Let $x_1 , x_2 , \dots , x_n$ be $n$ integers. If $k > n>1$ is an integer, prove that the only solution to $x_1^2 + x_2^2 + \dots + x_n^2 = kx_1 x_2\dots x_n$ is $x_1 = x_2 = · · · = x_n = 0.$
Here is my progress. I used vietta jumping.
- Note that $x_1^2 + x_2^2 + \dots + x_n^2 \ge 0. $ Hence there are even number of $x_i$'s such that $|x_i|=-x_i.$ Let them be $a_1,a_2,\dots, a_{2k}.$
- Note that we can always replace all the $a_i\rightarrow -a_i$ and it won't effect our equation.
- So WLOG, we can assume $x_1, x_2 , \dots , x_n$ to be positive ( if anyone is $0$ we get $x_1 = x_2 = · · · = x_n = 0.$
) - Suppose for some fixed $k$ let $$x_1>x_2>\dots>x_n$$ be the solution such that $(x_1+x_2+\dots +x_n)$ is minimal.
- Then let $$f(t)=t^2-ktx_2\dots x_n+x_2^2+ \dots + x_n^2.$$ Note that $x_1$ is a root. Let the other root be $w.$
- Then since $$w=kx_2\dots x_n -x_1\implies w\in \Bbb Z.$$ And since $$w=\frac{x_2^2+ \dots + x_n^2}{x_1}\implies w>0.$$ So $w$ is a positive integer.
- Now $$f(x_2)=x_2^2-kx_2^2x_3\dots x_n+x_2^2 + \dots + x_n^2\le x_2^2(n-kx_3\dots x_n).$$
- Now as $k>n\implies f(x_2)$ is negative.
- So by IVT, we get $w< x_2<x_1.$
- Which is a contradiction as we are getting $(w+x_2+\dots+x_n)<(x_1+\dots+x_n).$ And we had assumed $(x_1+x_2+\dots +x_n)$ is minimal.
- So $$x_1>x_2\dots >x_n\implies x_1 = x_2 = · · · = x_n = 0.$$
- Now if we have $x_1=x_2.$ We get $$2x_1^2+x_2^2+\dots +x_n^2=kx_1^2x_3\dots x_n$$
- Now this implies $x_1^2|x_2^2+\dots +x_n^2.$
I don't know how to proceed with this case, since it's not symmetric, so I can't use vietta jumping. I tried using modulo $l$ but did not progress as $n$ can we anything. Bounding wont work too as $k$ can be anything. Any hints?
Best Answer
Almost there! Here's a loosely stated argument$-$
Assume the contrary; WLOG let $x_1\geq x_2\geq \cdots\geq x_n\geq 1$. As you already showed $f(x_2)\geq 0$, by using minimality argument, we have $$0\overset{k>n}{>}x_2^2(n-k\prod_{i=3}^{n}x_i)=2x_2^2-kx_2^2\prod_{i=3}^{n}x_i+(n-2)x_2^2\geq 2x_2^2-kx_2^2\prod_{i=3}^{n}x_i+\sum_{i=3}^{n} x_i^2\overset{f(x_2)\geq 0}{\geq} 0$$which is absurd.$\tag*{$\blacksquare$}$