rogue wrote: Is it true that for every $n\ge 2021$ there exist $n$ integer numbers such that the square of each number is equal to the sum of all other numbers, and not all the numbers are equal? (O. Rudenko) Let $\{x_i\}_{i=1}^n$ be the searched numbers and $S=\sum_{i=1}^nx_i$ Constraint is $x_i^2=S-x_i$ and so all $x_i$ are solutions of $X^2+X-S=0$ So the set is made of two integers (since not all equal) $-1-k$ and $k$ such that $k(k+1)=S$ Suppose we have $m$ numbers ($m\in\{1,2,...,n-1\}$) $-1-k$ and $n-m$ numbers $k$ We have $S=m(-1-k)+(n-m)k=k(k+1)$ And so $m=\frac{-k^2+(n-1)k}{2k+1}$ which is $-4m=2k+1-2n+\frac{2n-1}{2k+1}$ If $2n-1$ is a prime number, this implies $2k+1\in\{-2n+1,-1,1,2n-1\}$ $2k+1=-2n+1$ would imply $m=n$, which does not fit. $2k+1=-1$ would imply $m=n$, which does not fit. $2k+1=+1$ would imply $m=0$, which does not fit. $2k+1=2n-1$ would imply $m=0$, which does not fit. And so $\boxed{\text{At least no such numbers when }2n-1\text{ is prime}}$