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 using only elementary algebra or high school math
 A: $$(x+\frac{1}{x})(x^2+\frac{1}{x^2})\Rightarrow x^3+\frac{1}{x^3}=2$$
$$(x+\frac{1}{x})(x^3+\frac{1}{x^3})\Rightarrow x^4+\frac{1}{x^4}=-1$$
$$(x+\frac{1}{x})(x^4+\frac{1}{x^4})\Rightarrow x^5+\frac{1}{x^5}=-1$$
$$(x+\frac{1}{x})(x^5+\frac{1}{x^5})\Rightarrow x^6+\frac{1}{x^6}=2
\\ \vdots$$
$$x^{99}+\frac{1}{x^{99}}=2$$
and so you can obtain the result.
A: 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.
A: 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 
$$x^{99}+\frac{1}{x^{99}}=e^{i{66\pi}}+e^{-i{66\pi}}=2$$
A: $\displaystyle (x^n + \frac 1 {x^n})(x + \frac 1 x) = x^{n+1} + x^{n-1} + \frac 1 {x^{n-1}} + \frac 1 {x^{n+1}} \\
\displaystyle \Rightarrow x^{n+1} + \frac 1 {x^{n+1}} = -(x^n + \frac 1 {x^n}) - (x^{n-1} + \frac 1 {x^{n-1}})$
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.
