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Thread: How to start this ?

  1. #1
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    How to start this ?

    Let ϵ=cos2π/673+isin2π/673 . If a0,a1,,a672 are non zero integers show that the relation a0+a1ϵ+⋯+a672ϵ672=0 holds if and only if
    a0=a1==a672
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  2. #2
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    Re: How to start this ?

    the if direction is pretty straightforward

    let $a_k=c,~k = 0, 1, 2, \dots,~c\in \mathbb{C}$

    $\begin{align*}
    &\sum \limits_{k=0}^{672}~a_k \epsilon^k = \\

    &c \dfrac{1-\epsilon^{673}}{1- \epsilon} = \\

    &c \dfrac{1-e^{i2\pi \frac{673}{673}}}{1-\epsilon} = \\

    &c\dfrac{1-1}{1-\epsilon}= \\

    &0

    \end{align*}$

    Going the other direction I haven't figured out yet.
    Last edited by romsek; Dec 13th 2018 at 01:54 PM.
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  3. #3
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    Re: How to start this ?

    The minimal polynomial of $\displaystyle e^{2\pi i/p}$ for prime $p$ is

    $\displaystyle x^{p-1}+x^{p-2}+\text{...}+x +1$
    Last edited by Idea; Dec 15th 2018 at 08:50 AM.
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  4. #4
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    Re: How to start this ?

    Quote Originally Posted by Idea View Post
    The minimal polynomial of $\displaystyle e^{2\pi i/p}$ for prime $p$ is

    $\displaystyle x^{p-1}+x^{p-2}+\text{...}+x +1$
    so you're saying that the set of elements $\epsilon^k = e^{i\dfrac{2\pi}{673}k},~k=0,1,\dots 672$ form a finite field.

    and that $\sum \limits_{k=0}^{672} \epsilon^k =0$

    and the fact that it's minimal means that it's got no roots over the field? i.e. it can't be reduced?

    so there is no solution to $\sum \limits_{k=0}^{672}~a_k \epsilon^k = 0,~\text{where one or more of the }a_k=0$ ?

    Is this correct?
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  5. #5
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    Re: How to start this ?

    Let $\displaystyle g(x)=1+x+\text{...}+x^{p-1}$

    $g$ is the minimal polynomial of $\displaystyle \epsilon =e^{2\pi i/p}$ means

    $g$ is the nonzero polynomial of smallest degree such that $\displaystyle g(\epsilon )=0$

    we are given a polynomial $f$ same degree as $g$ and such that $\displaystyle f(\epsilon )=0$ namely

    $\displaystyle f(x)=a_0+a_1x+\text{...}+a_{p-1}x^{p-1}$

    $f$ must be a constant multiple of $g$

    to see this divide $f$ by $g$ , the remainder $r(x)$ is of degree $< p-1$ and $r(\epsilon )=0$

    so $r(x)$ must be identically 0

    example ($p=3$)

    $\displaystyle a_0+a_1x+a_2x^2=\left(1+x+x^2\right)a_2+\left(a_1-a_2\right)x+\left(a_0-a_2\right)$
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