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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.

Bill Dubuque
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Marcus F
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  • You must mean $a_i$ not $d_i$, right? –  Mar 08 '21 at 02:55
  • Yes i corrected it – Marcus F Mar 08 '21 at 02:59
  • This is a generalization of the euclidean algorithm. See here https://math.stackexchange.com/a/1613742/825910 –  Mar 08 '21 at 03:02
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    For $k=2$ this is called Bezout's identity: for any two integers $a,b$ there exist integers $x,y$ such that $ax + by = \text{gcd}(a,b)$. There are a number of proofs of this statement (a quick google search should do). After this notice that gcd$(a,b,c)$ = gcd$($gcd$(a,b)$,$c)$ so you can proceed by induction. – Tanny Sieben Mar 08 '21 at 03:03
  • @John what method of proof is provided in the answer for the post you linked? – Marcus F Mar 08 '21 at 03:04
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    I'm not sure it has its own name, I'd call it just direct proof. Also @TannySieben you are correct, I managed to forget the name Bezout after seeing it just the other day haha –  Mar 08 '21 at 03:07

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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 $\sum_{i=1}^{k+1} b_i a_i = d_{k+1}$.

fleablood
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