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I try to prove that statement using only Bachet-Bézout theorem (I know that it's not the best technique). So I get $k$ useful equations with $n_1$ then $(k-1)$ useful equations with $n_2$ ... then $1$ useful equation with $n_{k-1}$. I multiply all these equations to obtain $1$ for one side. For the other side I'm lost (there are too many terms) but I want to make appear the form $n_1 L_1+...+n_k L_k$.

Supposing the existence of all the integers we need :

$\underbrace{(a_1n_1+a_2n_2)(a_1n_1+a_3n_3)...(a_1n_1+a_kn_k)}_{\textit{k equations}} \underbrace{(b_2n_2+b_3n_3)...(b_2n_2+b_kn_k)}_{\textit{(k-1) equations}}...\underbrace{(\mu_{k-1} n_{k_1}+\mu_{k} n_k)}_{\textit{1 equation}}=1$

Maybe we can reduce the number of useful equations or start an induction to identify a better form for the product.

Thanks in advance !

Maman
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  • Not true without the additional assumption that $k\ge2.$ – bof May 02 '17 at 01:38
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    But this is just to say that if each are relatively prime then they all together are relatively prime which is to say if each share no prime factors then they all together share no prime factors. This is obvious. – Jacob Wakem May 02 '17 at 01:52
  • It will be easier to use other methods, but I posted one based on Bachet-Bezout below, since the problem is asking for an answer based on it. :) – Lazy Lee May 02 '17 at 01:59
  • @Alephnull of course the fast natural answer is obvious but using just the equations of Bachet-Bézout it's difficult to see :( – Maman May 02 '17 at 14:05

3 Answers3

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You are making this way too complicated.

There exists integers $x_1,x_2$ such that $n_1x_1+n_2x_2=1$. Letting $x_3=x_4=\cdots=x_k=0$. Then $\sum_{i=1}^{k} n_ix_i=1$ so the $x_i$ must be relatively prime.

Basically, $\gcd(x_1,x_2,\dots,x_k)\mid\gcd(x_1,x_2)$.

Thomas Andrews
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  • Thank you but I'm just a big fan of Bézout. So I wanted to write it formally with every equations but it's difficult to identify the form needed ! – Maman May 02 '17 at 14:04
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    But there is no reason to need all the variables, and this uses Bezout, too. Basically, the real theorem is if $\gcd(x_i,x_j)=1$ for some $i,j$, then $\gcd(x_1,\dots,x_k)=1$. So your theorem, as stated, is overkill. – Thomas Andrews May 02 '17 at 14:07
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    (FWIW, I like Bézout identity proofs, too. Here's an old question of mine: https://math.stackexchange.com/questions/463052/b%C3%A9zouts-identity-proof-that-if-a-b-c-1-then-axbxycz-1-has-integer-solut – Thomas Andrews May 02 '17 at 14:10
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The condition $\forall i \ne j \in \{1,2 \ldots, k \}, \gcd (n_i,n_j)=1$ is too strong. Only two $i \ne j$ so $\gcd(n_i,n_j)=1$ is enough to imply $\gcd (n_1, \ldots, n_k)=1$.

Indeed, let $d=\gcd (n_1, \ldots, n_k)$ then $d \mid n_i, d \mid n_j$ so $d \mid \gcd (n_i,n_j)=1$ so $d=1$. Thus, $\gcd (n_1, \ldots, n_k)=1$.

Tengu
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According to the Bachet-Bezout Lemma, the gcd of two integers $gcd(a,b)=d$ is the smallest positive integer that can be written as $ax+by=d$ for some integers $x,y$. We split the discussion into two parts:

If $k$ is even: Then there exists integers $\{x_i\}$ such that $$n_1x_1+n_2x_2=1$$$$n_3x_3+n_4x_4=1$$$$...$$$$n_{k-1}x_{k-1}+n_kx_k=1$$ Just multiply the first equation by $\frac{k}{2}$ and subtract all the rest to get $$\frac{k}{2}\cdot n_1+\frac{k}{2}\cdot n_2+\sum_{i=3}^k (-x_i)\cdot n_i=1 \implies gcd(n_1,n_2...,n_k)=1$$

If $k$ is odd: Arrange $\{n_i\}$ such that $n_1\neq 1$ (this is possible, otherwise all $n_i=1$ and there's nothing to prove). We prove that there exists $\{x_i\}$ such that $$n_1x_1+n_2x_2 + n_3x_3=1$$$$n_4x_4+n_5x_5=1$$$$...$$$$n_{k-1}x_{k-1}+n_kx_k=1$$ The second to last equations exist for some $x$ by Bachet-Bezout, while the first can be achieved with $(x'_2,x'_3)$ such that $$n_2x'_2+n_3x'_3=1$$ and setting $(x_1,x_2,x_3)=(1,(-n_1+1)x'_2,(-n_1+1)x'_3)$. Hence, similar to above, we multiply the first equation by $\frac{k-1}{2}$ then subtract the rest to get $$\frac{k-1}{2}\cdot x_1\cdot n_1+\frac{k-1}{2}\cdot x_2\cdot n_2+\frac{k-1}{2}\cdot x_3\cdot n_3+\sum_{i=r}^k (-x_i)\cdot n_i=1 \implies gcd(n_1,n_2...,n_k)=1$$

Lazy Lee
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