Determine the positive integers $n \ge 3$ such that, for every integer $m \ge 0$, there exist integers $a_1, a_2,..., a_n$ such that $a_1 + a_2 +...+ a_n = 0$ and $a_1a_2 + a_2a_3 + ...+a_{n-1}a_n + a_na_1 = -m$ Alexandru Mihalcu
Problem
Source: 2018 Romania JBMO TST 2.1
Tags: Sum, algebra, number theory
24.06.2020 12:03
bumb anyone?
05.05.2022 19:40
I just realized the question says $a_1+a_2+...+a_n=0$, I am wrong
08.11.2024 15:13
$n \le 4$ and $m = 2$ $\implies$ There is no solution. Let's prove this. $n = 4$ $\implies$ $(a_1 + a_3)(a_2 + a_4) = a_1a_2 + a_2a_3 + a_3a_4 + a_4a_1 = -2$ Let $(a_1 + a_3) = x$ $\implies$ $(a_2 + a_4) = -x$ $\implies$ $(a_1 + a_3)(a_2 + a_4) = -x^2$ $-2 $ ≠ $ -x^2$ $\implies$ No Solution $n = 3 \implies a_3 = -(a_1 + a_2)$ $\implies$ $a_1a_2 - (a_1 + a_2)^2 = a_1a_2 + a_2a_3 + a_3a_1 = -2$ $\implies$ $a_1a_2 - (a_1 + a_2)^2 = -(a_1^2 + a_1a_2 + a_2^2) = -2$ And $a_1$ or $a_2 = 0$ $\implies$ $a_1^2 + a_1a_2 + a_2^2$ must be perfect square $\implies$ No Solution $a_1a_2 > 0$ $\implies$ $-(a_1^2 + a_1a_2 + a_2^2) \le -3$ $\implies$ No Solution $a_1a_2 < 0$ $\implies$ Let $|a_1| \ge |a_2|$ $\implies$ If $|a_1| = |a_2|$ $\implies$ $a_1^2 + a_1a_2 + a_2^2$ must be perfect square $\implies$ No Solution If $|a_1| > |a_2|$ $\implies$ $-(a_1^2 + a_1a_2 + a_2^2) \le -(|a_1| + a_2^2) \le -3$ $\implies$ No Solution $n = 5$ $\implies$ $a_1 = 0, a_2 = m -1, a_3 = 0, a_4 = -m, a_5 = 1$ $\implies$ Provides for each $m$ integer $n \ge 6$ $\implies$ $a_1 = 0, a_2 = m - 1, a_3 = 0, a_4 = -m, a_5 = 1$ and $a_6 = a_7 = ... = a_n =0$ $\implies$ Provides for each $m$ integer And the solution ends.
19.11.2024 15:39
This solution is the best solution I've ever seen.