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Here $x$ is the polynomial and $n,k$ are both integers. Why is this true or is not true?

I tried to prove it by induction.

Base case: $(q^{k} - 1) / (q^{k} - 1) = 1$.

Inductive step: $(q^{ck} - 1) / (q^{k} - 1) = f(x)$, then prove that $(q^{(c+1)k} - 1) / (q^{k} - 1) = f'(x)$. And here is where I stuck, how can I express such $f'(x)$, that is to say, I need to solve $(q^k - 1) * f(x) + q^{k} = (q^k - 1) * g(x) = f'(x) $.

Yuqi Hu
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    Put $y=x^k$ to give $y^n-1$ and $y-1$ respectively. Do you know how to do that? – Mark Bennet Jan 27 '21 at 15:59
  • Welcome to MSE. You'll get a lot more help if you show that you have made a real effort to solve the problem yourself, even if you haven't made much progress. What are your thoughts? What have you tried? How far could you get? Where are you stuck? This question will likely be closed if you don't add more context. Please respond by editing the question body. Clarifications do not belong in the comments. – Kyan Cheung Jan 27 '21 at 16:00
  • Prove that roots of the second polynomial are roots of the first polynomial – Damien Jan 27 '21 at 16:02

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One way to prove it is by using the ciclotomic equation (you can see a prove to this equality in this same site [here][1]): that given $x,y\in\mathbb{R}$ and $n\in\mathbb{N}$, then $$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\cdots xy^{n-2}+y^{n-1}).$$ If we pick $x$ as your $x^k$, and $y$ as $1$, then we get that $$x^{nk}-1=\left(x^k\right)^{n}-\left(1\right)^n = \left(x^k-1\right)\left(\left(x^k\right)^{n-1}+\left(x^k\right)^{n-2}+\cdots + x + 1\right),$$ so from this we conclude that $x^{kn}-1$ is divisible by $x^k-1$.

[1]: Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$)

Alejandro Bergasa Alonso
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