Competition problem — Algebra — Maths Olympiad Prep

Algebra Difficulty 4.9 AIME Prove it Romania

Let $S = x_1x_2 + x_3x_4 + \dots + x_{2015}x_{2016}$ , where $x_1, x_2, \dots, x_{2016} \in \{\sqrt{3} - \sqrt{2}, \sqrt{3} + \sqrt{2}\}$ . Is the equality $S = 2016$ possible?

Solution

The answer is in the affirmative.
The terms of the sum can be: $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2}) = 1$ , $(\sqrt{3}+\sqrt{2})^2 = 5+2\sqrt{6}$ or $(\sqrt{3}-\sqrt{2})^2 = 5-2\sqrt{6}$ . If there are $a$ terms equal to $1$ , $b$ terms equal to $5+2\sqrt{6}$ and $c$ terms equal to $5-2\sqrt{6}$ , then $a, b, c$ need to satisfy $a+b+c = 1008$ , $a+(5+2\sqrt{6})b+(5-2\sqrt{6})c = 2016$ . The last equality can be written $a+5b+5c-2016 = \sqrt{6}(2c-2b)$ . As $\sqrt{6}$ is irrational, it follows that $b=c$ and $a+5b+5c = 2016$ . Finally we obtain $a=756, b=c=126$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.