General deduction of expression $x^n + y^n + z^n $ provided some of its previous powers are known.

Question

If $x^3 -x +1 = 0$ and $\alpha$ , $\beta$ , $\gamma$ are its roots, find $\alpha^2+\beta^2+\gamma^2$ ; $\alpha^3+\beta^3+\gamma^3$ ; $\alpha^4+\beta^4+\gamma^4$ and $\alpha^5+\beta^5+\gamma^5$

My approach:

1. Find $\alpha^2+\beta^2+\gamma^2$

$$\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma + \gamma\alpha)$$

Putting the values, we get $$\alpha^2+\beta^2+\gamma^2 = 2$$

2. Find $\alpha^3+\beta^3+\gamma^3$

$$\alpha^3+\beta^3+\gamma^3 = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma - \gamma\alpha) + 3\alpha\beta\gamma$$

Here we substitute the value of $\alpha^2+\beta^2+\gamma^2$ and get $\alpha^3+\beta^3+\gamma^3=3$

3. Find $\alpha^4+\beta^4+\gamma^4$

$$\alpha^4+\beta^4+\gamma^4 = (\alpha^2+\beta^2+\gamma^2)^2 -2[(\alpha\beta+\beta\gamma + \gamma\alpha)^2 -2\alpha\beta\gamma(\alpha+\beta+\gamma)]$$

Here we substitute the value of $\alpha^2+\beta^2+\gamma^2$ and get $\alpha^4+\beta^4+\gamma^4=2$

But I am stuck at $\alpha^5+\beta^5+\gamma^5$. How do I factorize/break it in terms of known expressions? Moreover, is there any general algorithm for breaking down these expressions into known lower-level powers since those formulae are pretty hard to remember and also lengthy to derive by trial and error?

Answer 1

Multiply the equation by $x^2$: $$x^5-x^3+x^2=0$$ and plug $\alpha, \beta, \gamma$: $$\begin{cases}\alpha^5-\alpha^3 +\alpha^2=0\\ \beta^5-\beta^3+\beta^2=0\\ \gamma^5-\gamma^3+\gamma^2=0\\ \end{cases}$$ and add them: $$\alpha^5+\beta^5+\gamma^5=(\alpha^3+\beta^3+\gamma^3)-(\alpha^2+\beta^2+\gamma^2)=3-2=1.$$ Note: To find $x^3+y^3+z^3$, you can use the same trick.

Answer 2

There is a quite direct way using linear recurrencies.

$x^3-x+1$ is the characteristic polynomial of the linear recurrence

$$a_{n+3} - a_{n+1}+a_n=0$$

This has the general solution

$$a_n = A\alpha^n + B\beta^n + C\gamma^n \text{ for } n \geq 0$$

You are looking for the special solution where

$$A = B = C = 1$$

Now, you have $$a_0 = A+B+C = 3$$ $$a_1 \stackrel{Vieta}{=} \alpha+\beta+\gamma = 0$$ $$a_2 = a_1^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \stackrel{Vieta}{=} 0-2\cdot(-1) = 2$$

Now, you can start the recursion:

$$a_3 = \alpha^3 + \beta^3 + \gamma^3 = a_1 - a_0 = -3$$ $$a_4 = \alpha^4 + \beta^4 + \gamma^4 = a_2 - a_1 = 2$$ $$ \ldots $$