Let $ [n] $ be the set $ \{1,2,\ldots n\}$.


A *summoid* is a subset $ A \subset [n] $ of the form $ \{a,b,a+b\} $ (you can choose a better name, if it doesn't exist already).


Now, I developed by accident this simple result:



> 
> There is no partition of $ [9] $ into (disjoint) summoids.
> 
> 
> 


I want to ask the following questions:


1. Is it true for general $ [3k] $ when $ k > 1 $? According to a computer program I found that it is true for $ [12] $, but it seems my method for $ [9]$ can't be applied to the general case (maybe my program for $[12]$ isn't trustworthy).
**According to the comment below by R. van Dobben de Bruyn**, there is no such partition when $ k \equiv 2,3 \pmod 4$.
**According to the comment below by Gerhard Paseman**, there is a counterexample for $ k = 5 $ that extends also to $ n = 3 \cdot 4^k $ and to $ n = 3 \cdot (1+4^k) $.
2. Is it useful for something?
3. Is there any study of such results?



Partition of [3n] into summoids
I have the following results:


* N = 12: 
(1, 11, 12)
(3, 7, 10)
(4, 5, 9)
(2, 6, 8)
* N = 15:
(1, 14, 15)
(3, 10, 13)
(4, 8, 12)
(5, 6, 11)
(2, 7, 9)
* N = 24:
(1, 23, 24)
(2, 20, 22)
(5, 16, 21)
(6, 13, 19)
(7, 11, 18)
(8, 9, 17)
(3, 12, 15)
* N = 27:
(1, 26, 27)
(2, 23, 25)
(4, 20, 24)
(7, 15, 22)
(8, 13, 21)
(9, 10, 19)
(6, 12, 18)
(3, 14, 17)
(5, 11, 16)


and so on.


My guess would be that there are always (a lot of) solutions for any $k \equiv 0, 1\mod 4$.
In my opinion, this is some kind of Goldbach-ish problem. It is likely to get some probabilistic heuristic, but may be difficult to prove.




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For reference, here I add the number of different solutions for each $k$.


* N = 12: 8 solutions;
* N = 15: 21 solutions;
* N = 24: 3040 solutions;
* N = 27: 20505 solutions;


and so on.




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EDIT:


Thanks to OEIS, we now have some reference.


In this paper: <http://oeis.org/A104429/a104429.pdf>, the existence of solutions is discussed in detail in section I.3 (starting from page 22).


