Proving the equations $x_1+\dots+x_n=0$, ..., $x_1^n+\dots+x_n^n=0$ have a unique solution

Question

Let equations of the form $\left\{\begin{matrix} x_{1}+x_{2}+...+x_{n}=0\\ x^{2}_{1}+x^{2}_{2}+...+x^{2}_{n}=0\\ .........\\ x^{n}_{1}+x^{n}_{2}+...+x^{n}_{n}=0 \end{matrix}\right.$.

Proof: $(0,0,...,0)^{'}$ is its unique solution in $\mathbb{C^{n}}.(\mathbb{C}\: {\buildrel\rm def \over=}\left\{a+bi|a,b\in\mathbb{R} \right \}$ )

I found a simple way to prove this statement:Using mathematical induction for $n$. but there must have some other ways to prove the same conclusion. Can you give me some other methods ? Any of your help will be appreciated!

The question seems designed in a way where induction cannot work. Because induction works if you want to prove something "for all $n$" which is not the case here. You have $n$ variables, and $n$-th powers. — Landon Carter, Mar 30 '15 at 07:17
possible duplicate of Does $\lambda_1^n+ \lambda_2^n+ \dots +\lambda_k^n =0 $ for all $n$ imply that $\lambda_1= \lambda_2= \dots= \lambda_k = 0 $? — Travis Willse, Mar 30 '15 at 08:09
This question is slightly stronger than the ostensible duplicate in that here we assume only the weaker hypothesis the sum $x_1^k + \cdots + x_n^k$ holds only for $k \leq n$, but the answers there only use that much. — Travis Willse, Mar 30 '15 at 08:12

score 1 · Answer 1 · answered Mar 30 '15 at 09:46

It is well known that the elementary symmetric functions on $n$ variables: $$ E_1=x_1+\ldots+x_n,\quad E_2 = \sum_{i\neq j}x_i x_j,\quad \ldots\quad E_n = \prod_{i=1}^{n} x_i $$ as well as the power sums: $$ P_1=E_1,\quad P_2 = \sum_{i=1}^{n} x_i^2,\quad\ldots\quad E_n=\sum_{i=1}^{n}x_i^n $$ give a base of the ring of the symmetric functions on $n$ variables. For a proof, look for the Newton-Girard formulas on Wikipedia. Such formulas also imply that, if $P_1=P_2=\ldots=P_n=0$, then $E_1=E_2=\ldots=E_n=0$, so, if we take $p(t)$ as the monic polynomial having degree $n$ and roots in $t=x_i$ (accounted with multiplicity), by Vieta's formulas it follows that: $$ p(t) = t^n, $$ from which $x_1=x_2=\ldots=x_n = 0$.

Proving the equations $x_1+\dots+x_n=0$, ..., $x_1^n+\dots+x_n^n=0$ have a unique solution

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