# Solve the generalised diophantine equation: $x_1^2 + x_2^2 + \dots + x_n^2 = kx_1 x_2\dots x_n$

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 $$w< x_2
• Which is a contradiction as we are getting $$(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?

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

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.$$\tag*{\blacksquare}$$

• :O thanks! I am still verifying it, but it looks correct :O P.S. Look who posted an answer :O Oct 6, 2021 at 23:40

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.

• page 189, number (15) zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf Oct 5, 2021 at 12:49
• i see, how it's coming but i don't know where it's useful :( Oct 5, 2021 at 23:39