KöMaL (Metresis) 1894, Problem 8

Problem:
If $2 \cos \vartheta = u+\dfrac{1}{u}$, prove that $2 \cos n \vartheta = u^n+\dfrac{1}{u^n}$ for every positive integer $n$.

Solution 1:
We prove the statement by induction on $n \geq 2$. If $n=2$ we have $$2 \cos 2\vartheta =4 \cos^2 \vartheta - 2=\left(u+\dfrac{1}{u} \right)^2 - 2=u^2+\dfrac{1}{u^2}.$$ Assume that the statement is true for every $2 \leq k \leq n$. Then, $$\begin{eqnarray*} 2 \cos (n+1)\vartheta &= &2[2\cos n \vartheta \cos \vartheta - \cos (n-1)\vartheta]\\ &= & 2\left[\left(u^n+\dfrac{1}{u^n}\right)\dfrac{1}{2}\left(u+\dfrac{1}{u}\right)-\dfrac{1}{2}\left(u^{n-1}+\dfrac{1}{u^{n-1}}\right)\right] \\ & = & u^{n+1}+\dfrac{1}{u^{n+1}}, \end{eqnarray*}$$ and the statement follows.

Solution 2:
From the given equality, we obtain $u^2-(2\cos \vartheta) u + 1=0$, and solving this equation in $u$ we get $u=\cos \vartheta \pm i \sin \vartheta=e^{\pm i \vartheta}$. Then, for every positive integer $n$
$$2 \cos n \vartheta = e^{\pm in\vartheta}+e^{\mp in \vartheta}=u^n+u^{-n}=u^n+\dfrac{1}{u^n}$$ as we wanted to prove.