# How to show that $x_{k+1}+x_{k+2} + \cdots + x_n < 2m$?

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.

• I believe it's possible to assume $m\geq 1$ is a real number, and apply Lagrange multipliers. – LeechLattice Jan 26 at 10:28
• @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? – Jianrong Li Jan 26 at 12:00

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.