$a_{n}$ geometric sequence, prove that : $a_{1}+a_{2}+a_{3}+...+a_{n}\|\ a_{1}^{k}+a_{2}^{k}+a_{2}^{k}+...+a_{n}^{k}$ with $(n,k)=1$

Question

Problem : Let $a_{n}$ be a geometric sequence of the integer numbers $a_{n}\in \mathbb Z$ for all $n\in \mathbb N$. Prove that:

$$a_{1}+a_{2}+a_{3}+...+a_{n} \mid a_{1}^{k}+a_{2}^{k}+a_{2}^{k}+...+a_{n}^{k}$$ with $(n,k)=1$

My attempt :

$$a_{1}+a_{2}+a_{3}+...+a_{n}=a\left(1+r+r^{2}+...+r^{n-1}\right)=a\frac{r^{n}-1}{r-1}$$

Then :

$$a_{1}^{k}+a_{2}^{k}+a_{2}^{k}+...+a_{n}^{k}=a^{k}\left(1+r^{k}+r^{2k}+...+r^{k(n-1)}\right)=a^{k}\frac{r^{nk}-1}{r^{k}-1}$$

Now I'm going to prove :

$$a\frac{r^{n}-1}{r-1} \mid a^{k}\frac{r^{nk}-1}{r^{k}-1}$$

But I don't know how I complete from here ?

Answer 1

Show that the greatest common divisor of $r^k-1$ and $r^n-1$ is $r-1$

Answer 2

Note that

$$r^{nk}-1 = {r^{n}}^{k}-1^k = (r^n-1)({r^n}^{k-1} + {r^n}^{k-2}\dots+ 1)$$

and

$$r^k-1 = (r-1)(r^k+r^{k-1}\dots +1)$$

This can be found by GP sum formula or from the (related) expansion of $(x^n-y^n)$.