A simple problem with a simple and a nonsimple solution

Question

I have this seemingly simple problem to solve. Its statement is straightforward: "If $x+y+z=3,x^2+y^2+z^2=5, x^3+y^3+z^3=7$ show that $x^4+y^4+z^4=9,x^5+y^5+z^5\neq 11$".

There is a highschool approach, that is I expanded $(x+y+z)^2, (x+y+z)^3$ and managed to get $$xyz=-2/3, xy+yz+zx=2$$ etc, which I guess I could employ to expand $(x+y+z)^4, (x+y+z)^5$ and get some results (I guess).

The thing is I don't think this is the proper approach. Is there a better way than this tedious, long calculation (if it is a solution that is)

Say we could divide the polynomials $$f=x^4+y^4+z^4,g=x^3+y^3+z^3, h=x^2+y^2+z^2,k=x+y+z$$ Wouldn't this give us what we want by using the remainder polynomial?

Thanks in advance for your time.

Answer 1

$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+x^2z^2+y^2z^2)=$$ $$=25-2((xy+xz+yz)^2-2xyz(x+y+z))=$$ $$=25-2\left(4-2\left(-\frac{2}{3}\right)3\right)=9.$$

$$x^5+y^5+z^5=(x^2+y^2+z^2)(x^3+y^3+z^3)-\sum_{sym}x^3y^2=$$ $$=35-(x^2y^2+x^2z^2+y^2z^2)(x+y+z)+xyz(xy+xz+yz)=$$ $$=35-8\cdot3-\frac{2}{3}\cdot2=\frac{29}{3}$$

Answer 2

For your question, assume a cubic equation whose roots are $x,y$ and $z$.

Now, for a polynomial

$$P(x)=a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$$

$$\color{blue}{\sum_{i=0}^{r-1} a_iS_{r-i}+ra_r=0}$$

Where ;

$S_r=\sum \alpha_i^r$

And, $\alpha_i$ are roots of polynomial $P(x)=0$

Apply this formula for $r=0,1,2,3...m$ till you get value of $S_m$

If you run out of $a_r$ (when $r>n$) multiply the whole polynomial by $x$ , this will introduce on more root of $P(x)=0$ , but the new root would be $x=0$ so it would have no effect on your $S_m$ ( Since $0^m$ i.e. $0$ will be added to $S_m$ which will have no effect)

Answer 3

If you are in the market for more sophisticated ways... you have found $x, y, z$ to be the roots of $P(t)=t^3-3t^2+2t+\frac23=0$. Now if $\alpha$ is a root of $P(t)$, we seek the minimal polynomial of $\alpha^m$ (for $m=4, 5$).

So if $P_4(r)$ is the polynomial with roots $x^4, y^4, z^4$, then it is given by the resultant (why?) $$P_4(r) = Res(r-t^4, P(t))=r^3-9r^2+\tfrac{536}9r-\tfrac{16}{81}$$ so $x^4+y^4+z^4=9$.

$\parallel$ly $$P_5(r) = Res(r-t^5, P(t))=-r^3+\tfrac{29}3r^2-\tfrac{496}3r-\tfrac{32}{243}$$ so $x^5+y^5+z^5=\frac{29}3 \neq 11$.