Colour the integers 1,2,...,1978 with 6 colours. Prove that there are three integers x,y,z in the set that are of the same colour, and x+y=z.
I have just joined this site, I'm informed in combinatorics, and have just started looking into Ramsey Theory. In my research I found this problem, and would be curious to see what approaches anyone would have for the proof.
I would like to give more context, although I was told this question verbally from an acquaintance when asking for problems in Ramsey Theory and I can't seem to find it anywhere on the internet.
Any thoughts or solutions would be exceedingly helpful, or if anyone could recognize the problem or it's source.
This is IMO 1978/6. A solution is provided here: https://www.math.cmu.edu/~ploh/docs/math/mop2009/combin-gems.pdf (page 4).
A summary: 1978 is larger than the 6-color Ramsey number for triangles. Let $G$ be a complete graph with vertex set ${0, 1, ..., 1978}$, and color the edge $ij$ with the color of $|i-j|$. Then there exists a monochromatic triangle, so some $|i-j|, |j-k|, |k-i|$ all have the same color. Two of those must sum to the third.