# Is the trivial solution the only solution? [closed]

Let n be a positive integer, and c_1, c_2, ... c_n be (unkown) real numbers .

Consider the system

$$c_1+c_2+ ... +c_n=0,$$

$$c_1^2+c_2^2+ ... +c_n^2=0,$$

$$c_1^3+c_2^3+ ... +c_n^3=0,$$

$$.... .... ....$$

$$c_1^n+c_2^n+ ... +c_n^n=0.$$

The question is: Is the trivial solution the only solution to the system?

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## closed as off topic by Steve Huntsman, Andres Caicedo, Federico Poloni, Benjamin Steinberg, algoriAug 23 '12 at 15:27

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en.wikipedia.org/wiki/Vandermonde_matrix. Voting to close as not research-level (for which see the FAQ). –  Steve Huntsman Aug 23 '12 at 14:55
The problem is more interesting over the complex numbers, for there sums of squares obey different constraints than for real numbers. Gerhard "Has Solution For Many Equations" Paseman, 2012.08.23 –  Gerhard Paseman Aug 23 '12 at 15:17
I didn't note the comment. How do you use the Vandermonde determinant to show that all $c_j$ are zero? –  Pietro Majer Aug 23 '12 at 15:34

Yes, assuming we are in a field. A first consequence is that all the symmetric functions of $(c_1,\dots,c_n)$ are zero (see e.g. this wiki article ). But this can be written as an identity of polynomials $$x^n=\prod_{j=1}^n(x-c_j)$$ which obviously implies $c_j=0$ for all $j$.
or, alternatively, from $\prod_{j=1}^n c_j=0$ one gets that at least one $c_j$ vanishes; and iterating, that they all vanish. –  Pietro Majer Aug 23 '12 at 15:32