Prove a $gcd(x^a-1,x^b-1)=x^{gcd(a,b)}-1$ using knowledge of polynomials and not number theory

Asked May 05 '22 at 14:46

Active May 05 '22 at 16:12

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Prove the following:

$U_n=\{Z \in \mathbb{C} : Z^n = 1\}$. Then, show that $U_a \cap U_b = U_{gcd(a,b)}$

Hence prove that, $gcd(x^a-1,x^b-1)=x^{gcd(a,b)}-1$

I started out with if $α \in U_a$, then $α = e^{\frac{2ikπ}a}$ where $0<k<a$ and if $β \in U_b$, then $β = e^{\frac{2ilπ}b}$ where $0<l<b$. So, if $α=β$, then $e^{\frac{2ikπ}a}=e^{\frac{2ilπ}b}$ $\implies \frac kl=\frac ba$

Then i claimed that : For every $k, 0<k<gcd(a,b)$, we find $l$ such that $\frac kl=\frac ba$

But then I am not able to prove my claim. Please help me.

asked May 05 '22 at 14:46

Gloona

This post has really many answers. I suppose there is one, which you can use. – Dietrich Burde May 05 '22 at 14:59
I am not getting an answer by using the first part and that is what I want. – Gloona May 05 '22 at 15:07
The first part is the above duplicate, see here. – Dietrich Burde May 05 '22 at 16:10
1

See the "alternative" argument using "order" in this answer in the linked dupe. – Bill Dubuque May 05 '22 at 16:13

Prove a $gcd(x^a-1,x^b-1)=x^{gcd(a,b)}-1$ using knowledge of polynomials and not number theory

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