Integer Combination of Coprime Integers/General Result

Theorem
Let $a_1, a_2, \ldots, a_n$ be integers.

Then $\gcd \left\{ {a_1, a_2, \ldots, a_n}\right\} = 1$ there exists an integer combination of them equal to $1$:
 * $\exists m_1, m_2, \ldots, m_n \in \Z: \displaystyle \sum_{k \mathop = 1}^n m_k a_k = 1$

Proof
First let $\exists m_1, m_2, \ldots, m_n \in \Z: \displaystyle \sum_{k \mathop = 1}^n m_k a_k = 1$.

Let $\gcd \left\{ {a_1, a_2, \ldots, a_n}\right\} = d$.

Then $\displaystyle \sum_{k \mathop = 1}^n m_k a_k$ has $d$ as a divisor.

That means $d$ is a divisor of $1$.

Thus $\gcd \left\{ {a_1, a_2, \ldots, a_n}\right\} = 1$.

It remains to be shown that if $\gcd \left\{ {a_1, a_2, \ldots, a_n}\right\} = 1$, then $\exists m_1, m_2, \ldots, m_n \in \Z: \displaystyle \sum_{k \mathop = 1}^n m_k a_k = 1$.

The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $P \left({n}\right)$ be the proposition:
 * $\gcd \left\{ {a_1, a_2, \ldots, a_n}\right\} = 1 \iff \exists m_1, m_2, \ldots, m_n \in \Z: \displaystyle \sum_{k \mathop = 1}^n m_k a_k = 1$

Basis for the Induction
$P \left({2}\right)$ is the case:
 * $\gcd \left\{ {a_1, a_2}\right\} = 1 \iff \exists m_1, m_2 \in \Z: m_1 a_1 + m_2 a_2 = 1$

This is demonstrated in Integer Combination of Coprime Integers.

Thus $P \left({2}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\gcd \left\{ {a_1, a_2, \ldots, a_r}\right\} = 1 \iff \exists m_1, m_2, \ldots, m_r \in \Z: \displaystyle \sum_{k \mathop = 1}^r m_k a_k = 1$

from which it is to be shown that:
 * $\gcd \left\{ {a_1, a_2, \ldots, a_r, a_{r + 1} }\right\} = 1 \iff \exists m_1, m_2, \ldots, m_r, m_{r + 1} \in \Z: \displaystyle \sum_{k \mathop = 1}^{r + 1} m_k a_k = 1$

Induction Step
This is the induction step:

Let $\gcd \left\{ {a_1, a_2, \ldots, a_r, a_{r + 1} }\right\} = 1$.

Let $\gcd \left\{ {a_1, a_2, \ldots, a_r}\right\} = d$.

Dividing through by $d$, we have:
 * $\gcd \left\{ {\dfrac {a_1} d, \dfrac {a_2} d, \ldots, \dfrac {a_d} d}\right\} = 1$

By the induction hypothesis:
 * $\exists m_1, m_2, \ldots, m_r \in \Z: \displaystyle \sum_{k \mathop = 1}^r m_k a_k = 1$

whence multiplying through by $d$:
 * $\exists m'_1, m'_2, \ldots, m'_r \in \Z: \displaystyle \sum_{k \mathop = 1}^r m'_k a_k = d$

where $m'_k = m_k d$ throughout.

Then by hypothesis:
 * $\gcd \left\{ {d, a_{r + 1} }\right\} = 1$

From the basis for the induction:
 * $\exists p, q \in \Z: p d + 1 a_{r + 1} = 1$

We take:
 * $m_1 = m'_1 p, m_2 = m'_2 p, \ldots, m_r = m'_r p$

and:
 * $m_{r + 1} = q$

and thus we have:


 * $\gcd \left\{ {a_1, a_2, \ldots, a_r, a_{r + 1} }\right\} = 1 \iff \exists m_1, m_2, \ldots, m_r, m_{r + 1} \in \Z: \displaystyle \sum_{k \mathop = 1}^{r + 1} m_k a_k = 1$