Gratest common divisor problem

Question

Let $a_1, a_2,...,a_n \in\mathbb Z$ not all equal to zero prove that:

$gcd(a_1,a_2,...,a_n)=gcd(a_1,gcd(a_2,...,a_n)$

I tried using induction but didn't get anywhere.

Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. — Shaun, May 06 '23 at 19:00
Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. — Community, May 06 '23 at 19:06

score -1 · Answer 1 · answered May 06 '23 at 19:12

You could try to prouve this in two parts. First, prove that: $$ gcd(a_1, a_2, …,a_n) | gcd(a_1, gcd(a_2, a_3,…, a_n)) $$ Then prove that: $$ gcd(a_1, gcd(a_2, a_3, …, a_n)) | gcd(a_1, a_2, …, a_n) $$

It must be good, you can try with $n=3$ to notice a pattern.

Gratest common divisor problem

1 Answers1