Linear Congruence

$ax \equiv b \pmod{m}$

If we find a multiplicative inverse $\bar{a}$ to $a \pmod{m}$, i.e. $\bar{a}\cdot a \equiv 1 \pmod{m}$, then $ax \equiv b \pmod{m}$ becomes $x \equiv \bar{a}\cdot b \pmod m$

Theorem

Let $a,m \in \Z$ with $m\geq 2$. If $gcd(a,m) = 1$, then the inverse of $a \pmod{m}$ exists and is unique.