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If $p$ is a prime number and $ a_0,a_1,\ldots , a_{p - 1} $ are rational numbers satisfying $$ a_0 + a_1 \alpha + a_2 \alpha ^2 + \ldots + a_{ p - 1} \alpha ^{ p - 1} = 0 $$ where $$ \alpha = \cos(\frac{2 \pi}{p}) + i \cdot \sin(\frac{2 \pi}{p} ) = e^{\frac{2i\pi}{p}} $$ then $a_0 = a_1 = \ldots = a_{p -1}$.

I think that it is helpful to consider:

(i) $ g(x) = 1+ X^1 + \ldots + X^{p - 1}$ roots(particularly, $\alpha$ and its conjugate) and the fact that it is irreducible.

(ii) $f(x) = a_0+ a_1X^1 + \ldots + a_{p - 1}X^{p - 1}$.

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Hint:

If $p$ is prime, $1+X+\dots+X^{p-1}$ is the $p$-th cyclotomic polynomial and it is irreducible. Actually it is the minimal polynomial of $\alpha$ and its conjugates.

What can you deduce for $f(X)$?

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  • $\begingroup$ Well, we know that $(x-\alpha)(x-\beta) \mid \gcd(f(X),g(X)) $ since both $f$ and $g$ are zero for $X=\alpha,\beta$, where $\beta$ is the conjugate of $\alpha$ $\endgroup$ Jul 13, 2018 at 15:48
  • $\begingroup$ You forget $f(X)$ is the generator of $\{g(X)\in\mathbf Q[X]\mid g(\alpha)=0\}$. $\endgroup$
    – Bernard
    Jul 13, 2018 at 15:55
  • $\begingroup$ I am not sure I understand correctly. What do you mean by $ f(X) $ generator of $h(X)$, where $h(\alpha)=0$? $\endgroup$ Jul 13, 2018 at 16:08
  • $\begingroup$ Not of $g(X)$ – a generator of the set of $g(X)$s which vanish at $\alpha$ (this is an ideal in $\mathbf Q[X]$). $\endgroup$
    – Bernard
    Jul 13, 2018 at 16:12

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