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Sixteenth STARS OF MATHEMATICS Competition

Algebra Difficulty 5.9 AIME, harder Prove it Romania

Let $n \in \mathbb{N}$ , $n \ge 4$ and $a_1, a_2, \dots, a_n$ be real numbers such that
$a_k^3 = a_{k+1}^2 + a_{k+2}^2 + a_{k+3}^2$
for all $k \in \{1, 2, \dots, n\}$ – indices are considered modulo $n$ . Show that $a_1 = a_2 = \dots = a_n$ .

Solution

Suppose there exists $i \in \{1, 2, \dots, n\}$ for which $a_i = 0$ . Clearly $a_{i+1} = 0$ , and consequently $a_1 = a_2 = \dots = a_n = 0$ .

Suppose now that $a_i \ne 0$ for all $i \in \{1, 2, \dots, n\}$ . Choose $p, q \in \{1, 2, \dots, n\}$ such that $a_q \le a_i \le a_p$ for all $i \in \{1, 2, \dots, n\}$ .

Now $a_p^3 = a_{p+1}^2 + a_{p+2}^2 + a_{p+3}^2 \le 3a_p^2$ and $a_q^3 = a_{q+1}^2 + a_{q+2}^2 + a_{q+3}^2 \ge 3a_q^2$ yield $a_p \le 3 \le a_q$ , therefore $a_1 = a_2 = \dots = a_n = 3$ .

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