I have to proof:
If $(a, m) = 1,$ then $ax\equiv{b}\pmod{m}$ has ONE solution in $\mathbb{Z}/m\mathbb{Z}$.
I did the following:
We have: $ax\equiv{b}\pmod{m}$. It follows: $b = ax + nm$, with $n$ being any integer. If $(a, m) = 1,$ it follows that $1|(ax + nm)$. But I got stuck and I don't know how to conclude that there's only ONE solution or if I even went on the right path.
Thanks for reading and helping.
EDIT:
After reading your comments I got here:
First, since we have $(a,m)=1$ then we have the linear expression: $ax + mn = 1$ where $mn$ can be interpreted as the residue and it is unique given the linear combination. Then, we can state: $$ax \equiv 1 \pmod{mn}$$
The expression means that $mn |(1-ax)$, and since $mn$ is unique, it follows that it has only one solution.
Thanks again for reading and correcting me.