<s>[INST] Algebraic Solving Contest Problem The problem is as follows
If $x^2+x-1=0$, compute all possible values of $\frac{x^2}{x^4-1}$
This was a no-calculator 10 min for 2 problem format contest.
I started by using quadratic formula, but the answers I got were too ugly to be plugged into the equation before time ran out. I tried several algebraic manipulations, but after losing time on the other problem, i found no quick answer to this. I presume there is one step i'm missing to make that leap. Can this be done easily?
 [/INST] Let $a, b$ be the roots of $x^2+x-1=0$.  Then $a+b = -1, ab = -1$.  For simplicity,  consider now the reciprocals of the values we wish to find, i.e. $\dfrac{a^4-1}{a^2} = k_1$ and $\dfrac{b^4-1}{b^2} = k_2$, then 
$$k_1 + k_2 = a^2+b^2-\left(\frac1{a^2}+\frac1{b^2} \right), \quad k_1 k_2 = \frac{(a^4-1)(b^4-1)}{(ab)^2}$$
Now $a^2+b^2 = (a+b)^2-2ab = 3$, so $k_1 + k_2 = 3-3=0$.
Also $(a^4-1)(b^4-1) = 1+(ab)^4-(a^4+b^4) = 1+1-(3^2-2) = -5$, so $k_1 k_2 = -5$.
So $k_1, k_2$ are roots of $y^2-5=0 \implies \{k_1, k_2\} = \{\sqrt5, -\sqrt5\}$.  Take reciprocals now.
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