GCD from Congruence Modulo m

Theorem
Let $$a, b \in \Z, m \in \N$$. 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:
 * $$a \equiv b \left({\bmod\, m}\right) \implies \gcd \left\{{a, m}\right\} = \gcd \left\{{b, m}\right\}$$.

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

Thus:
 * $$a = b + k m$$

and the result follows directly from GCD with Remainder.