I need to proof the following statement:
Let $a, b, n \in \Bbb{Z}$ with $ n≥ 2, gcd(a,n)=1$. Proof that if $s_{1},s_{2}$ are solutions to $ax\equiv b \pmod{n}$, then $s_{1}\equiv s_{2} \pmod{n}$.
I don't know where to start my proof. I do know that if you get any solution, then by adding the modulo you get equivalent solutions. Then, there are n possible solutions. But I don't think my argument is correct.
Note that if $s_1,s_2$ are solutions, $as_1\equiv as_2\equiv b\pmod n$, so $as_1\equiv as_2\pmod n$.
Hence, $n\mid a(s_2-s_1)$. But, since $\gcd(a,n)=1$,
$$n\mid s_2-s_1$$ so $s_1\equiv s_2\pmod n$.
If $a\in\Bbb Z$ is such ${\rm gcd}(a,n)=1$ then $a$ has a multiplicative inverse modulo $n$, i.e. there exists $a^*\in\Bbb Z$ such that $aa^*\equiv1\bmod n$.
Thus, if you have a congruence $$ ab_1\equiv ab_2\bmod n $$ simply multiplying both hands by the multiplicative inverse $a^*$ gives $$ b_1\equiv b_2\bmod n. $$ This is false if ${\rm gcd}(a,n)\neq1$, e.g. $2\cdot1\equiv2\cdot4\bmod6$ but $1\not\equiv4\bmod6$.
