Solution of Linear Congruence/Existence
Theorem
Let $a x \equiv b \pmod n$ be a linear congruence.
$a x \equiv b \pmod n$ has at least one solution if and only if:
- $\gcd \set {a, n} \divides b$
that is, if and only if $\gcd \set {a, n}$ is a divisor of $b$.
Proof
Consider the linear congruence $a x \equiv b \pmod n$.
Suppose $\exists x_0 \in \Z: a x_0 \equiv b \pmod n$.
Then $\exists y_0 \in Z: a x_0 - b = n y_0$ by definition of congruence.
Thus $x = x_0, y = y_0$ is a solution to the linear Diophantine equation $a x - n y = b$.
On the other hand, if $x = x_0, y = y_0$ is a solution to the linear Diophantine equation $a x - n y = b$, then it follows that $a x \equiv b \pmod n$.
Hence:
- 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 $a x - n y = b$.
From Solution of Linear Diophantine Equation:
The linear Diophantine equation $a x - n y = b$ has at least one solution if and only if:
- $\gcd \set {a, n} \divides b$
Hence the result.
$\blacksquare$