So there's this set of polynomial expressions with degree n=3:
$$ \left\{ \begin{array}{c} x_1 + x_2 + x_3 = a \\ x_1^2 + x_2^2 + x_3^2 = b \\ x_1^3 + x_2^3 + x_3^3 = c \end{array} \right. $$
How to find atleast one set of x1,x2,x3 values, knowing that all variables and constants (a,b,c) are positive integers?
Thank you.
Update:
Alright, with the help in the comments I was able to transform this set into a single polynomial expression that can be solved in various ways.
Using Newton's identities: $$ \begin{array}{} x_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2 - 2(x_1x_2 + x_2x_3 + x_1x_3) \\ x_1^3 + x_2^3 + x_3^3 = (x_1 + x_2 + x_3)^3 - 3(x_1x_2 + x_2x_3 + x_1x_3)(x_1 + x_2 + x_3) + 3x_1x_2x_3 \end{array} $$
We can substitute expressions in the set:
$$ \left\{ \begin{array}{} x_1 + x_2 + x_3 = a \\ (x_1 + x_2 + x_3)^2 - 2(x_1x_2 + x_2x_3 + x_1x_3) = b \\ (x_1 + x_2 + x_3)^3 - 3(x_1x_2 + x_2x_3 + x_1x_3)(x_1 + x_2 + x_3) + 3x_1x_2x_3 = c \end{array} \right. $$
$$ \left\{ \begin{array}{} x_1 + x_2 + x_3 = a \\ x_1x_2 + x_2x_3 + x_1x_3 = \frac{a^2 - b}{2} \\ (x_1 + x_2 + x_3)^3 - 3(x_1x_2 + x_2x_3 + x_1x_3)(x_1 + x_2 + x_3) + 3x_1x_2x_3 = c \end{array} \right. $$
$$ \left\{ \begin{array}{} x_1 + x_2 + x_3 = a \\ x_1x_2 + x_2x_3 + x_1x_3 = \frac{a^2 - b}{2} \\ x_1x_2x_3 = \frac{c + 3a\frac{a^2-b}{2} - a^3}{3} \end{array} \right. $$
To simplify, let's assume (since the right halves of the last set are all constants):
$$ \begin{array}{} p_1 = a \\ p_2 = \frac{a^2 - b}{2} \\ p_3 = \frac{c + 3a\frac{a^2-b}{2} - a^3}{3} \end{array} $$
So we get:
$$ \left\{ \begin{array}{} x_1 + x_2 + x_3 = p_1 \\ x_1x_2 + x_2x_3 + x_1x_3 = p_2 \\ x_1x_2x_3 = p_3 \end{array} \right. $$
Now, using Viete's theorem for general polynomial of degree 3:
$$ P(x) = a_3t^3 + a_2t^2 + a_1t + a_0 $$
With the following properties:
$$ \begin{array}{} t_1 + t_2 + t_3 = -\frac{a_2}{a_3} \\ t_1t_2 + t_2t_3 + t_1t_3 = \frac{a_1}{a_3} \\ t_1t_2t_3 = -\frac{a_0}{a_3} \end{array} $$
Applying to our set:
$$ \left\{ \begin{array}{} p_1 = -\frac{a_2}{a_3} \\ p_2 = \frac{a_1}{a_3} \\ p_3 = -\frac{a_0}{a_3} \end{array} \right. $$
And assuming that $a_3 = 1$, the final polynomial will be the following:
$$ t^3 - p_1t^2 + p_2t - p_3 = 0, $$
where the three roots of $t$ are variables $x_1, x_2$ and $x_3$.
(Using Buchberger's algorithm might be a good way to solve it a well, but I was struggling with it and decided to do it this way for now.)
Take the resultant of $x_1 + x_2 + x_3 - a$ and $x_1^2 + x_2^2 + x_3^2 - b$ with respect to $x_3$, the resultant of $x_1 + x_2 + x_3 - a$ and $x_1^3 + x_2^3 + x_3^3 - c$ with respect to $x_3$, and the resultant of those two resultants with respect to $x_2$. You get the square of a cubic polynomial in $x_1$ that must be $0$:
$$ -6\,{x_{{1}}}^{3}+6\,a{x_{{1}}}^{2}+ \left( -3\,{a}^{2}+3\,b \right) x _{{1}}+{a}^{3}-3\,ab+2\,c $$
Since it's a cubic with real coefficients, there is at least one real root. The discriminant is $$\Delta = -216\,{a}^{6}+1944\,{a}^{4}b-1728\,{a}^{3}c-4536\,{a}^{2}{b}^{2}+7776 \,abc+648\,{b}^{3}-3888\,{c}^{2} $$ If $\Delta > 0$, there are three distinct real roots; if $\Delta < 0$, there is only one real root. If $\Delta = 0$, there is at least one real root of multiplicity $> 1$. However, even when there is a real root for $x_1$ the solutions for $x_2$ and $x_3$ might not be real. By symmetry, in any solution the values of $x_2$ and $x_3$ are also possible values of $x_1$. Thus if we want real solutions for $x_1, x_2, x_3$ we need $\Delta \ge 0$.