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.
$\blacksquare$
Sources
- George E. Andrews: Number Theory (1971): $\S 2.2$: Corollary $2.2$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $2.4$