bumb anyone?

Can someone check this?

I just realized the question says $a_1+a_2+...+a_n=0$, I am wrong

$n \le 4$ 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.

This solution is the best solution I've ever seen.