## Problem 1

Consider the sum $S=x_{1} x_{2}+x_{3} x_{4}+\ldots+x_{2015} x_{2016}$, where $x_{1}, x_{2}, \ldots, x_{2016} \in\{\sqrt{3}-\sqrt{2}, \sqrt{3}+\sqrt{2}\}$. Is it possible to have $S=2016$?

Cristian Lazăr
## Solution.

Answer: Yes.

The terms of the sum are of the form $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=1$, $(\sqrt{3}+\sqrt{2})^{2}=5+2 \sqrt{6}$, or $(\sqrt{3}-\sqrt{2})^{2}=5-2 \sqrt{6}$.

Let's say there are $a$ terms of the first type, $b$ terms of the second type, and $c$ terms of the third type.

We have $a+b+c=1008$ and $a+b(5+2 \sqrt{6})+c(5-2 \sqrt{6})=2016$, which is equivalent to $a+5 b+5 c-2016=\sqrt{6}(2 c-2 b)$. From here, we deduce that $b=c$.

We obtain $a+2 b=1008$ and $a+10 b=2016$. Therefore, $a=756$ and $b=c=126$.