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.

$\blacksquare$

Sources

• 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-2}$ Divisibility: Corollary $\text {2-2}$
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Some Properties of $\Z$: Exercise $2.4$
• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 3$: Corollary
• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $13 \ \text {(a)}$
• 1982: Martin Davis: Computability and Unsolvability (2nd ed.) ... (previous) ... (next): Appendix $1$: Some Results from the Elementary Theory of Numbers: Lemma $2$