rogue wrote: Find all collections of $63$ 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 $\{x_i\}_{i=1}^{63}$ be the searched numbers and $S=\sum_{i=1}^{63}x_i$ 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 ($m\in\{1,2,...,62\}$) $-1-k$ and $63-m$ numbers $k$ We have $S=m(-1-k)+(63-m)k=k(k+1)$ And so $m=\frac{-k^2+62k}{2k+1}$ which is $-4m=2k-125+\frac{125}{2k+1}$ And so $2k+1\in\{\pm 1,\pm 5,\pm 25,\pm 125\}$ $2k+1\in\{-125,-1,1,125\}$ give $m=0$ or $m=63$, which does not fit (all numbers are equal) $2k+1=-25$ gives $k=-13$ and $m=39$ and solution $\boxed{\text{S1 : }\{\overbrace{12,...,12}^{39\times},\overbrace{-13,...,-13}^{24\times}\}}$ $2k+1=-5$ gives $k=-3$ and $m=39$ and solution $\boxed{\text{S2 : }\{\overbrace{2,...,2}^{39\times},\overbrace{-3,...,-3}^{24\times}\}}$ $2k+1=5$ gives $k=2$ and $m=24$ and solution $S2$ above $2k+1=25$ gives $k=12$ and $m=24$ and solution $S1$ above

Interesting problem