
Problem 3. A positive integer $n$ and a Figure 4 . real number $\varphi$ are given in a way that $\cos \varphi=$ $\frac{1}{n}$. Find all positive integers $k$, for which the number $\cos k \varphi$ is an integer.


Solution. Case 1. $n=1$. Then $\cos \varphi=1$ and $\varphi=2 m \pi(m=0, \pm 1, \ldots)$. For all $k \in \mathbb{N}$ the number $\cos k \varphi$ is an integer.

Case 2. $n=2$. Then $\cos \varphi=\frac{1}{2}$ and $\varphi= \pm \frac{\pi}{3}+2 m \pi(m=0, \pm 1, \ldots)$. It is clear that for all $k \in \mathbb{N}$, which are divisible by 3 , the number $\cos k \varphi$ is an integer.

![](https://cdn.mathpix.com/cropped/2024_06_03_e679746950d7d3832835g-209.jpg?height=472&width=748&top_left_y=360&top_left_x=1231)

Case 3. $n \geq$ 3. We shall prove that for

all $k \in \mathbb{N}$ the number $\cos k \varphi$ is not an integer. Let $n$ be odd. We have $\cos \varphi=\frac{1}{n}, \cos 2 \varphi=$ $2 \cos ^{2} \varphi-1=\frac{2-n^{2}}{n^{2}}$ and $\left(2-n^{2}, n\right)=1$. We shall prove by induction that $\cos k \varphi=\frac{a}{n^{k}}$, where $(a, n)=1$. Assume that the assertion is true for all integers from 1 to $k$. We shall check it for $k+1$. From

$$
\cos (k+1) \varphi+\cos (k-1) \varphi=2 \cos k \varphi \cos \varphi
$$

it follows that $\cos (k+1) \varphi=\frac{2}{n} \cdot \frac{a}{n^{k}}-\frac{b}{n^{k-1}}=\frac{2 a-b n^{2}}{n^{k+1}}$, where $\cos k \varphi=\frac{a}{n^{k}}$ and $\cos (k-1) \varphi=$ $\frac{b}{n^{k-1}}$. We have $(a, n)=1$ and $(b, n)=1$ according to the inductive assumption. It is clear that $\left(2 a-b n^{2}, n\right)=1$ and this ends the proof. The case when $n$ is even is analogous. Now cos $k \varphi$ is expressed by a fraction which denominator is equal to $2 p^{k}$ with $n=2 p$, while the nominator has no common divisor with $p$.

Answer: $\quad$ if $n=1 \quad \forall k \in \mathbb{N}$;

if $n=2 \quad k=3 q$, where $q \in \mathbb{N}$;

if $n \geq 3$ there is no solution.

## Grade 11
