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Let be a lacunary sequence (so there exists some with for all ).
Is it true that there must exist a finite colouring of with no monochromatic solutions to ?
Source: erdosproblems.com/894
An accepted solution exists. The statement is true.
Proved. Theorem 1.1 of Peres and Schlag [PeSc10], cited from the arXiv version and published in the Bulletin of the London Mathematical Society in 2010 (refereed; page references follow the preprint), gives, for every lacunary with ratio at least , , a with and hence, by Katznelson's reduction (the proof of his Theorem 1.1, [Ka01] p. 212; restated in the same paper), a proper coloring of the graph with at most colors; the site's bound is this. The same paper gives a self-contained elementary coloring with colors, , on its p. 3, and shows that the power of in the bound cannot be improved. The earlier solutions of the Diophantine question behind the reduction (de Mathan and Pollington, 1979--80) are the site's account of Problem 464; Pollington's theorem gives a second route to finiteness through the reduction, without an explicit dependence on . Katznelson's own Theorem 1.1 (p. 211), "If is lacunary then ", is the original answer, proved from his Theorem 1.2 by the reduction and again, elementarily, with colors where (p. 212). The frontmatter standing is derived from the claim pages: Peres and Schlag's theorem (claim page (Peres and Schlag, 2007)) and Katznelson's (claim page (Katznelson, 2001)) are accepted full claims on their refereed publications and on the site's commentary, which credits both papers.