Erken wrote: Find all natural numbers $ n$,such that there exist $ x_1,x_2\dots,x_n,y\in\mathbb{Z}$ and $ x_1 + x_2 + \dots + x_n = 0$ $ ny^2 = x_1^2 + x_2^2 + \dots + x_n^2$. I think you must add $ x_k\neq 0$, else it is trivial for any $ n>0$

Why?I add that $ y\neq 0$.Is it correct now?

Obviosly if $ n$ is solution, then $ n+2$ is solution too ($ x_{n+1}=y=-x_{n+2}$ work). n=2 solution, therefore any even n solution. It is not hard to prove, that n=1 and n=3 had not non trivial solution. I think n=5 had solution.

Could you show a solution to $ n=5$.

Erken wrote: Could you show a solution to $ n = 5$. $ 1+1+1+1+(-4)=0$ $ 1^2+1^2+1^2+1^2+(-4)^2=5\times 2^2$