If $(b,n)=1$, $n\mid(ad-bc)$ and $n\mid(a-b)$ then $n\mid (c-d)$.

Question

Pretty straightforward. I am stuck on a problem, and would love it if someone could give me a small hint or nudge in the right direction.

The problem is $(b,n)=1$ and $n\mid(ad-bc)$ and $n\mid(a-b)$ implies that $n\mid(c-d)$.

What I have so far is that because $b$ and $n$ are relatively prime, $n\nmid b$, and since $n\mid (a-b)$, $n\nmid a$, because if not $b$ would be divisble by $n$. However, this is where I get stuck.

Any suggestions?

Answer 1

Hint: $ad-bc=(a-b)d-(c-d)b$.

Answer 2

Using rules of moduar-arithmetic we have: $$ad-bc\equiv 0\pmod n$$ but $a\equiv b\pmod n$ therefore replacing $$bd-bc\equiv 0\pmod n$$ from here because $(b,n)=1$, $c\equiv d\pmod n$