. Consider the Fibonacci sequence defined by $F_{0}=0, F_{1}=1$, and for all $n \geqslant 1$, $F_{n+1}=F_{n}+F_{n-1}$. Determine all pairs of integers $(k, m)$ satisfying $m>k>0$ for which the sequence $x_{n}$ defined by $x_{0}=\frac{F_{k}}{F_{m}}$ and

$$
x_{n+1}=\frac{2 x_{n}-1}{1-x_{n}} \text { if } x_{n} \neq 1, \quad x_{n+1}=1 \text { if } x_{n}=1
$$

contains the number 1.

### 2.2 The solutions

## Geometry

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
. Suppose the pair $(k, m)$ is a solution, and let $n$ be the smallest index for which $x_{n}=1$. For all $s>1$ as soon as $k>1$. Thus, if $m>k+1$, we also have $\frac{F_{m}}{F_{k}}>2$ and therefore the desired equality could not be satisfied. Therefore, we must necessarily have $k=m+1$. Moreover, it is easily shown by induction that two consecutive terms of the Fibonacci sequence are coprime. It follows that the equality $\frac{F_{2 t+1}}{F_{2 t+2}}=\frac{F_{k}}{F_{m}}$ implies $k=2 t+1$ and $m=2 t+2$. In the end, the solutions to the problem are the pairs of the form $(2 t+1,2 t+2)$ with $t \geqslant 0$.

## 3 Test

For the tests, students were divided into three groups based on their level: the first group had to solve exercises 1, 2, and 3, the second group exercises 2, 3, and 4, and finally the last group exercises 3, 4, and 5.

### 3.1 The geometry test