To prove that if $ab\equiv ac\bmod n$ and $(a,n)=1$ then $b\equiv c\bmod n$ To prove that if $ab \equiv ac\bmod n$ and $(a,n)=1$ then $b\equiv c \bmod n$
I write as
$1=ar+ns$
$ac=ab+nq$
I have to prove $c=b+nr$
How do I manipulate equations to reach conclusion
Thanks
 A: I think your notation is a little wrong. I think what you're asking is:

If $$ab \equiv ac \mod{n}$$
  and $(a,n)=1$, then $b \equiv c \mod{n}$

As you suggest, write $a(b-c)=nq$ and $1=ar+ns$.  
Multiplying through by $r$, we have $ar(b-c)=nqr$, so $(1-ns)(b-c)=nqr$.  
It follows that $b-c=n(qr+s(b-c))$, so $b-c$ is divisible by $n$.
A: $$ac=ab+nq$$
means $a(c-b)=nq$, in other words $n$ divides $a(c-b)$. But
$$1=ar+ns$$
so either $n=\pm1$ or $n$ does not divide $a$.
If $n=\pm1$ then as $b=c+(b-c)(1)=c+(c-b)(-1)$, we have $b\equiv c\pmod n$.
If $n$ does not divide $a$, but $n$ divides $a(c-b)$, then $n$ must divide $c-b$, i.e. again $b\equiv c\pmod n$.
A: As $\gcd(a,n)=1$, we have a Bézout's relation: $\; ua+vn=1$, so $ua\equiv 1\mod n$. 
Now $\quad ab\equiv ac\implies uab\equiv uac\iff 1\cdot b\equiv 1\cdot c\mod n$.
A: n divides ab-ac or a(b-c).
n cannot divide a.
That implies n divides (b-c)
Edit: since $ n | a *(b-c)$ , prime factorisation of $  a * (b-c)  $ should contain or include prime factorisation of $ n$.
Since $a, n$ are co-prime, $a$ does not contain any prime factors of $n$, thus $ b-c $ should contain all the prime factorisation of $n$, meaning $ n | b-c $
