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Let $\mathbb{F}_p$ be a finite field (for a prime integer $p$), and let $x_1, \ldots, x_{2p-1}$ be any sequence of elements from it. Prove that I can chose $p$ elements from this sequence such as their sum is zero.

I think there can be useful probability method. But I can't it.

Jyrki Lahtonen
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Andrey Komisarov
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    What do you think should be the approach to the problem? Where are you getting stuck? – gt6989b May 10 '19 at 20:39
  • @gt6989b probability method is when you have a set and you know that random element from it has property A with probability >0, so you can say that in a there are element with property A. – Andrey Komisarov May 10 '19 at 20:43
  • I know what the probabilistic method is :) what I don't understand is where does the probability enter in your problem. How are you setting up the sample space? – gt6989b May 10 '19 at 20:52
  • @gt6989b I think will exist a way to prove that probability of sum of random subsequence with p element is zero is >0 for each 2p-1-sequence – Andrey Komisarov May 10 '19 at 20:57
  • but maybe it won't – Andrey Komisarov May 10 '19 at 20:58
  • How do you suggest to go about proving this? I am also wondering if you replace the field $\left(\mathbb{F}_p,+,\cdot\right)$ by the group $\left(\mathbb{Z}_p,+\right)$, how will that change the problem? – gt6989b May 10 '19 at 20:58
  • yes Zp is better, but it not seems important – Andrey Komisarov May 10 '19 at 21:01
  • My hunch is, this is a Pigeonhole principle problem. I would think about taking the first $p-1$ elements $x_1, \ldots, x_{p-1}$ and thinking what other one can you add from the last $p$ elements that would make it work... But I am not sure about it – gt6989b May 10 '19 at 21:05
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    But it is not true that for every first p-1 you can chose last element – Andrey Komisarov May 10 '19 at 21:15
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    I replaced the probability tag with [tag:combinatorics] because when we are dealing with certainties probability does not feel right. At the same time I removed the sequences tag, because in my opinion in a sequence the order of the elements is important but here it is not. This is just semantics (language?), hopefully no problem. Anyway, the problem rang a bell, so I went to look for it, a found a duplicate. The solution needs Chevalley-Warning which is not trivial (but not exceedingly hard either). – Jyrki Lahtonen May 11 '19 at 04:50

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