a problem on a special root of $x^{11}-1=0$

Asked Mar 07 '13 at 15:41

Active Mar 07 '13 at 15:48

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I have came across the following problem

if $\alpha$ be a special root of the equation $x^{11}-1=0$ , then prove that $$(\alpha+1)(\alpha^2+1)......(\alpha^{10}+1)=1$$

totally stuck on it. how to solve this.please help me somebody.

asked Mar 07 '13 at 15:41

user59908

231

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What does "special root" mean? – Chris Eagle Mar 07 '13 at 15:41
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i guess other than $1$ – Aang Mar 07 '13 at 15:43
special roots means -" it is not a root of the equation $x^k-1$ where $k<11$" – user59908 Mar 07 '13 at 15:47
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Since $11$ is prime, in this case $x=1$ is the only root which is not special. – Berci Mar 07 '13 at 15:49
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The usual name for such a root is a primitive $11$-th root of unity, I guess. – Andreas Caranti Mar 07 '13 at 15:50

1 Answers1

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If $\alpha$ is a 'special root', in this case meaning that $\alpha\ne 1$, then using $\alpha^{11}-1=0$, prove that $\alpha^n$ is also a special root of this equation for all $n=1,2,..,10$.
Considering these $11$ (distinct!) roots, conclude that the polynomial $f(x)=x^{11}-1$ factors as $f(x)=(x-1)(x-\alpha)(x-\alpha^2)\ldots(x-\alpha^{10})$.
Substitue $x=-1$ on both sides.

answered Mar 07 '13 at 15:48

Berci

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