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Problem 828

AMC 12 late, AIME early

Number theory Difficulty 4.5 Prove it NMO Selection Tests for the Junior Balkan Mathematical Olympiad · Romania

Let $n$ be a positive integer and consider the integers $x_1, x_2, \dots, x_n$ , $y_1, y_2, \dots, y_n$ such that

a) $x_1 + x_2 + \dots + x_n = y_1 + y_2 + \dots + y_n = 0$ ;

b) $x_1^2 + y_1^2 = x_2^2 + y_2^2 = \dots = x_n^2 + y_n^2 = 0$ .

Prove that $n$ is an even number.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Set $a = x_1^2 + y_1^2$ . If $a$ is odd, the numbers $x_i$ and $y_i$ do not have the same parity, so $x_i + y_i$ is odd. Since $\sum (x_i + y_i) = 0$ , it follows that $n$ is even.

Suppose $a = 4k+2$ . Then $x_i$ and $y_i$ are both odd. The equality $x_1+x_2+\dots+x_n = 0$ implies $n$ is even.

Finally, if $a = 4k$ , then $x_i$ and $y_i$ are even. The numbers $a_i = \frac{x_i}{2}$ and $b_i = \frac{y_i}{2}$ satisfy the initial conditions. Furthermore, $a_1^2 + b_1^2 = \frac{a}{4}$ . Repeating the argument for $\frac{a}{4}$ instead of $a$ , after a finite number of steps we end up in a previous case.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.