Suppose $f:\mathbb{Z}^+\longrightarrow X$ is a function, with $X$ a finite set. Is it true that there are $a,b\in\mathbb{Z}^+$ such that $f(a)=f(b)=f(a+b)$.
Coloring of positive integers
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$\begingroup$
combinatorics
elementary-set-theory
coloring
ramsey-theory
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0This is known as [Schur's theorem](http://en.wikipedia.org/wiki/Schur%27s_theorem). – 2015-10-19
1 Answers
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I believe, though I am not sure, this is what you are looking for: http://books.google.com/books?id=hUGxm9RdTeUC&pg=PA4&lpg=PA4&dq=schur+coloring+of+positive+integers&source=bl&ots=IrYWq_eBSY&sig=wkWhM0nQfCAbAGPCZFoYDHwRYFE&hl=en&sa=X&ei=sWaTUKGjOITk0QHSq4HoBQ&ved=0CE0Q6AEwBw#v=onepage&q=schur%20coloring%20of%20positive%20integers&f=false
Please correct me if I am wrong. This is not something I am familiar with, but is nonetheless very interesting.
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0Thought I'd include the proof on the site for completeness. Let g:\{(i,j)\in\mathbb{Z}^+\times\mathbb{Z}^+:i
such that $g(i,j)=f(j-i)$. By Ramsey's Theorem there are i such that $g(i,j)=g(j,k)=g(i,k)$, i.e. $f(j-i)=f(k-j)=f(k-i)$. The result follows with $a=j-i$ and $b=k-j$.$~$ Thanks for finding this. That book looks like it has some nice historical context also. – 2012-11-03