# Sum of roots: Vieta's Formula

The roots of the equation $x^4-5x^2+2x-1=0$ are $\alpha, \beta, \gamma, \delta$. Let $S_n=\alpha^n +\beta^n+\gamma^n+\delta^n$ Show that $S_{n+4}-5S_{n+2}+2S_{n+1}-S_{n}=0$ I have no idea how to approach this. Could someone point me in the right direction?

• You didn't show enough of an effort in trying the problem yourself so I just gave you a hint, but it should be a pretty big hint. Basically look at the coefficients in the equation, substitute, and factor. – Ahaan S. Rungta Dec 22 '14 at 14:48
• I really didn't know what to show, because I thought about this for about half an hour and couldn't come up with anything! And thank you, a hint is exactly what I was looking for. – user140161 Dec 22 '14 at 14:51

Here is a hint: multiply your equation by $x^n$. When you substitute $\alpha$ in the modified equation, what do you find? How can you relate that to the equation for $S_n$?

$$x^4-5x^2+2x-1=0$$

Multiply both sides by $x^n$

$$x^n(x^4-5x^2+2x-1)=x^n\cdot 0=0$$

$$x^{n+4}-5x^{n+2}+2x^{n+1}-x^n=0$$

Subsituting the roots, We have the four eq that :

$$\color{blue}{\alpha^{n+4}}-5\color{crimson}{\alpha^{n+2}}+2\color{purple}{\alpha^{n+1}}-\alpha^n=0$$ $$\color{blue}{\beta^{n+4}}-5\color{crimson}{\beta^{n+2}}+2\color{purple}{\beta^{n+1}}-\beta^n=0$$ $$\color{blue}{\gamma^{n+4}}-5\color{crimson}{\gamma^{n+2}}+2\color{purple}{\gamma^{n+1}}-\gamma^n=0$$ $$\color{blue}{\delta^{n+4}}-5\color{crimson}{\delta^{n+2}}+2\color{purple}{\delta^{n+1}}-\delta^n=0$$

$$(\color{blue}{\alpha^{n+4}+\beta^{n+4}+\gamma^{n+4}+\delta^{n+4}})-5(\color{crimson}{\alpha^{n+2}+\beta^{n+2}+\gamma^{n+2}+\delta^{n+2}})+2(\color{purple}{\alpha^{n+1}+\beta^{n+1}+\gamma^{n+1}+\delta^{n+1}})-(\alpha^n+\beta^n+\gamma^n+\delta^n)=0$$

Hint. \begin {align*} S_{n+4} - 5S_{n+2} + 2S_{n+1} - S_n &= \alpha^{n+4} + \beta^{n+4} + \gamma^{n+4} + \delta^{n+4} - 5 \cdot \left( \alpha^{n+2} + \beta^{n+2} + \gamma^{n+2} + \delta^{n+2} \right) + 2 \cdot \left( \alpha^{n+1} + \beta^{n+1} + \gamma^{n+1} + \delta^{n+1} \right) - \left( \alpha^n + \beta^n + \gamma^n + \delta^n \right). \end {align*}Now, see what you can factor out. The coefficients should have given you this suspicion.

• I don't see what I can factor out – user140161 Dec 22 '14 at 15:21
• Okay I think I got it..here's what I did. I first wrote the sum expression as $S_{n+4} - 5S_{n+2} + 2S_{n+1} - S_n= x^{n+4}-5x^{n+2}+2x^{n+1}-x^n$, then factored out $x^n$ and replaced $x^4-5x^2+2x-1$ with $0$. Is this what you were trying to get me to do? – user140161 Dec 22 '14 at 16:11
• Correct; you've got it! Glad I could help. :) – Ahaan S. Rungta Dec 22 '14 at 16:38

since $$S_{n+4}=xS_{n+3}-yS_{n+2}+zS_{n+1}-wS_{n}$$ use Vieta's Formula $$x=\alpha+\beta+\gamma+\delta=0$$ $$y=\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=-5$$ $$z=\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=-2$$ $$w=\alpha\beta\gamma\delta=-1$$ and so $$S_{n+4}-5S_{n+2}+2S_{n+1}-S_{n}=0$$

• How did you come up with $S_{n+4}=xS_{n+3}-yS_{n+2}+zS_{n+1}-wS_{n}$? – user140161 Dec 22 '14 at 15:22
• I really don't understand what you did, at all. – user140161 Dec 22 '14 at 15:58

Banging in your formula for $S_n$ and focusing on only one component I.e $\alpha$

$$\alpha^{n+4} -5\alpha^{n+2} + 2\alpha^n -\alpha^n = \alpha^n\left(\alpha^4 -5\alpha^2 +2\alpha -1\right) = \alpha^n\cdot 0$$ The bracket term is Zero because it is the root of the fundemental equation.