Can anyone help me in determining the value of a,b and c?

Question

given 3 equations, $$ \left\{ \begin{array}{cccc} a^2&+b^2&+c^2 &=& x \\ a^3&+b^3&+c^3 &=& y\\ a^4&+b^4&+c^4 &=& z\\ \end{array} \right. $$ where a,b,and c are unknown integers, and $x,y,z$ are known. I want to calculate the value of $a,b$ and $c$ in terms of $x,y,z$ . I tried it but failed to reduce the dependency among the variables.

Answer 1

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, $$9u^2-6v^2=x,$$ $$27u^3-27uv^2+3w^3=y$$ and $$81u^4-108u^2v^2+18v^4+12uw^3=z,$$ which gives $$81u^4-108u^2\cdot\frac{9u^2-x}{6}+18\left(\frac{9u^2-x}{6}\right)^2+12u\cdot\frac{y-27u^3+27u\cdot\frac{9u^2-x}{6}}{3}=z$$ or $$27u^4-18xu^2+8yu+x^2-2z=0,$$ which is a quartic equation, which we can solve for concrete values of $x$, $y$ and $z$.

In the general case, it's very ugly and we can use here the Ferrari method. https://en.wikipedia.org/wiki/Quartic_function

After solving if this equation we can get values $v^2$, $w^3$ and by solving the cubic equation we can get values of $a$, $b$ and $c$.

Actually, if we need to find real solutions, so we need $$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0.$$