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Find $n$ such that the polynomial function $1+X+X^2$ divides $P_n = (1+X^4)^n-X^n$.

What I've written is $$ \left(1+X^4\right)^n - X^n = \left(1+X+X^2\right)Q\left(X\right) + R\left(X\right) $$ where $R\left(X\right) = aX+b$. Finding $a$ and $b$ as function of $n$ 'd lead me to the result by putting $a=b=0$.

What I've tried is to evaluate the equation in $e^{2i\pi/3}$ and $e^{-2i\pi/3}$ which are the roots of $1+X+X^2$ then $$ P_n\left(e^{2i\pi/3}\right) = R\left(e^{2i\pi/3}\right) \text{ and also }P_n\left(e^{-2i\pi/3}\right) = R\left(e^{-2i\pi/3}\right) $$ However my problem 'd be to simplify as far as I can $P_n\left(e^{2i\pi/3}\right)$ : $$P_n\left(e^{2i\pi/3}\right) = \left(1+e^{8 i \pi/3}\right)^n-\left(e^{2i\pi/3}\right)^n=\left(1+e^{2 i \pi/3}\right)^n-\left(e^{2i\pi/3}\right)^n = \sum_{k=0}^{n-1}\left(1+e^{2i\pi/3}\right)^{n-k}\left(e^{2i\pi/3}\right)^k $$ Then, yes I can make the sum of a geometrical sequence terms appear, but it looks still not friendly to manipulate. Is there a better way to go to solve the problem ? I think the result is $n$ being a multiple of $6$

Thanks for the help

Atmos
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  • Related manipulations of third roots of unity. – Jyrki Lahtonen Feb 13 '22 at 12:58
  • @JyrkiLahtonen Your link is an exact dupe of the link I posted a minute prior on your answer. These dupes (and minor variants) are everywhere... – Bill Dubuque Feb 13 '22 at 13:04
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    Reposting the comment so that dust can settle. Evaluation at $\omega=e^{2\pi i/3}$ is a good idea (+1). We have $\omega^3=1$ and $$1+\omega+\omega^2=0 \implies 1+\omega=-\omega^2.$$ Additionally you need to use $\omega^4=\omega$. With these hints you can surely evaluate $(1+X^4)^n-X^n$ at $X=\omega$, and find out when it simplifies to zero. – Jyrki Lahtonen Feb 13 '22 at 13:10
  • @JyrkiLahtonen Probably the "n mins ago" time was out of sync before page refresh. I don't see any need to waste further time. The key ideas are already in said answer (and its link). – Bill Dubuque Feb 13 '22 at 13:15
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    @JyrkiLahtonen Your tip just made it so easy, thanks for that, appreciated it – Atmos Feb 13 '22 at 14:32
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    You are welcome. Sorry about needing to close this as a duplicate. I was on a fence about it, but another user agreed. FWIW the upvote is mine. – Jyrki Lahtonen Feb 13 '22 at 15:07

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