Let X be a subset of {1,2,...,2014}.
How would I show that if |X| ≥ 64, then there exist at least two different pairs {x,y} and {u,v} of distinct elements of X which |x-y|=|u-v|?
I'm not too sure how to go about this problem. Any help is appreciated.
On
Order the elements of $X$, $x_1<x_2<\dots<x_{64}$.
If all the gaps between consecutive elements ($x_{i+1}-x_i$) were different, we'd have that the sum of the gap sizes is at least $1+2+\cdots+63=\frac{63\cdot 64}{2}>2014$. Hence at least two pairs of the consecutive elements must have the same gap size.
Since $|X|\ge 64$, there exist $\frac12 \times 64\times 63 =\frac12 \times 4032=2016$ unordered pairs. And any pair satisfies $0 <|x-y|\le 2013$ since $X \subset \{1,2.\cdots,2014\}$ .
As we kmow $2013< 2016$, therfore there must exist two pairs $\{x,y\}$ and $\{u,v\}$ of distinct elements of $X$ for which $|x-y|=|u-v|$.