Prove that $mc = mb + 8k \implies c=b+8k$ in congruence

Question

If $m$ is an odd number, then $$mc = mb + 8k \implies c=b+8k$$ and where $k\in\mathbb Z$.

I know that $$c=b +8k \implies cq= bq + 8k$$ but I do not know how I can prove the congruence above

Answer 1

Since $gcd(8,m)=1$, we have that $m$ is invertible modulo $8$. Hence $mc\equiv mb\bmod 8$ implies $c\equiv b\bmod 8$. In other words, in the group $G=(\mathbb{Z}/8)^{\times}$, we can multiply the equation $mc=mb$ with $m^{-1}$, so that $c=b$ follows in $G$.