<s>[INST] 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
 [/INST] $\displaystyle (x^n + \frac 1 {x^n})(x + \frac 1 x) = x^{n+1} + x^{n-1} + \frac 1 {x^{n-1}} + \frac 1 {x^{n+1}} \\
\displaystyle \Rightarrow x^{n+1} + \frac 1 {x^{n+1}} = -(x^n + \frac 1 {x^n}) - (x^{n-1} + \frac 1 {x^{n-1}})$
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.
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