The question is very simple and I apologize for that, but I am not an expert of this kind of problem. Given the polynomial $$ P(x_1,\ldots,x_{2n})=x_1^2+\ldots+x_n^2-x_{n+1}^2-\ldots-x_{2n}^2,$$ I would like to know if there are non trivial integer roots $(y_1,\ldots, y_{2n})$ such that $$y_1+\cdots+y_{n}=y_{n+1}+\cdots+ y_{2n}.$$ With non trivial I mean the ones like $$y_1=y_{n+1},\ldots,y_{n}=y_{2n},$$ or their permutations.
$\begingroup$
$\endgroup$
3
-
5$\begingroup$ See en.wikipedia.org/wiki/… for a more general problem. $\endgroup$– Richard StanleyDec 13, 2016 at 13:54
-
$\begingroup$ You may also be interested by Tito Pieza's excellent blog: sites.google.com/site/tpiezas/019 $\endgroup$– WolfgangDec 14, 2016 at 8:58
-
$\begingroup$ See also mathoverflow.net/questions/103044/question-on-sums-of-squares/… $\endgroup$– WolfgangDec 14, 2016 at 9:47
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Fix large $N$ and consider all $n$-tuples $(x_1,\dots,x_n)\in \{1,\dots,N\}^n$. There are $N^n$ such $n$-tuples, at least $N^n/n!$ tuples modulo permutations, and for them the pairs $(x_1+\dots+x_n,x_1^2+\dots+x_n^2)$ take at most $n\cdot N\cdot n\cdot N^2=n^2N^3$ possible values. Thus by pigeonhole principle some value is obtained at least $N^{n-3}/(n^2\cdot n!)$ times. This is greater than 1 if $n>3$ and $N$ is chosen large enough.
answered Dec 13, 2016 at 16:05
$\begingroup$
$\endgroup$
Yes, there are. For instance, (1,4,6,7,2,3,5,8)
The general principle behind this solution is that $$ n^2+(n+1)^2=((n-1)^2+(n+2)^2)-4. $$ Combining two such collections on the opposite sides always gives you a solution.
answered Dec 13, 2016 at 15:14