Exercise 1.14

Proof. Let $a_1,a_2,\dots ,a_r\in \mathbb{Z}$.

Note $\gcd (a_1,a_2,\dots ,a_r)=a_1\wedge a_2\wedge \cdots \wedge a_r$. The following result generalizes Ex. 6 :

Let $d=a_1\wedge a_2\wedge \cdots \wedge a_r$.

$\text{∙}$ If $a_1{x}_1+a_2{x}_2+\cdots +a_r{x}_r=c$, as $d\mid a_1,\dots ,d\mid a_r$, $d\mid a_1{x}_1+a_2{x}_2+\cdots +a_r{x}_r=c$.

$\text{∙}$ Conversely, if $d\mid c$, then $c=d{c}^{\prime },c^{\prime }\in \mathbb{Z}$.

As $\mathit{d\mathbb{Z}}=a_1\mathbb{Z}+a_2\mathbb{Z}+\cdots +a_r\mathbb{Z}$, so $d=a_1{m}_1+a_2{m}_2+\cdots +a_r{m}_r,m_1,m_2,\dots ,m_r\in \mathbb{Z}$. $c=d{c}^{\prime }=a_1(m_1{c}^{\prime })+\cdots a_r(m_r{c}^{\prime })=a_1{x}_1+\cdots +a_r{x}_r$, where $x_i=m_i{c}^{\prime },i=1,2,\dots ,r$. □