The polynomial $x^{2k}+1+(x+1)^{2k}$ is not divisible by $x^2+x+1$. Find the value of $k\in \mathbb{N}$.
I tried finding out the roots of $x^2+x+1$ which were $\dfrac{-1±\sqrt{3}i}{2}$ but in vain. I got no result other than making the polynomial more complicated.
Here's what I got : $$\left(\frac{-1±\sqrt{3}i}{2}\right)^{2k}+1+\left(\frac{-1±\sqrt{3}i}{2}+1\right)^{2k}.$$
Now, I don't know what to do next.
Any help would be appreciated.
Let $\omega=\frac12(-1+i\sqrt3)$. Then $\omega^2=\overline\omega$ and $\omega^3=1$. Also $1+\omega=\frac12(+1+i\sqrt3)=\exp(\pi i/3)$ so that $(1+\omega)^2=\exp(2\pi i/3)=\omega$. So you are interested in the quantity $$a_k=\omega^{2k}+1+\omega^k$$ (and its complex conjugate $\overline{a_k}$). As $\omega^3=1$, then $a_{k+3}=a_k$: the sequence $a_k$ repeats with period $3$. So all you need to compute are $a_0$, $a_1$ and $a_2$ etc.
Using the remainder theoram, putting $x^2 + x + 1 = 0$ in the polynomial $P(x) =x^{2k} + 1 + (x + 1)^{2k}$ we will get the remaimder. $$ P(x) = 1 + x^{2k} + (x^2 + 2x + 1)^k = 1 + x^{2k} + x^k $$ We can also prove that if $x^2 + x + 1 = 0$, $x^3 = 1$. We can now clearly see that for $k$ is not equal to $3q$, $P(x)=0$ But given it should not be divisible by $x^2 + x + 1$, Hence $k$ must be $3q$.
