# Is it possible to extend phi to higher order polynomials?

Below, $x=\phi$ when $n=2$:

$$x^n-\sum_{i=1}^{n}x^{n-i}=0$$

($\phi$ being the golden ratio)

Is there a way to express $x$ in terms of $\phi$ for $n>2$?

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I hope you realize you can simplify that sum right ? I dont know if that helps much though. – mick Sep 2 '12 at 22:28
The positive roots $>1$ of $x^{n+1}-2x^n+1$.. looking at limiting ratios of successive terms defined by linear recurrence relations $a_{m+1}=a_m+\cdots+a_{m-n+1}$ are we? Relevant: mathworld.wolfram.com/Fibonaccin-StepNumber.html – anon Sep 2 '12 at 22:37
What do you mean by "express"? – Qiaochu Yuan Sep 2 '12 at 22:40

When $n=5$, we are talking about the roots of $x^5-x^4-x^3-x^2-x-1=0$. I suspect that, like most polynomials of degree 5, this one has Galois group $S_5$. If that's the case, then these roots can't be expressed in terms of the four arithmetical operations and square roots, cube roots, fifth roots, etc. It would follow that there's no expression in terms of the golden ratio (and arithmetic operations, and square roots, etc.).
It doesn't get any better for $n\gt5$.