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Let $x_n$ be the $n$ roots of unity.

Does there exist a closed expression for

$$ F_n:=\prod_{i=1}^n(1+x_i+x^{n-1}_i)?$$

Interestingly, if $n=1$, then $F_n=3$ and if $n=2$ then $F_2=-3$ and if $n=3$ then $F_3=0.$ So maybe it just oscillates between these numbers, but how could one prove something like this?

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Note that ${x_i}^{n-1}=\frac{1}{x_i}$ because ${x_i}^n=1, \forall 1 \leq i \leq n$. So, rewriting, $F_n = \displaystyle\prod_{i=1}^{n}\dfrac{{x_i}^2+x_i+1}{x_i}=\prod_{i=1}^n\dfrac{(x_i-\omega)(x_i-{\omega}^2)}{x_i}$, where $\omega$ satisfies $x^2+x+1=0$ i.e. cube root of unity. Since $x_i, 1 \leq i \leq n$ is an $n$th root of unity, it is a root of the polynomial $z^n-1=0$. So, $x^n-1=\displaystyle\prod_{i=1}^n(x-x_i)$. Thus, $F_n=\displaystyle\prod_{i=1}^n\dfrac{(x_i-\omega)(x_i-{\omega}^2)}{x_i}=\dfrac{\prod_{i=1}^n(\omega-x_i)\prod_{i=1}^n({\omega}^2-x_i)}{(-1)^n\prod_{i=1}^n(-x_i)}=\dfrac{f(\omega)f({\omega}^2)}{(-1)^nf(0)}$, where $f(z)=z^n-1=\displaystyle\prod_{i=1}^n(z-x_i)$. So, $F_n=\dfrac{((\omega)^n-1)(({\omega}^2)^n-1)}{(-1)^{n+1}}=(-1)^{n+1}(1-\omega^n)(1-\omega^{2n})$. Can you proceed?

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    $\begingroup$ Very nice! – You could mention that $\omega \ne 1$ is a third root of unity (although one can infer it easily from the context). $\endgroup$
    – Martin R
    May 31 at 11:41
  • $\begingroup$ Ah, yes. Thank you for pointing it out! $\endgroup$ May 31 at 11:44
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Yes, a truncated form will exist. Note that if $x_k$ is $n^{th}$ root of unity, then we have: $$x_k=e^{\frac {2k\pi i}{n}}$$ Hence, using $e^{i\theta}=\cos \theta +i\sin \theta$, we have $$F_n=\prod_{k=1}^n (2\cos (\frac {2k\pi}{n})+1)=2^{2n}\prod_{k=1}^n \cos(\frac {\pi}{6}+ \frac {k\pi}{n})\prod_{k=1}^n \cos (-\frac {\pi}{6} +\frac {k\pi}{n})$$ Now, both the products are evaluated using a similar technique, which is as follows: We have, $$\sin nx =2^{n-1} \prod_{k=0}^{n-1} \sin (x+\frac {k\pi}{n})$$ This result comes from Euler's formula itself. Now, substitute $x=t+\frac {\pi}{2}$ to get a cosine product, after changing limit of product too to match the question($k=0$ term should be divided while $k=n$ term should be multiplied externally). Then put $t=\frac {\pi}{6}$ in the formula to obtain: $$\prod_{k=1}^n \cos(\frac {\pi}{6}+\frac {k\pi}{n})=-\frac {\sin(\frac {2n\pi}{3})}{2^{n-1}}$$ Similarly evaluate the other term, your final answer should be: $$F_n= 4\sin(\frac {n\pi}{3})\sin(\frac {2n\pi}{3})$$

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  • $\begingroup$ Simplified: $$ F_n = \begin{cases} 0, & n \bmod 3 = 0 \\ 3, & n \bmod 6 =1,5 \\ -3, & n \bmod6 = 2,4 \end{cases} $$ $\endgroup$
    – GohP.iHan
    May 31 at 13:28

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