Since for prime $p$, $p^2-1$ is divisible by $8,$ $x^{p^2-1}-1$ is divisible by $x^8-1$.

Question

I was reading Dummit and Foote, and in there, there is a line that goes like this:

Since for prime $p$, $p^2-1$ is divisible by $8,$ $x^{p^2-1}-1$ is divisible by $x^8-1$.

I get that $p^2-1$ is divisible by $8.$ But how does that fact imply that $x^{p^2-1}-1$ is divisible by $x^8-1$?

Does this answer your question? Prove：$x^d-1 \mid x^n-1$ iff $d \mid n$. See also e.g. https://math.stackexchange.com/questions/3889183 or https://math.stackexchange.com/questions/609900 etc. — Anne Bauval, Jul 23 '23 at 17:18

score 2 · Answer 1 · edited Jul 23 '23 at 21:02

2

$$x^{ab}-1=(x^a-1)(x^{a(b-1)}+x^{a(b-2)}+x^{a(b-3)}+\cdots+x^{a}+x^0)$$

edited Jul 23 '23 at 21:02

user26857

answered Jul 13 '18 at 05:27

vadim123

score 2 · Accepted Answer · edited Jul 13 '18 at 05:32

2

If $$p^2-1=8k$$ then

$$x^{p^2-1}-1=x^{8k}-1$$

Let $y=x^8$, we have $y^k-1=(y-1)\sum_{r=0}^{k-1}y^r$. That is $y-1$ divides $y^k-1$.

That is $x^8-1$ dividis $x^{8k}-1=x^{p^2-1}-1$.

edited Jul 13 '18 at 05:32

Math Lover

answered Jul 13 '18 at 05:30

Siong Thye Goh

2 Answers2