An algebra problem by Rakshit Joshi

From ${\color{#3D99F6}z^2-z+1=0}$ $\implies z^3-z^2+z=0$ $\implies z^3-({\color{#3D99F6}z^2-z+1})=-1$ $\implies z^3-({\color{#3D99F6}0})=-1$ $\implies {\color{#D61F06}z^3=-1}=e^{\pi i}$ $\implies z= e^\frac {\pi i}3 = e^\frac {2\pi i}6$ , the sixth root of unity. This means that $z^6 = 1$ and $z^n = z^{n \text{ mod 6}}$ . Now, we have:

$\begin{aligned} S & = \sum_{n=1}^{24} \left(z^n + z^{-n}\right)^2 & \small {\color{#3D99F6}\text{Since }z^n = z^{n \text{ mod 6}}} \\ & = 4 \sum_{n=1} ^6 \left(z^n + z^{-n}\right)^2 \\ & = 4 \sum_{n=1}^6 \left(z^{2n} + 2 + z^{-2n}\right) \\ & = 48 + \sum_{n=1}^6 z^{2n} + \sum _{n=1}^6 z^{-2n} & \small {\color{#3D99F6}\text{Since }z^n = z^{n \text{ mod 6}}} \\ & = 48 + 2\sum_{n=1}^3 z^{2n} + 2\sum_{n=1}^3 z^{-2n} \\ & = 48 + 2\left(z^{2} + {\color{#D61F06}z^{4}} + z^{6} \right) + 2\left(z^{-2} + z^{-4} + z^{-6} \right) \\ & = 48 + 2\left(z^{2} + {\color{#D61F06}z(z^{3})} + 1 \right) + 2\left(z^6z^{-2} + z^6z^{-4} + 1 \right) \\ & = 48 + 2\left(z^{2} + {\color{#D61F06}z(-1)} + 1 \right) + 2\left({\color{#D61F06}z^4} + z^2 + 1 \right) \\ & = 48 + 4\left({\color{#3D99F6}z^2 -z + 1} \right) \\ & = 48 + 4\left({\color{#3D99F6}0} \right) \\ & = \boxed{48} \end{aligned}$

The equation $z^2-z+1$ can be manipulated into $z+z^{-1}=1$ . I will use that fact to derive the following relationship: $z^n+z^{-n}=(1)(z^n+z^{-n})=(z+z^{-1})(z^n+z^{-n})=z^{n+1}+z^{n-1}+z^{-(n-1)}+z^{-(n+1)}$ $\Rightarrow z^{n+1}+z^{-(n+1)}=z^n+z^{-n}-(z^{n-1}+z^{-(n-1)})$ This gives us a recursive formula. We know that $z^n+z^{-n}$ is equal to $1$ for $n=1$ ; and it is easily determined that it equals $-1$ for $n=2$ . Using the recursion, we get the sequence $1,-1,-2,-1,1,2,1,-1,...$ (where each term after the second is the difference of the previous two). We can see that it repeats every six terms, so we can calculate the sum of the squares of the first six terms, and then multiply the result by $4$ (since there are twenty-four terms). Hence the answer is $4(1+1+4+1+1+4)=48$ .