Stars of Mathematics Competition

The $x_i$ are either all equal to $1+\frac{2m}{n}=\frac{(m+1{)}^2}{m^2+1}$ or exactly one is equal to $m+1$ and the other are all equal to $1+\frac{1}{m}$. The verification offers no difficulty and is hence omitted.

Leaving aside the trivial case where the $x_i$ are all equal, consider a solution $x_1,x_2,\dots ,x_n$ whose entries are not all equal. Let $s=x_1^2+x_2^2+\cdots +x_n^2$ and notice that each $x_i$ is a root of the quadratic polynomial $2m{X}^2-sX+s$. Since the $x_i$ are not all equal, and each $x_i$ is positive (in fact, at least $1$), the roots $u$ and $v$ of this quadratic polynomial are distinct positive real numbers satisfying $2muv=s$. Let $k$ be the number of indices $i$ such that $x_i=u$, so $x_i=v$ for the remaining $n-k$ indices. We may and will assume that $k\ge \frac{n}{2}$; and since the $x_i$ are not all equal, $k\le n-1=m^2$. Write $2muv=s=k{u}^2+(n-k)v^2\ge 2uv\sqrt{k(n-k)}$, so $k(n-k)\le m^2$, and recall that $k\ge \frac{n}{2}$, to infer that $k\ge \frac{1}{2}(n+\sqrt{n^2-4m^2})=m^2$. Further, the condition $k\le m^2$ forces $k=m^2$ which in turn forces $v=mu$, by the preceding. Finally, since the $x_i$ add up to $n+2m=(m+1{)}^2$, it follows that $u=1+\frac{1}{m}$ and $v=m+1$, and the solution has the form stated in the first paragraph.