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I heard about this problem and I don't know how to solve it.

Can we divide the first $40$ positive integers in $4$ arrays such that for every array choosing any $x, y, z$ (not necessarily distinct) from that array we have $x+y \not= z$ ?

I think that the answear is no, but I don't know to prove it. I found a demonstration that we can't divide the first 16 numbers in 3 arrays using Dirichlet's Principle, but we can't apply that at this problem.

Solution for $16$ numbers in $3$ arrays

In at least an array (let's say $A$) we have $6$ numbers, let's call them $a_1< a_2 < ... < a_6$. All the diferences $a_i-a_1, i>1$ must be in the other 2 arrays . In one of the others arrays we will have 3 differences $b_1, b_2, b_3$ (let' say they are in $B$).Then the diferences $b_2-b_1, b_3-b_2, b_3-b_1$ must be in the third array and this is not possible.

razvanelda
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  • Note that the we can also look at the differences instead of the sums because $x+y=z$ is equivalent to $x=z-y$. I have no concrete idea, but I think the pigeonhole-principle can prove that we cannot divide the $40$ numbers in the required way. – Peter Aug 30 '17 at 10:14
  • @Peter This is how you prove for $16$ numbers in $3$ arrays, I tried to apply this method for $40$ numbers but it doesn't work. – razvanelda Aug 30 '17 at 10:33
  • Are you sure ? We might need a more complicated variant of the pigeonhole-principle. Unfortunately, I am not an expert of this topic. If there is a possibility to do it that way, you will have an answer soon, since at this site , there are several experts, as I saw when similar questions were asked. – Peter Aug 30 '17 at 10:36
  • @Peter I am not sure, I am not an expert at this topic too :)) – razvanelda Aug 30 '17 at 10:37
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    @razvanelda can you share us this demonstration for 16 numbers? – Jaroslaw Matlak Aug 30 '17 at 10:40
  • @JaroslawMatlak did it – razvanelda Aug 30 '17 at 10:52
  • @razvanelda Why can the three differences not be in the third array ? – Peter Aug 30 '17 at 11:01
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    @Peter $(b_2-b_1)+(b_3-b_2)=(b_3-b_1)$ – razvanelda Aug 30 '17 at 11:03
  • @razvanelda Ah, I see. If we have $4$ arrays things are much more complicated, but I guess we can still apply the main idea. The first step is noting that at least one of the arrays contains at least $10$ numbers, so $9$ differences, just try to continue as in the given proof, maybe it turns out that a similar proof exists for $4$ arrays. – Peter Aug 30 '17 at 11:05
  • @Peter Nope, it doesn't work – razvanelda Aug 30 '17 at 11:11

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