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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 $f(a)=f(b)=f(a+b)$.

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up vote 4 down vote accepted

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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Nice find!${}{}$ –  wj32 Nov 2 '12 at 6:30
    
Thought I'd include the proof on the site for completeness. Let $$g:\{(i,j)\in\mathbb{Z}^+\times\mathbb{Z}^+:i<j\}\longrightarrow X$$ such that $g(i,j)=f(j-i)$. By Ramsey's Theorem there are $i<j<k$ 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. –  gamel Nov 3 '12 at 5:18
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