1 \tag 1$
together yield
$[a]([b] - [c]) = [a][b] - [a][c] = 0; \tag 2$
thus,
$[b] - [c] = 1 \cdot ([b] - [c]) = [a][b]([b] - [c])$
$= [b][a]([b] - [c]) = [b] \cdot 0 = 0, \tag 3$
whence
$[b] = [c]. \tag 4$
One can of course argue directly from (1), which gives
$[a][b] = 1 = [a][c],  \tag 5$
and so
$[b][a][b] = [b][a][c],  \tag 6$
or
$[b] = 1 \cdot [b] = [a][b][b]$
$= [b][a][b] = [b][a][c] = [a][b][c] = 1 \cdot [c] = [c]. \tag 7$