Set of Integer Combinations equals Set of Multiples of GCD

Theorem
The set of all integer combinations of $a$ and $b$ is precisely the set of all integer multiples of the GCD of $a$ and $b$:


 * $\gcd \left\{{a, b}\right\} \mathop \backslash c \iff \exists x, y \in \Z: c = x a + y b$

Necessary Condition
Let $d = \gcd \left\{{a, b}\right\}$.

Then $d \backslash c \implies \exists m \in \Z: c = m d$.

So:

Thus $\gcd \left\{{a, b}\right\} \backslash c \implies \exists x, y \in \Z: c = x a + y b$.

Sufficient Condition
Suppose $\exists x, y \in \Z: c = x a + y b$.

From Common Divisor Divides Integer Combination, we have $\gcd \left\{{a, b}\right\} \backslash \left({x a + y b}\right)$.

It follows directly that $\gcd \left\{{a, b}\right\} \backslash c$ and the proof is finished.