Exercise 8. Let $a$ and $b$ be two real numbers. We define the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ by $a_{0}=a, b_{0}=b$ and for all $n$ a natural number, $a_{n+1}=a_{n}+b_{n}$ and $b_{n+1}=a_{n} b_{n}$. Determine all pairs $(a, b)$ such that $\mathrm{a}_{2022}=\mathrm{a}_{0}$ and $\mathrm{b}_{2022}=\mathrm{b}_{0}$.
## Solution to Exercise 8

Suppose that $(\mathrm{a}, \mathrm{b})$ is a solution to the statement. We calculate

$$
\sum_{i=1}^{2022} a_{i}=\sum_{i=1}^{2022}\left(a_{i-1}+b_{i-1}\right)=\sum_{i=0}^{2021} a_{i}+\sum_{i=0}^{2021} b_{i}
$$

But we have

$$
\sum_{i=0}^{2021} a_{i}=a_{0}+\sum_{i=1}^{2021} a_{i}=a_{2022}+\sum_{i=1}^{2021} a_{i}=\sum_{i=1}^{2022} a_{i}
$$

so we can rewrite the previous equation as

$$
\sum_{i=0}^{2021} b_{i}=0
$$

Similarly, we have

$$
0=\sum_{i=1}^{2022} b_{i}=\sum_{i=1}^{2022} a_{i-1} b_{i-1}=\sum_{i=0}^{2021} a_{i} b_{i}
$$

To make the products $a_{i} b_{i}$ appear, we square the equation $a_{i+1}=a_{i}+b_{i}$ to get $a_{i+1}^{2}=$ $a_{i}^{2}+b_{i}^{2}+2 a_{i} b_{i}$. We sum this from $i=0$ to 2021 to get

$$
\sum_{i=1}^{2022} a_{i}^{2}=\sum_{i=0}^{2021} a_{i}^{2}+\sum_{i=0}^{2021} b_{i}^{2}+2 \sum_{i=0}^{2021} a_{i} b_{i}
$$

But again,

$$
\sum_{i=1}^{2022} a_{i}^{2}=\sum_{i=0}^{2021} a_{i}^{2}
$$

so

$$
\sum_{i=0}^{2021} b_{i}^{2}=0
$$

A sum of squares of real numbers is necessarily non-negative, and can only be zero if all the $b_{i}$ are zero. This implies in particular that $b=0$. Conversely, if $a$ is a real number, the pair $(a, 0)$ works because we show by induction that for all $n$, $a_{n}=a$ and $b_{n}=0$, and then $a_{2022}=a=a_{0}$.

Comment from the graders: The exercise was very difficult, was rarely attempted, and even less often successfully completed. However, several students manage to obtain that $\mathrm{b}_{0}+\cdots+\mathrm{b}_{2021}=0$, which is a good start.