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Jan 6, 2014 at 23:53 vote accept Bjørn Kjos-Hanssen
Jan 6, 2014 at 18:49 answer added Will Sawin timeline score: 4
Jan 5, 2014 at 23:28 comment added Bjørn Kjos-Hanssen Thanks for interesting answers, feel free to turn the comments into an official answer.
Jan 5, 2014 at 21:56 comment added Will Sawin Certainly we can take $x_i<2N+4$, where $N$ is the number of hyperplanes. This is because there are $(N+1)^k$ points in the cube that are rational numbers of denominator $N+2$, and each hyperplane includes at most $(N+1)^{k-1}$ of them. so one of them must not be on any hyperplane. The number of hyperplanes is one less than the number of ways to put $4k$ balls in $k+1$ boxes, which is $\left(\begin{array}{c} 5k \\ k \end{array} \right)$. This gives a bound. By putting some effort into optimization, this can probably be improved dramatically.
Jan 5, 2014 at 21:02 comment added Bjørn Kjos-Hanssen Nice. I wonder if bounds on $ x_i $ follow from this?
Jan 5, 2014 at 20:07 comment added Will Sawin It follows from the fact that there are rational points in every nonempty open subset of $\mathbb R$. Consider the set of $x_1,\dots,x_k$ such that $1<x_1<2,\dots, 1<x_k<x_2$ and $\sum_{i=1}^k a_i x_i$ does not equal $\sum_{i=1}^k 2 x_i$ for each sequence $a_1,a_k$ satisfying $\sum_{i=1}^k a_i \leq 4k$ and one of the $a_i\neq 2$. This is an open cube minus finitely many hyperplanes, hence a nonempty open set. It has a rational point. Clear denominators to get integer solutions to your equation.
Jan 5, 2014 at 18:13 history edited Bjørn Kjos-Hanssen CC BY-SA 3.0
added 38 characters in body
Jan 5, 2014 at 16:47 history asked Bjørn Kjos-Hanssen CC BY-SA 3.0