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I want to show that if $n_1, n_2, \ldots, n_k$ are integers such that $\gcd(n_1, n_k)=\cdots=\gcd(n_{k-1}, n_k)=1$, then $\gcd(n_1n_2\cdots n_{k-1}, n_k)=1$. I want to preferably use Bezout's Lemma, but I need help refining my argument. Here it is:

From Bezout we know that for $i=1, \ldots, k-1$ there exist integers $a_i$ and $b_i$ such that $a_in_i+b_in_k=1$. Then multiplying these equations together gets us $$(a_1n_1+b_1n_k)\cdots (a_{k-1}n_{k-1}+b_{k-1}n_k)=1.$$ Now, I know that this product will be in the form $$a_1a_2\cdots a_{k-1}(n_1n_2\cdots n_{k-1})+Bn_k=1,$$ since every other term in this large product has $n_k$ as a factor.

Is there any way for me to make this explanation more formal? I feel it is a little hand-wavy. Please don't suggest to me alternate methods of proof; I am compiling notes and I know there is a much easier proof using Euclid's lemma, but Euclid's lemma is introduced after this.

Arturo Magidin
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    Isn't this easier than that? If $d$ divides all of the $n_i$ then $d$ divides $n_i$ and $n_k$ for any $i$ so $d$ must be $1$. – John Douma Aug 04 '24 at 00:07
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    You can prove it explicitly for $k=3$ and then use induction, replacing $a_1$ and $a_2$ with $a_1a_2$, using the base to show the new list still satisfies the desired property. – Arturo Magidin Aug 04 '24 at 00:16
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    Just use induction. If $\gcd(a,c)=\gcd(b,c)=1$ then $\gcd(ab,c)=1$ is easy to prove. And via induction you won't not to prove anything more. – fleablood Aug 04 '24 at 00:16
  • That's Euclid's Lemma, which is applicable for prime divisors $p$, not general $d$. However, I am not looking for a proof using that, since that is introduced afterward. – A R Aug 04 '24 at 00:17
  • It's not Euclid's Lemma. You can prove it for $k=3$ using your method, explicitly writing everything out and associating. – Arturo Magidin Aug 04 '24 at 01:09
  • The Bezout-based proof essentially inlines a Bezout based proof of EL = Euclid's Lemma (cf. Remark here), so it is better to prove EL first then invoke it by name, rather than repeat its proof in this instance.' – Bill Dubuque Aug 04 '24 at 01:23
  • Re: more formal: $!\bmod n!:\ \color{#c00}{a_i n_i\equiv 1},\Rightarrow, \prod_i a_in_i\equiv 1^k\equiv 1,$ by the Congruence Product Rule (inductive extension). More conceptually: all $,a_1,$ are $\rm\color{#c00}{invertible}$ thus so too is their product (cf. here in 2nd dupe). $\ \ $ – Bill Dubuque Aug 04 '24 at 01:40
  • Or use polynomials: $,f(x) := \prod (a_i+b_ix) = \overbrace{\prod a_i}^{f(0)} + x, g(x),,$ i.e. $,x\mid f(x)-f(0),,$ by the Factor (or Remainder) Theorem, by eval at $,x=0,,$ i.e. work $!\bmod x,$ (the polynomial analog of the modular proof in my prior comment). $\ \ $ – Bill Dubuque Aug 04 '24 at 02:22

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