Solution of Linear Congruence/Unique iff Coprime to Modulus

Theorem

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

$a x \equiv b \pmod n$ has a unique solution if and only if $\gcd \set {a, n} = 1$.

Proof

Sufficient Condition

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Necessary Condition

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:

$\gcd \set {a, n} = 1$

Let $x = x_0, y = y_0$ be 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 $a x \equiv b \pmod n$.

$\blacksquare$

Sources

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• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 2.3$: Congruences: Theorem $4$