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When reading a proof of why $x^3+y^3=z^3$ has no nontrivial integer solutions I came across following identity:

$$ y^3 = z^3-x^3 = (z-x)(z-\omega x)(z-\omega^2 x) \qquad \text{where } \omega = e^{2\pi i /3} \quad \text{i.e.}\quad \omega^3 = 1$$

Expanding the RHS results in:

$$ z^3-(1+\omega+\omega^2)z^2x+(\omega+\omega^2+\omega^3)zx^2-\omega^3x^3 = z^3-x^3,$$

since obviously $\omega+\omega^2+\omega^3 = 1+\omega +\omega^2 = 0$. Well then I thought about how obvious that is. I mean geometrically it is obvious that the sum of all $n$-th unity roots must equal $0$, but is there an analytical proof? I was not able to come up with one straight away.

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    $\begingroup$ Given any polynomial, the second coefficient is the sum of the roots of the polynomial. If we take $p(X) = X^n - 1$, then its roots are the $n^{th}$ roots of unity, and the second coefficient is the coefficient of $X^{n-1}$, which is $0$ as long as $n>1$. This is really Vieta's formula as mentioned below, but I find it easier to state it this way. $\endgroup$ Aug 9 '14 at 13:05
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I think I just found one more time the answer myself just after submitting the question, it is so simple...

Let $\omega = e^{2 \pi i / n}$ which implies $\omega^n = 1$.

$$ 1 + \omega + \omega^2 + \ldots + \omega^{n-1} = \frac{\omega^n-1}{\omega-1} = 0 $$

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Also consider $$\omega S=\omega(1+\omega+\omega^2)=\omega+\omega^2+\omega^3=\omega+\omega^2+1=S.$$ Unless $\omega=1$, $S=0$.

You needn't know the summation formula for geometric progressions.

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  • $\begingroup$ You can easily extend this to a proof for the geometric progression. $\endgroup$
    – Christoph
    Jun 8 '20 at 9:18
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    $\begingroup$ This answer is the best one. Purely visual and conceptual. No algebra. $\endgroup$
    – Smithey
    Apr 1 at 19:24
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Nongeometricrally, nth-roots of unity are the solutions to the equation $x^n - 1 = 0$. The $x^n$ coeff is $1$ and the $x^{n-1}$ coeff is $0$, so the sum of the roots is zero.

Geometrically, the n-th roots of unity are equally spaced vectors around a unit circle, so their sum is the center of the circle, which is $0 + 0i$.

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Let $S$ denote the sum of the $n$ roots of unity. We have

$$\exp\bigg(\frac{2{\pi}i}{n}\bigg)S = \exp\bigg(\frac{2{\pi}i}{n}\bigg) \sum_{a=0}^n \exp\bigg(\frac{2{\pi}ia}{n}\bigg)$$

$$= \sum_{a=0}^n \exp\bigg(\frac{2{\pi}i(a+1)}{n}\bigg)$$

$$= \sum_{b=0}^n \exp\bigg(\frac{2{\pi}ib}{n}\bigg), \; b=a+1$$

$$= S$$

Because $a+1$ is just a cyclic shift of the roots, the sum still contains the same terms. So we have shown that

$$\exp\bigg(\frac{2{\pi}i}{n}\bigg)S = S$$

and therefore $S = 0$.

Terras, A. (1999). The Discrete Fourier Transform on the Finite Circle ℤ/nℤ. In Fourier Analysis on Finite Groups and Applications (London Mathematical Society Student Texts, pp. 30-45). Cambridge: Cambridge University Press. doi:10.1017/CBO9780511626265.004

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You know that $$\omega = e^{\frac{2\pi i}{n}}$$

$$\Longrightarrow \omega^n = 1$$

Now if your sum (the one of all the roots up to n-1) is S, you can pose :

$$1+\omega S = \omega^n = 1$$

$$1+\omega S = 1$$

$$\omega S = 0$$

$$S = \frac{0}{\omega} = 0$$

Seems pretty ghetto but i'm fairly sure it's legit.

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This is just a more general rewriting of Yves Daoust's answer:

$$ \begin{align*} S&=\sum_{k=0}^{n}\omega^{k}\\\\ \omega\cdot S&=\sum_{k=0}^{n}\omega^{k+1}\\\\ &=\omega+\sum_{k=1}^{n-1}\omega^{k+1}+\omega^{n+1}\\\\ &=1+\omega+\sum_{k=2}^{n}\omega^{k}=\sum_{k=0}^{n}\omega^{k}\\\\ &=S\\\\ \implies S&=0 \end{align*} $$

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