GCD from Congruence Modulo m

Theorem
Let $$a, b \in \mathbb{Z}, m \in \mathbb{N}$$. Let $$a$$ be congruent to $$b$$ modulo $$m$$.

Then the GCD of $$a$$ and $$m$$ is equal to the GCD of $$b$$ and $$m$$.

That is: $$a \equiv b \left({\bmod\, m}\right) \Longrightarrow \gcd \left\{{a, m}\right\} = \gcd \left\{{b, m}\right\}$$.

Proof
We have $$a \equiv b \left({\bmod\, m}\right) \Longrightarrow \exists k \in \mathbb{Z}: a = b + k m$$.

Thus $$a = b + k m$$, and the result follows directly from GCD with Remainder.