Congruence modulo $n$ problem

Question

Let $ab\equiv cd\pmod{n}, a\equiv c\pmod{n}, (a,c)=1$. Show that $b\equiv d\pmod{n}$. I have to somehow arrive at $n\mid (d-b)$.
I have $n\mid (cd-ab), n\mid (c-a)$ at my disposal and perhaps the fact that $a,c$ have no common factor. $$n\mid (cd-ab)\land n\mid (c-a)\Longrightarrow n\mid a(d-b)$$ which is almost what I want. I just don't see how $(a,c)=1$ comes into play here. $(a,n)=1$ would make more sense (Euclid's lemma) or am I missing something?

Answer 1

As you said the deduction we need is $(a,n)=1$. What we know is $a\equiv c\pmod n$ (which implies $a=c+nk$ for some $k$) and $(a,c)=1$. Now let $d=(a,n)$. Then $d\mid a-nk$, so $d\mid c$. Since $(a,c)=1$, we get $d=1$. We can also get $(c,n)=1$, thus $a$ and $c$ are invertible modulo $n$.

Answer 2

If $\color{blue}{n|a}$, then $\color{red}{a\equiv c \pmod n \Rightarrow n|c}$.

So $n$ is a common factor of $a$ and $c$.

But it is given that $\color{brown}{gcd(a,c)=1}$.

Hence, contradiction $\Rightarrow \color{violet}{n \not | \, \,a}$.

Hope this helps you.