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Let $k \le n$ be positive integers and let $m$ be a positive integer. Assume that $x_1, \ldots, x_n$ are non-negative integers and \begin{align} & x_1^2 + x_2^2 + \cdots + x_n^2 - (k-2) m^2=2, \\ & x_1 + \cdots + x_n = k m, \\ & x_1 \ge x_2 \ge \cdots \ge x_n. \end{align} How to show that $x_{k+1}+x_{k+2} + \cdots + x_n < 2m$?

It is easy to see that the result is true for $k=1,2$.

In the case of $k=3$, we have \begin{align} & x_1^2 + x_2^2 + \cdots + x_n^2 = m^2 + 2, \\ & x_1 + \cdots + x_n = 3 m, \\ & x_1 \ge x_2 \ge \cdots \ge x_n. \end{align} We have to estimate the solutions of the above equations. Are there some method to do this? Thank you very much.

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  • $\begingroup$ I believe it's possible to assume $m\geq 1$ is a real number, and apply Lagrange multipliers. $\endgroup$ – LeechLattice Jan 26 at 10:28
  • $\begingroup$ @LeechLattice, thank you very much. I tried to use your method to prove the case of $k=3$, $n=7$, and it works. But when I tried the case of $k=4$, $n=7$, the critical values are $$ [x_1 = 0, x_2 = 0, x_3 = 0, x_4 = 1, x_5 = 1, x_6 = 1, x_7 = 1, l_1 = 1/2, l_2 = 0, m = 1], \\ [x_1 = 0, x_2 = 0, x_3 = 0, x_4 = -1, x_5 = -1, x_6 = -1, x_7 = -1, l_1 = -1/2, l_2 = 0, m = -1].$$ They don't satisfy the constrains $x_1 \ge x_2 \ge \cdots \ge x_n$. In some other cases, it is also possible that the critical values are not integers. Do you know how to solve these problems? $\endgroup$ – Jianrong Li Jan 26 at 12:00
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I am afraid it is not true. Test the situation when $x_2=x_3=\ldots=1$, equations read as $x_1^2+(n-1)=(k-2)m^2+2$, $x_1+(n-1)=km$, that gives $x_1^2-x_1=(k-2)m^2-km+2$. Let's think that both $k$ and $m$ are large (say greater than 1000). Then $x_1$ is something like $\sqrt{k}m$, $x_{k+1}+\ldots+x_n=n-k=km+1-k-x_1$ is something like $(k+o(k))m\gg 2m$.

It remains to find the solution of $x_1^2-x_1=(k-2)m^2-km+2$ with large $k$ and $m$. It reads as $x_1^2-x_1+m^2-2=k(m^2-m)$. For fixed $m$ it is solvable if $x^2-x+m^2-2$ may be divisible by $m^2-m$ (then it may be divisible for large $x$ that makes $k$ also large). Modulo $m$ we may choose $x\equiv 2\pmod 2$. Modulo $m-1$ we need $x^2-x-1\equiv 0$, so just take $m-1=a^2-a-1$ for some large $a$. Now combine solutions modulo $m$ and $m-1$ by Chinese remainders theorem.

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