10.2. Find all sets of natural numbers $x_{1}, x_{2}, \ldots, x_{20}$ such that

$$
x_{i+2}^{2}=\operatorname{LCM}\left(x_{i+1}, x_{i}\right)+\operatorname{LCM}\left(x_{i}, x_{i-1}\right)
$$

for $i=1,2, \ldots, 20$, where $x_{0}=x_{20}, x_{21}=x_{1}, x_{22}=x_{2} . \quad$ (P. Kozlov)
Answer. $x_{1}=x_{2}=\ldots=x_{20}=2$.

Solution. From the condition, it follows that all $x_{i}$ are greater than 1, and $x_{i+2}^{2}$ is divisible by $x_{i}$ for $i=1,2, \ldots, 20$ (here and below $x_{j+20}=$ $=x_{j}=x_{j-20}$ for $\left.j=1, \ldots, 20\right)$.

Let $x_{k}$ be the largest of the numbers $x_{1}, \ldots, x_{20}$, and let $p$ be a prime divisor of the number $x_{k-5}$. Since $x_{k-3}^{2}$ is divisible by $x_{k-5}$, and $x_{k-1}^{2}$ is divisible by $x_{k-3}$, then $x_{k-3}$ and $x_{k-1}$ are divisible by $p$. Therefore, both $\operatorname{LCM}\left(x_{k-5}, x_{k-4}\right)$ and $\operatorname{LCM}\left(x_{k-4}, x_{k-3}\right)$ are divisible by $p$, so $x_{k-2}^{2}$ is divisible by $p$. Thus, the numbers $x_{k-3}, x_{k-2}, x_{k-1}$ are all divisible by $p$, so their pairwise GCDs are at least $p$. Therefore,

$$
\begin{gathered}
x_{k}^{2}=\operatorname{LCM}\left(x_{k-1}, x_{k-2}\right)+\operatorname{LCM}\left(x_{k-2}, x_{k-3}\right)= \\
=\frac{x_{k-1} x_{k-2}}{\operatorname{GCD}\left(x_{k-1}, x_{k-2}\right)}+\frac{x_{k-2} x_{k-3}}{\operatorname{GCD}\left(x_{k-2}, x_{k-3}\right)} \leqslant \\
\leqslant \frac{x_{k-1} x_{k-2}}{p}+\frac{x_{k-2} x_{k-3}}{p} \leqslant \frac{2 x_{k}^{2}}{p}
\end{gathered}
$$

Since $p \geqslant 2$, such a chain of inequalities can only hold if $p=2$ and all inequalities turn into equalities. In particular, $x_{k-2}=x_{k-1}=x_{k}$ and GCD $\left(x_{k-2}, x_{k-1}\right)=$ $=p=2$. Thus, $x_{k}=2$, and then all $x_{i}$ are equal to 2 (since $x_{k}$ is the largest of them, and all these numbers are greater than 1).

It remains to note that the set $x_{1}=x_{2}=\ldots=x_{20}=2$ satisfies the condition.