Non-negative integer solutions of a system of equations $\sum_{i=1}^{n} x_i^2 = 4k-6, \sum_{i=1}^{n} x_i = 2k$

Question

Fix $k \ge 3$, $n \ge 2k$. Consider the following system of equations: \begin{align} \sum_{i=1}^{n} x_i^2 = 4k-6, \\ \sum_{i=1}^{n} x_i = 2k. \end{align} It seems that the only non-negative integer solutions of the system of equations are (up to permutations of indices): \begin{align} x_1 = \ldots = x_{k-3} =2, x_{k-2}=x_{k-1}=x_k=x_{k+1}=x_{k+2}=x_{k+3}=1, x_{k+1} = \ldots = x_{n} =0. \end{align} Are these all the non-negative integer solutions? What are all the non-negative integer solutions of the system of equations? Thank you very much.

Edit: thanks for the answers. Sorry I realized that I need to add the condition $x_i \le 2$, $i \in \{1,2,\ldots,n\}$. Maybe after we add this condition, the non-negative integer solution is unique.

Answer 1

Here is another solution:

$x_1=\dots=x_9=1, x_{10}=3$ and $k=6$.

Added: There are infinitely many solutions if one allows $x_i$'s to be greater than $2$. If $x_i\in\{0,1,2\}$, clearly the only solution (up to permutations) is the one mentioned in the question: Let $a$ be the number of $2$'s and $b$ be the number of $1$'s among $x_i$'s. Then we have $4a+b=4k-6$ and $2a+b=2k$. Solving for $a$ and $b$ yields $a=k-3$ and $b=6$.

Answer 2

These are not all the non-negative integer solutions. We need to saticfy the condition \begin{align} \sum_{i=1}^{n} x_i(x_i-2) = -6. \end{align}

One may take arbitrary $x_1, \ldots, x_l$ and $x_{l+1}=\cdots=x_{l+m}=1$. Then for some large $m$ we shall have \begin{align} \sum_{i=1}^{l+m} x_i(x_i-2) = -6, \\ \sum_{i=1}^{l+m} x_i = 2k. \end{align} After that one may take $x_{l+m+1}=\cdots=x_{n}=0$. And if $n$ is large enough, then $n\ge 2k$ and \begin{align} \sum_{i=1}^{n} x_i(x_i-2) = -6, \\ \sum_{i=1}^{n} x_i = 2k. \end{align}