# 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 $b$:

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

## Proof

### Necessary Condition

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

Then:

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

So:

 $\ds \exists p, q \in \Z: \,$ $\ds d$ $=$ $\ds p a + q b$ Bézout's Identity $\ds \leadsto \ \$ $\ds c$ $=$ $\ds m d$ $\ds$ $=$ $\ds m p a + m q b$ $\ds$ $=$ $\ds \paren {m p} a + \paren {m q} b$ $\ds \leadsto \ \$ $\ds \exists x, y \in \Z: \,$ $\ds c$ $=$ $\ds x a + y b$

Thus:

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

$\Box$

### 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.

$\blacksquare$