<s>[INST] How to solve the equations? $\begin{cases}x_1+x_2+x_3+x_4 = 6,\\
 x_1^2+x_2^2+x_3^2+x_4^2 = 10,\\
 x_1^3+x_2^3+x_3^3+x_4^3 = 18,\\
 x_1^4+x_2^4+x_3^4+x_4^4 = 34.\end{cases}$
I have found a solution of $(1,1,2,2)$. And I tried to solve it by MAPLE. 
All solutions are in the form of $(1,1,2,2)$ (and any possible arrangement of $(1,1,2,2)$). But how can I prove that the equation has no more solution except these 6 solutions?
How about a general $(a_1,a_2,a_3,a_4)$? 
$\begin{cases}x_1+x_2+x_3+x_4 = a_1,\\
 x_1^2+x_2^2+x_3^2+x_4^2 = a_2,\\
 x_1^3+x_2^3+x_3^3+x_4^3 = a_3,\\
 x_1^4+x_2^4+x_3^4+x_4^4 = a_4.\end{cases}$
Do they have an unique solution? Or not? 
Do we have a general method to deal with it?
 [/INST] As Rene comments, generally speaking, to solve equations of this kind, you should first transform it to
\begin{cases}x_1+x_2+x_3+x_4 = c_1,\\
 x_1 x_2 + x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 = c_2,\\
 x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 = c_3,\\
 x_1x_2x_3x_4=c_4.\end{cases}
Then, by Vieta's formulas, you immediately get that $x_1, x_2, x_3, x_4$ are $4$ roots of the equation
$$
  (x-x_1)(x-x_2)(x-x_3)(x-x_4) = x^4-c_1x^3+c_2x^2-c_3x^3+c_4=0
$$
Therefore all your problems are solved.
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