Solution of Linear Congruence/Unique iff Coprime to Modulus

Theorem
Let $a x \equiv b \pmod n$ be a linear congruence.

If $\gcd \left\{{a, n}\right\} = 1$, then $a x \equiv b \pmod n$ has a unique solution.

Proof
From Solution of Linear Congruence: Existence:
 * the problem of finding all integers satisfying the linear congruence $a x \equiv b \pmod n$

is the same problem as:
 * the problem of finding all the $x$ values in the linear Diophantine equation $a x - n y = b$.

Let:
 * $\gcd \left\{{a, n}\right\} = 1$

Let $x = x_0, y = y_0$ is one solution to the linear Diophantine equation:
 * $a x - n y = b$

From Solution of Linear Diophantine Equation, the general solution is:
 * $\forall k \in \Z: x = x_0 + n k, y = y_0 + a k$

But:
 * $\forall k \in \Z: x_0 + n k \equiv x_0 \pmod n$

Hence $x \equiv x_0 \pmod n$ is the only solution of $a x \equiv b \pmod n$.