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I was going through this paper. On Page 6, it's written

Corollary 2.3. For any positive integers $a_1$, $a_2$, $\ldots$, $a_n$, there exist integers $x_1$, $x_2$, $\ldots$, $x_n$, such that $$a_1x_1+a_2x_2+\cdots+a_nx_n = gcd(a_1, a_2, \ldots, a_n)$$

So, I was wondering if it is true for all $n≥0$ i.e. is it universal, or are there any constraints?

Bill Dubuque
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Dhananjay Dahiya 2k19ae020
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  • https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity#For_three_or_more_integers – JMoravitz Dec 08 '20 at 15:26
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    That statement is clear. No constraints. – TonyK Dec 08 '20 at 15:28
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    This is just Bézout's identity, isn't it? – Divide1918 Dec 08 '20 at 15:29
  • Yes, true for all $n\geq 1$. I don't know what you might mean for the case of $n=0$. As for constraints, it can be used in any euclidean domain. – JMoravitz Dec 08 '20 at 15:29
  • @JMoravitz if i have xi and ai in range [0, p) can i find all the possible permutations of xi which makes this equation a1.x1+a2.x2+⋯+an.xn divisible by p. p is some prime no. and ai is known – Dhananjay Dahiya 2k19ae020 Dec 08 '20 at 15:42
  • The proof shows that it is true for all naturals $n$ and all integers $a_i$. The same proof works for any Euclidean domain, i.e. a domain that enjoys division with smaller remainder, e.g. polynmials over a field. – Bill Dubuque Dec 08 '20 at 17:24

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