$\sum (\sqrt{x_k-k^2}-k)^2=0$ implies $x_k=2k^2$?

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Let $x_1,x_2,\ldots,x_n$ be reals numbers such that $$\sum_{k=1}^n k\sqrt{x_k-k^2}=\frac12\sum_{k=1}^n x_k$$ Find all possible $n$-tuples of solution.

So, I got the following solution from the proposer of this problem. The given equation is equivalent to $$\sum_{k=1}^{n} (\sqrt{x_k-k^2}-k)^2=0$$ Hence we have $\sqrt{x_k-k^2}=k$ for all $k$. Thus $x_k=2k^2$ $\forall k$ is the only possible solution.

But I seriously doubt this solution. The fact that $x_k$ is real does not necessarily imply $\sqrt{x_k-k^2}$ is real. So sum of squares being zero does not imply each term is zero in this case.