Problem 1079

Solution to Exercise 10

First, note that if a sequence $\left(u_{n}\right)$ is $t$-periodic, then for all $n$, $u_{n}=u_{n+t}=u_{n+2t}$, for example. Thus, we can prove by induction that for all natural numbers $k$ and $n$, $u_{n+kt}=u_{n}$: the sequence is $kt$-periodic for all $k$.

Let $n$ be a natural number. Then, since the sequence is 2022-periodic,

$$u_{n+1}=u_{n+1+2 \times 2022}=u_{n+4045}=u_{n+5 \times 809}$$

But the sequence is 5-periodic, so this also equals $u_{n}$. Thus, for all natural numbers $n$, we have $u_{n+1}=u_{n}$. We deduce by induction that for all $n$, $u_{n}=u_{0}$, and therefore the sequence $\left(u_{n}\right)$ is constant.

Grader's Comment: The exercise was successfully completed by almost everyone who attempted it, using three main methods:

- Showing that $u_{0}=u_{1}=u_{2}=u_{3}=u_{4}$.
- Showing that $u_{n}=u_{n+1}$.
- Using Bézout's theorem to show $u_{0}=u_{n}$.

The few students who did not get 7 points mostly used somewhat strong properties without justifying them (even a quick justification would have sufficed).