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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!

snulty
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Huanzhen Zhang
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  • Apply $au+bv = \gcd(a,b)$ (Bezout identity) $r$ times – reuns Sep 26 '17 at 16:59
  • Proving Bezout's identity (which everyone should do before they use it) is a bit irritating. But it's a proof by ruduction. If $a < b$ then $b= wa + r$ where $r < a$ and so if $r > 0$ then $\gcd(b,a)|r$ and we can repeat indefinitely. If time we repeat, $a= zr + r_2$ ,we will get smaller and smaller remainders. As there are only finitely many remainders will eventually get $r_{n-1}= v_n*r_n + 0; r_n=\gcd(a,b)$. Then the values $u,v$ so that $au+bv = \gcd(a,b)$ can be calculate via the series of attempts. You can think of it as a sort of proof by induction. – fleablood Sep 26 '17 at 17:17
  • @fleablood It's not too irritating if we do it conceptually by importing elementary facets of ideas from group / ideal theory. Namely it is easy to prove that a nonempty set $S$ of integers $\rm\color{#c00}{closed\ under\ subtraction}$ has the form $,S = d\Bbb Z,$ where $d$ is the least nonnegative element in $,S,$ or, equivalently $,d = \gcd S.,$ This viewpoint is presented in detail in my answer in the linked dupe. It's essence is clarified when one studies (cyclic) groups and (principal) ideals in abstract algebra. – Bill Dubuque Sep 26 '17 at 17:27
  • "irritating" is subjective. And if one hasn't studied group/ideal theory... It's not really "irritating" per say, but ... well, I always found proofs where we assume least elements exist and reduction ad finitim to be ... well, "irratating" in the sense that as a student it feels uncertain that we are allowed to do that. – fleablood Sep 26 '17 at 17:38

1 Answers1

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HINT.- Choose $a_1,a_2\in \mathbb Z$ such that $a_1m_1=k$ and $a_2m_2=l-k$ so you get $a_1m_1+a_2m_2=l$. Iterate the procedure keeping in mind at the end that if $(m,n)=1$ then $\mathbb Z=m\mathbb Z+n\mathbb Z$ by Bézout's identity.

Piquito
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