<s>[INST] Polynomial system If there are 3 numbers $x,y,z$ satisfying
$f=x+y+z=3$ ,  $g=x^2+y^2+z^2=5$ , $h=x^3+y^3+z^3=7$ then prove that they also satisfy 
$x^4+y^4+z^4=9$ but not $x^5+y^5+z^5=11$
I dont know how to tackle this to be honest, i have started trying to write $x^4+y^4+z^4-9$ as $r_1\times f + r_2\times g+ r_3\times h$ where $r_1,r_2,r_3$ in $C[x,y,z]$ but then i realised that i cant possibly do that by hand.This was an exercise on my first computational algebra course so we havent really learnt anything much yet. I would like a hint if possible.
 [/INST] I give here an answer advocating the simplicity of using Newton (or Newton-Girard) formulas. Using the notations of the Wikipedia article in all generality:
Let $e_0=1, e_1=x+y+z, e_2=xy+yz+zx, e_3=xyz.$
Let $p_1=x+y+z$, $p_2=x^2+y^2+z^2$, $p_3=x^3+y^3+z^3$.
Then:
$$\begin{cases}
e_1&=&p_1\\
2e_2&=&p_1e_1-p_2\\
3e_3&=&p_1e_2-p_2e_1+p_3
\end{cases} \ \ \ \text{Using the given values:} \ \ \ \begin{cases}
e_1&=&3\\
2e_2&=&3e_1-5\\
3e_3&=&3e_2-5e_1+7
\end{cases}$$
from which $e_1=3, e_2=2, e_3=-\frac23$.
Using Vieta's formulas, $x,y,z$ are solutions of the third degree equation:
$$\tag{1}t^3-3t^2+2t+\frac23=0$$
(whose roots are one real, and two complex conjugate other roots).
Any root $t$ of (1), verifies $t^4=3t^3-2t^2-\frac23t$. When one writes this relationship for $t=x$, $t=y$ and $t=z$.
$$\tag{2}\begin{cases}x^4=3x^3-2x^2-\frac23x\\y^4=3y^3-2y^2-\frac23y\\z^4=3z^3-2z^2-\frac23z\\\end{cases}$$
Adding these equations, one obtains $x^4+y^4+z^4=9.$
Multiplying the equations in (2) resp. by $x$,$y$ and $z$, and adding again, one obtains $x^5+y^5+z^5=\dfrac{29}{3}$ ($\neq 11$!).
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