# Thread: How to start this ?

1. ## 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

2. ## 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.

3. ## Re: How to start this ?

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

$\displaystyle x^{p-1}+x^{p-2}+\text{...}+x +1$

4. ## Re: How to start this ?

Originally Posted by Idea
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?

5. ## 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 ($p=3$)

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