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 $b$:


 * $\gcd \set {a, b} \divides c \iff \exists x, y \in \Z: c = x a + y b$

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

Then:
 * $d \divides c \implies \exists m \in \Z: c = m d$

So:

Thus:
 * $\gcd \set {a, b} \divides c \implies \exists x, y \in \Z: c = x a + y b$

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

From Common Divisor Divides Integer Combination, we have:
 * $\gcd \set {a, b} \divides \paren {x a + y b}$

It follows directly that $\gcd \set {a, b} \divides c$ and the proof is finished.