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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?

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  • $\begingroup$ Why you must have strict inequality $x_1>x_2>\dots>x_n$ in the fourth bullet point? $\endgroup$ Oct 5, 2021 at 7:44
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    $\begingroup$ @user10354138 I wrote suppose we have that strict ineq. That's what I couldn't prove, the case when $x_1=x_2$ $\endgroup$ Oct 5, 2021 at 7:51
  • $\begingroup$ Do you allow $n=1$? If so any $x_1=k>1$ is a solution. $\endgroup$ Oct 5, 2021 at 9:30
  • $\begingroup$ @AdamBailey woopsie, I think the question is meant for $n>1.$ $\endgroup$ Oct 5, 2021 at 9:49
  • $\begingroup$ see zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf $\endgroup$
    – Will Jagy
    Oct 5, 2021 at 12:51

2 Answers 2

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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$}$

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    $\begingroup$ :O thanks! I am still verifying it, but it looks correct :O P.S. Look who posted an answer :O $\endgroup$ Oct 6, 2021 at 23:40
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Hint: For the case $x_1 = x_2\geq x_3 \geq \dots \geq x_n \geq 1$, rearrange as:

$$\frac{x_1^2}{x_1x_2\dots x_n}+\frac{x_2^2}{x_1x_2\dots x_n}+\dots + \frac{x_n^2}{x_1x_2\dots x_n}=k$$

and consider the range of possible values of each of the fractions on the left hand side.

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