Integer Combination of Coprime Integers
Theorem
Let $a, b \in \Z$ be integers, not both zero.
Then:
- $a$ and $b$ are coprime
- there exists an integer combination of them equal to $1$:
- $\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$
In such an integer combination $m a + n b = 1$, the integers $m$ and $n$ are also coprime.
General Result
Let $a_1, a_2, \ldots, a_n$ be integers.
Then $\gcd \set {a_1, a_2, \ldots, a_n} = 1$ if and only if there exists an integer combination of them equal to $1$:
- $\exists m_1, m_2, \ldots, m_n \in \Z: \ds \sum_{k \mathop = 1}^n m_k a_k = 1$
Proof
The proof can conveniently be divided into two parts:
Sufficient Condition
Let $a, b \in \Z$ be integers, not both zero.
Let $a$ and $b$ be coprime to each other.
Then there exists an integer combination of them equal to $1$:
- $\forall a, b \in \Z: a \perp b \implies \exists m, n \in \Z: m a + n b = 1$
Necessary Condition
Let $a, b \in \Z$ be integers, not both zero.
Let $a$ and $b$ be such that there exists an integer combination of them equal to $1$.
Then $a$ and $b$ are coprime:
- $\forall a, b \in \Z: \exists m, n \in \Z: m a + n b = 1 \implies a \perp b$
In such an integer combination $m a + n b = 1$, the integers $m$ and $n$ are also coprime.
Also known as
This result is sometimes known as Bézout's Identity, although that result is usually used for the more general result for not necessarily coprime integers.
Some sources refer to this result as the Euclidean Algorithm, but the latter as generally understood is the procedure that can be used to establish the values of $m$ and $n$, and for any pair of integers, not necessarily coprime.
Sources
- 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 $2 \text{-} 4$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Lemma $1$