# Problem T-1 

Given a pair $\left(a_{0}, b_{0}\right)$ of real numbers, we define two sequences $a_{0}, a_{1}, a_{2}, \ldots$ and $b_{0}, b_{1}, b_{2}, \ldots$ of real numbers by

$$
a_{n+1}=a_{n}+b_{n} \quad \text { and } \quad b_{n+1}=a_{n} \cdot b_{n}
$$

for all $n=0,1,2, \ldots$ Find all pairs $\left(a_{0}, b_{0}\right)$ of real numbers such that $a_{2022}=a_{0}$ and $b_{2022}=b_{0}$.

Answer: All pairs $\left(a_{0}, b_{0}\right)$ where $b_{0}=0$ and $a_{0}$ is arbitrary.


Solution. Consider a function $\Phi$ of two real variables defined by $\Phi(a, b)=(a-1)^{2}+(b-1)^{2}$. We claim that $\Phi\left(a_{n+1}, b_{n+1}\right) \geq \Phi\left(a_{n}, b_{n}\right)$ for all $n$, with equality if and only if $a_{n} b_{n}=0$. Indeed, setting $a_{n}=a$ and $b_{n}=b$ we verify

$$
\Phi\left(a_{n+1}, b_{n+1}\right)-\Phi\left(a_{n}, b_{n}\right)=(a b-1)^{2}+(a+b-1)^{2}-(a-1)^{2}-(b-1)^{2}=(a b)^{2} \geq 0
$$

This means that

$$
\Phi\left(a_{2022}, b_{2022}\right) \geq \Phi\left(a_{2021}, b_{2021}\right) \geq \cdots \geq \Phi\left(a_{0}, b_{0}\right)
$$

so equalities must occur everywhere. We distinguish two cases:

(a) If $b_{0}=0$ then $a_{i}=a_{0}$ and $b_{i}=0$ for all $i=1, \ldots, 2022$ and we get a valid solution for arbitrary real number $a_{0}$.

(b) If $b_{0} \neq 0$ then from $\left(a_{0} b_{0}\right)^{2}=0$ we infer $a_{0}=0$. Thus $a_{1}=b_{0} \neq 0, b_{1}=0$ and from this point on $a_{i}=b_{0}, b_{i}=0$ for all $i=1, \ldots, 2022$. Hence we do not get any solution here.

Remark. Other functions work too. For example a function $\Phi_{2}$ defined by

$$
\Phi_{2}(a, b)=a^{2}-2 a-2 b
$$

gives an even slightly more convenient $\Phi\left(a_{n+1}, b_{n+1}\right) \geq \Phi\left(a_{n}, b_{n}\right)$ with equality if and only if $b_{n}=0$.
