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.

KöMaL (Metresis) 1894, Problem 2

Problem:
Find all positive integers $N$ divisible only by $2$ and by $3$ such that $N^2$ has three times the number of divisors of $N$.

Solution:
Let $N=2^a 3^b$, where $a,b \in \mathbb{N}^*$. Then $N^2 = 2^{2a}3^{2b}$. So, it must be $$3(a+1)(b+1)=(2a+1)(2b+1) \iff ab-a-b-2=0 \iff (a-1)(b-1)=3.$$ Since $a-1 \geq 0$ and $b-1 \geq 0$, we obtain $$\left\{ \begin{array}{rcl} a-1 & = & 1 \\ b-1 & = & 3 \end{array} \right. \qquad \textrm{or} \qquad \left\{ \begin{array}{rcl} a-1 & = & 3 \\ b-1 & = & 1 \end{array} \right.$$ i.e. $a=2, b=4$ or $a=4, b=2$. Then all the positive integers required are $N_1=2^2 3^4=324$ and $N_2=2^4 3^2=144$.

KöMaL (Metresis) 1894, Problem 1

Problem: 
Find all the two digit numbers $\overline{ab}$ such that $\overline{ababab}+1$ is a perfect cube.

Solution:
Let $x=\overline{ab}$ be the two digit number. Then, $101010x+1=n^3$, for some positive integer $n$. Clearly, $10^6 \leq n^3 < 10^7$, i.e. $100 \leq n \leq 215$. Moreover,
$$101010x=n^3-1.$$ Since $101010=30\cdot3367$, it's easy to see that $n^3-1 \equiv 0 \pmod{30}$ if and only if $n \equiv 1 \pmod{30}$, i.e. $n-1$ is divisible by $30$. So $n \in \{121,151,181,211\}$ and an easy check shows that only $n=211$ works. In this case, $x=93$.