# Solution to two non polynomial equation

If $$x+\frac1x=-1$$

Find the value of

$$x^{99}+\frac{1}{x^{99}}$$

Is there any formal (traditional) way to solve these problems.

• – Matti P. Jan 27 at 12:37

$$(x+\frac{1}{x})(x^2+\frac{1}{x^2})\Rightarrow x^3+\frac{1}{x^3}=2$$ $$(x+\frac{1}{x})(x^3+\frac{1}{x^3})\Rightarrow x^4+\frac{1}{x^4}=-1$$ $$(x+\frac{1}{x})(x^4+\frac{1}{x^4})\Rightarrow x^5+\frac{1}{x^5}=-1$$ $$(x+\frac{1}{x})(x^5+\frac{1}{x^5})\Rightarrow x^6+\frac{1}{x^6}=2 \\ \vdots$$ $$x^{99}+\frac{1}{x^{99}}=2$$ and so you can obtain the result.

• Thanks, that was clean! – Howard Jan 27 at 12:54

yeah actually there is

$$x^2 + x + 1$$ has two solution of form $$a_1,a_2$$ such that $$a_1^2=a_2\ and \ a_2^2=a_1$$ (they are complex roots so thats pretty much possible).

whichever root you pick $$a_1^3=a_2^3=1$$. Now i think you will be able to solve now. They are actually the roots of $$x^3=1$$ and I would suggest you to check out properties of numbers satisfying $$x^n$$=1.

For $$x\in\mathbb{R}$$ is $$\left|x+\frac1x\right|\geq 2,$$ therefore the number $$x$$ that satisfies $$x+\frac1x=-1$$ has a non-zero imaginary part.

Set $$x=r(\cos\varphi + i\sin\varphi),$$ then $$-1=x+\frac1x=(r+\frac1r)\cos\varphi+i(r-\frac1r)\sin\varphi.$$ From the above we have $$\sin\varphi\neq0$$, giving $$r=\frac1r=1$$ and $$\cos\varphi=-\frac{1}{2}.$$
Consequently, $$x=e^{i{2\pi\over3}}$$ or $$x=e^{-i{2\pi\over3}}$$ and $$x^{99}+\frac{1}{x^{99}}=e^{i{66\pi}}+e^{-i{66\pi}}=2$$

• Thank you @Servaes for pointing out my typo. Fixed. – user376343 Jan 27 at 21:20

$$\displaystyle (x^n + \frac 1 {x^n})(x + \frac 1 x) = x^{n+1} + x^{n-1} + \frac 1 {x^{n-1}} + \frac 1 {x^{n+1}} \\ \displaystyle \Rightarrow x^{n+1} + \frac 1 {x^{n+1}} = -(x^n + \frac 1 {x^n}) - (x^{n-1} + \frac 1 {x^{n-1}})$$

If we denote $$x^n + \frac 1 {x^n}$$ by $$f(n)$$ then we have just shown that

$$f(n+1) = -f(n) - f(n-1)$$

Since you know that $$f(0) = 2$$ and $$f(1)=-1$$, this recursion allows you to find $$f(n)$$ for any $$n >0$$. If you work out $$f(n)$$ for $$n=2,3,4 \dots$$ you won't have to go very far before you see a pattern.