B1. Let the sequence $a_{n}$ satisfy the recursive relation $a_{n+1}=\frac{1+a_{n}}{1-a_{n}}$ for all natural numbers $n$. Determine all possible values of the first term $a_{1}$, for which the sequence $a_{n}$ will contain a term $\mathrm{z}$ with the value 2022.
B1. Using the recursive relation, we derive

$$
a_{n+2}=\frac{1+a_{n+1}}{1-a_{n+1}}=\frac{1+\frac{1+a_{n}}{1-a_{n}}}{1-\frac{1+a_{n}}{1-a_{n}}}=\frac{\left(1-a_{n}\right)+\left(1+a_{n}\right)}{\left(1-a_{n}\right)-\left(1+a_{n}\right)}=\frac{2}{-2 a_{n}}=-\frac{1}{a_{n}}
$$

Considering this relation, we also have

$$
a_{n+4}=-\frac{1}{a_{n+2}}=-\frac{1}{-\frac{1}{a_{n}}}=a_{n}
$$

The sequence $a_{n}$ is thus periodic with a period of 4, so it has only four different values, which are $a_{1}, a_{2}=\frac{1+a_{1}}{1-a_{1}}, a_{3}=-\frac{1}{a_{1}}$, and $a_{4}=-\frac{1}{a_{2}}=\frac{a_{1}-1}{a_{1}+1}$. We thus have four possibilities: $a_{1}=2022$, or $\frac{1+a_{1}}{1-a_{1}}=2022$, from which we get $a_{1}=\frac{2021}{2023}$, or $-\frac{1}{a_{1}}=2022$, from which we get $a_{1}=-\frac{1}{2022}$, or $\frac{a_{1}-1}{a_{1}+1}=2022$, from which we get $a_{1}=-\frac{2023}{2021}$. We must also verify that in these cases no term in the sequence is equal to 1, otherwise the sequence would not be well-defined by the given recursive relation. In all four cases, we get the sequence $\ldots, 2022,-\frac{2023}{2021},-\frac{1}{2022}, \frac{2021}{2023}, 2022, \ldots$, only starting at a different value.

2. method. From the recursive relation, we express $a_{n}=\frac{a_{n+1}-1}{a_{n+1}+1}$. Since some term in the sequence must be equal to 2022, using the derived formula, we calculate that the preceding terms must be $\frac{2021}{2023},-\frac{1}{2022},-\frac{2023}{2021}, 2022, \ldots$ From this, it follows that the sequence is periodic and that the term $a_{1}$ must be equal to $\frac{2021}{2023},-\frac{1}{2022},-\frac{2023}{2021}$, or 2022. In this case, the periodicity and the calculated values automatically imply that no term in the sequence is equal to 1.
3. method.
Derivation of the relation between $a_{n}$ and $a_{n+2}$ ..... 2 points.
Derivation of the relation between $a_{n}$ and $a_{n+4}$ ..... 2 points.
Correct calculation of at least two possible $a_{1}$ from $\left\{-\frac{2023}{2021},-\frac{1}{2022}, \frac{2021}{2023}\right\}$ ..... 1 point.
Correct calculation of all four possible $a_{1}$ ..... 1 point.
Verification that in all initial terms we get a well-defined sequence ..... 1 point.
Note. The trivial solution $a_{1}=2022$ is worth 0 points.
4. method.
Express $a_{n}$ in terms of $a_{n+1}$ ..... 1 point.
The idea of starting to calculate backwards from 2022 ..... 1 point.
Correct calculation of the previous two terms $\frac{2021}{2023}$ and $-\frac{1}{2022}$ ..... 1 point.
Correct calculation of $-\frac{2023}{2021}$ ..... 1 point.
Discovery that the sequence is periodic with a period of 4 ..... 2 points.
Description of all 4 possible first terms ..... 1 point.

Note. If the competitor realizes that the sequence is well-defined only in the case where no term is equal to 0, 1, or -1, they get 1 point (for verifying the well-definedness).