Polynomial system

Question

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.

Answer 1

$$2(xy+yz+zx) = (x+y+z)^2 - (x^2+y^2+z^2) = 4$$ $$-3xyz = (x+y+z)^3 - (x^3+y^3+z^3) - 3(xy+yz+zx)(x+y+z) = 27 - 25 = 2$$ Thus $x,y,z$ are the roots of $$t^3 -3t^2 + 2t + \frac{2}{3} = 0$$ Now \begin{align*} t^4 = 3t^3 -2t^2 -\frac{2}{3}t \end{align*} Putting $t=x,y,z$ in the above and adding, we get $$x^4 + y^4 + z^4 = 3 \sum x^3 - 2\sum x^2 - \frac{2}{3} \sum x = 21 - 10 -2 = 9$$ Also, $$t^5 = 3t^4 - 2t^3 - \frac{2}{3} t^2$$ and hence $$\sum x^5 = 3\sum x^4 - 2 \sum x^3 - \frac{2}{3} \sum x^2 = 27 - 14 - 10/3 = 29/3$$

Answer 2

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.

Answer 3

For notational ease, let's define a few variables: \begin{align*} j&=x^3y+x^3z+xy^3+xz^3+y^3z+yz^3\\ k&=x^2y^2+x^2z^2+y^2z^2\\ l&=x^2yz+xy^2z+xyz^2. \end{align*}

Now, let's start multiplying things out: \begin{align*} fh&=(x^4+y^4+z^4)+(xy^3+xz^3+yx^3+yz^3+zx^3+zy^3)\\ &=(x^4+y^4+z^4)+j\\ g^2&=(x^4+y^4+z^4)+2(x^2y^2+x^2z^2+y^2z^2)\\ &=(x^4+y^4+z^4)+2k\\ f^2g&=((x^2+y^2+z^2)+2(xy+xz+yz))(x^2+y^2+z^2)\\ &=(x^4+y^4+z^4)+2(x^2y^2+x^2z^2+y^2z^2)\\ &\qquad\qquad+2(x^3y+x^3z+x^2yz+xy^3+xy^2z+y^3z+xyz^2+xz^3+yz^3)\\ &=(x^4+y^4+z^4)+2j+2k+2l\\ f^4&=(x^4+y^4+z^4)+4(xy^3+xz^3+yx^3+yz^3+zx^3+zy^3)+6(x^2y^2+x^2z^2+y^2z^2)\\ &\qquad\qquad+12(x^3yz+xy^3z+xyz^3)\\ &=(x^4+y^4+z^4)+4j+6k+12l \end{align*}

Combining, we have \begin{align*} 21&=fh=(x^4+y^4+z^4)+j\\ 25&=g^2=(x^4+y^4+z^4)+2k\\ 45&=f^2g=(x^4+y^4+z^4)+2j+2k+2l\\ 81&=f^4=(x^4+y^4+z^4)+4j+6k+12l \end{align*}

This is now a system of equations where we can use linear algebra to solve. For example, subtracting $6$ times the third from the fourth gives $$ -189=-5(x^4+y^4+z^4)-8j-6k $$ Now, adding $3$ times the second to this gives $$ -114=-2(x^4+y^4+z^4)-8j $$ Finally, adding $8$ times the first to this gives $$ 54=6(x^4+y^4+z^4) $$ In other words, $x^4+y^4+z^4=9$.

Answer 4

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 $x^5+y^5+z^5=\dfrac{29}{3}$ ($\neq 11$!).