Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove or disprove:

For all functions $f:\Bbb Z^+ \to \{0,1\}$ , the following is true: There exists positive integers n and d such that: $$f(n)=f(n+d)=f(n+2d)=f(n+3d)$$

share|cite|improve this question
Is it homework ? – Lierre May 4 '12 at 16:31
No, it is not. I can show that their exists n and d such that f(n)=f(n+d)=f(n+2d). I wanted to know if the statement that I posted originally is true or not. – Amr May 4 '12 at 16:41
I would say yes, but so far, I have no proof. I think its even true for arbitrary length. (T. Tao would know of course ;)) – Lierre May 4 '12 at 16:43
I think it is true too. Moreover I think that that for all positive integers k there exists n and d such that f(n)=f(n+d)=f(n+2d)=f(n+3d)=......=f(n+kd) – Amr May 4 '12 at 16:46

1 Answer 1

The Szemerédi’s theorem applies here. It says that for all $0 < d < 1$, and $k$ positive integer, there exists $N$, such that every subset of $\{1, \dotsc, N \}$ with cardinal at least $dN$ contains an arithmetic progression with length $k$.

Take $d = 1/2$ and $k = 4$, and let $N = N(1/2, 4)$. Either $[1, N]\cap f^{-1}(1)$ as cardinal at least $N/2$, or $[1, N] \cap f^{-1}(0)$ has. So there is a arithmetic progression of length 4 in one of them. Qed.


It appears that this easy corollary of the Szemerédi’s theorem is in fact a special case (2 colors) of the easier van der Waerden's theorem (1927).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.