Let $m_1, m_2, \ldots , m_r$, $r \geq 2$, be non zero integers which do not have a common prime divisor. Show that there exists $a_1, \ldots , a_r \in \Bbb Z$, such that  $$\sum_{i=1}^r a_i \cdot m_i = 1$$  I am new in this area, any help would be appreciate!
Given $a_1, a_2, a_3, \ldots , a_n$ not all zero, show $\gcd(a_1, a_2, a_3, \ldots , a_n)$ is the least positive integer that can be expressed in the form $a_1x_1+a_2x_2+ \ldots +a_nx_n$. Also deduce $a_1x_1+a_2x_2+ \ldots +a_nx_n = b$ has solutions $\iff \gcd(a_1, a_2, a_3, \ldots , a_n)|b$.   I know that $ax+by=n$ has integer solutions $\iff \gcd(a,b)|n$. I'll need to use induction to show $a_1x_1+a_2x_2+ \ldots +a_nx_n = \gcd(a_1, a_2, a_3, \ldots , a_n)$ has solutions, and then I will be able to say that as $\gcd(a_1, a_2, a_3, \ldots , a_n) | a_1x_1+a_2x_2+ \ldots +a_nx_n $ for any $x_1,x_2, x_3 \ldots x_n$ then $c \leq \gcd(a_1, a_2, a_3, \ldots , a_n)$ and $c \geq \gcd(a_1, a_2, a_3, \ldots , a_n)$, and thus $c=\gcd(a_1, a_2, a_3, \ldots , a_n)$.
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