Integer Combinations Multiples of GCD

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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 $b$:

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


Proof

Sufficient Condition

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

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


So:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \exists p, q \in \Z: d\) \(=\) \(\displaystyle p a + q b\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Bézout's Lemma          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle c\) \(=\) \(\displaystyle m d\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle m p a + m q b\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left({m p}\right) a + \left({m q}\right) b\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \exists x, y \in \Z: c\) \(=\) \(\displaystyle x a + y b\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    


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

$\Box$


Necessary 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.

$\blacksquare$


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