75th NMO Selection Tests

The required sequences are all constant sequences of positive rational numbers. Clearly, any such satisfies $(\ast )$.

Let $(a_n{)}_{n\ge 1}$ be a sequence of positive rational numbers satisfying $(\ast )$. Then so does $(r{a}_n{)}_{n\ge 1}$, where $r$ is any positive rational number. Letting $r$ be the product of the denominators of $a_1$ and $a_2$, we may assume that $a_1$ and $a_2$ are integers. If $a_k$ and $a_{k+1}$ are integers, then so is $a_{k+2}$, as it is a root of the monic polynomial $X^2-a_{k+1}X+a_k^2-a_k{a}_{k+1}$ with integer coefficients. Inductively, the $a_n$ are all integers.

Write $(\ast )$ in the form $$(a_{k+1}-a_k)a_k=(a_{k+2}-a_{k+1})a_{k+2}.(\ast \ast )$$ If $a_1>a_2$, then $(\ast \ast )$ forces $(a_n{)}_{n\ge 1}$ to be a strictly decreasing sequence of positive integers and we reach a contradiction.

If $a_1<a_2$, then $(a_n{)}_{n\ge 1}$ is strictly increasing, by $(\ast \ast )$. $$a_{k+1}=\frac{a_k^2+a_{k+2}^2}{a_k+a_{k+2}}>\frac{(a_k+a_{k+2}{)}^2}{2(a_k+a_{k+2})}=\frac{1}{2}(a_k+a_{k+2})\text{for all }k\ge 1,$$ it follows that $a_{k+1}-a_k>a_{k+2}-a_{k+1}$ for all $k\ge 1$, so the $a_{n+1}-a_n$ form a strictly decreasing sequence of positive integers and we reach again a contradiction.

Finally, if $a_1=a_2$, then $(\ast \ast )$ forces $a_n=a_1$ for all $n\ge 1$, so the sequence $(a_n{)}_{n\ge 1}$ is indeed constant, as desired.