Showing $na \equiv 1 \pmod m$ and $n'a \equiv 1 \pmod m \implies n \equiv n' \pmod m$ for $(a,m) = 1$ We have $(a,m) = 1$ iff exists integer $n$ such that $na \equiv 1 \pmod m$. Prove  $na \equiv 1 \pmod m$ and $n'a \equiv 1 \pmod m$ implies $n \equiv n' \pmod m$.
For this question, I have so far
$na \equiv 1$ and
$n'a \equiv 1$
So, I have the equations 
$na -1 = mk$ and
$n'a - 1 = mk'$
Subtracting, I end up with
$a(n-n') = m(k-k')$ dropping $k$ because its a multiple. Then I just end up with 
$na \equiv n'a \pmod m$ which seems circular. Any help?
 A: So, $m\mid a(n-n')\implies m\mid (n-n')$ as $(a,m)=1$
A: Uniqueness of inverses $\rm\ n'\equiv n\ $ follows purely from associativity: 
$$\begin{eqnarray}\rm\ &\rm an&\!\!\equiv 1\equiv &\rm an'&\\ \rm \Rightarrow\ \  n' \equiv n'\!\!\!&(\rm an)& \  \equiv &\rm \!\!\!\!\!(n'a)&\rm \!n \equiv n\end{eqnarray}\qquad$$
Remark $\ $ Equivalently, canceling $\rm\:a\:$ from $\rm\:an \equiv 1\equiv an'\:$ yields $\rm\:n\equiv n',\:$ and canceling $\rm\:a\:$ can by achieved by left-multiplying through by any inverse of $\rm\:a.\:$ But we already know two inverses of $\rm\:a,\:$ both $\rm\:n,\,n',\:$ so multiplying by either will cancel $\rm\:a,\:$ yielding the sought equality of inverses. 
It is important to realize that to cancel $\rm\,a\,$ it is not necessary to use Bezout to find another inverse, or to use Euclid's Lemma, or any other special properties of $\rm\:\Bbb Z/m.\:$ Rather, the above uniqueness proof works universally in any monoid, i.e. any set with an associative product and $1\ (= $ neutral element), assuming only that the inverses are two sided, i.e. both $\rm\:n'a = 1 = an'\ $ (note that the above proof implicitly used $\rm\:an' = 1\:\Rightarrow\:n'a = 1,\:$ which is true in our commutative case).
A: Since $\gcd(a,m)=1$, by Bezout's Identity, we have $x$ and $y$ so that
$$
ax+my=1
$$
By the hypotheses, we have
$$
a(n-n')\equiv0\pmod{m}
$$
Multiply both sides by $x$ (the inverse of $a\bmod{m}$):
$$
\begin{align}
0
&\equiv ax(n-n')\\
&\equiv(1-my)(n-n')\\
&\equiv n-n'\pmod{m}
\end{align}
$$
A: $$n\equiv n\cdot n'a =n'na\equiv n'\cdot 1=n' \pmod{m}.$$
All you have to use is that if $x\equiv x'$ then $xy\equiv x'y$.
A: I have noticed you are asking a number of similar questions regarding congruence. Obviously you are getting great answers from some participants here who really know their stuff. 
I would encourage you to try to pull it all together so you can feel validly confident that you can handle these types of questions.
As I mentioned in an answer to one of your similar questions: take a look at "Ireland and Rosen" from the bottom of page 30 to the middle of page 33. It's basically three pages. It is extremely readable and will prepare you for the questions you are asking, and the more general question of solvability of $ax \equiv \pmod b$.
Just take it slowly and write out their presentation, trying to understand what's happening. If you get stuck, you know you can always ask here for assistance in working through it.
Once you get on top of it, you can look back and see how to do these questions in your own hand, and then appreciate all the interesting solutions you've received here.
I think it will be worth your effort and you can stand more securely in your math endeavors. Good luck.
