For positive integers $a_{1}, a_{2}\cdots, a_{k},$ define $\gcd(a_{1}, a_{2}\cdots, a_{k})$ to be the largest positive integer $d$ such that $d$ divides every $a_{i}$ and any positive integer $c$ that divides every $a_{i}$ also has to divide $d.$ Is it true that there are integers $m_{i},$ not necessarily positive, such that $d= \sum_{i= 1}^{k}m_{i}a_{i},$ right ?
How can i prove this ? I was thinking maybe one could use mathematical induction but i'm not quite sure.
First Prove that if $\gcd(a_1, ...., a_k) = d_k$ then $\gcd(a_1,a_2,.....,a_k, a_{k+1}) = \gcd(d_k, a_{k+1})=d_{k+1}$.
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Then we can prove our induction step: that if there exist $m_i$ so that $\sum_{i=1}^k m_i a_i = d_k$, and by Bezout, there are integers $N, P$ so that $Nd_k + Pa_{k+1} = d_{k+1}$--- so if we define $b_i = a_i*N$ for $i = 1...k$ and $b_{k+1} = P$ then we have $\sum_{i=1}^{k+1} b_i a_i = d_{k+1}$.