<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] Systematically work order by order
\begin{eqnarray*}
(x+y+z)^2= x^2+y^2+z^2+2(xy+yz+zx)
\end{eqnarray*}
right firstly we do not need to write everything out; We shall use the following short hand
\begin{eqnarray*}
(\sum x)^2= \sum x^2+2\sum xy
\end{eqnarray*}
so $\sum xy=2$
\begin{eqnarray*}
(\sum x)(\sum x^2)= \sum x^3+\sum x^2y
\end{eqnarray*}
so $\sum x^2y=8$
\begin{eqnarray*}
(\sum x)(\sum xy)= \sum x^2y+3xyz
\end{eqnarray*}
so $xyz=-2/3$.
\begin{eqnarray*}
(\sum xy)^2= \sum x^2y^2+2xyz(\sum x)
\end{eqnarray*}
so $\sum x^2y^2=8$
\begin{eqnarray*}
(\sum x^2)^2= \sum x^4+2(\sum x^2y^2)
\end{eqnarray*}
so $\sum x^4=9$ ... the first part is shown.
\begin{eqnarray*}
(\sum xy)(\sum x^3)= \sum x^4y+xyz(\sum x^2)
\end{eqnarray*}
so $\sum x^4y=52/3$
\begin{eqnarray*}
(\sum x) (\sum x^4)= \sum x^5+(\sum x^4y)
\end{eqnarray*}
so $\sum x^5=29/3$ ... the second part is shown.
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