Corners theorem: Difference between revisions

From Polymath Wiki
Jump to navigationJump to search
No edit summary
No edit summary
Line 3: Line 3:
'''Corners theorem''': (<math>({\Bbb Z}/3{\Bbb Z})^n</math> version) If n is sufficiently large depending on <math>\delta</math>, then any subset A of <math>{}(({\Bbb Z}/3{\Bbb Z})^n)^2</math> must contain a "corner" (x,y), (x+r,y), (x,y+r) with <math>r \neq 0</math>.
'''Corners theorem''': (<math>({\Bbb Z}/3{\Bbb Z})^n</math> version) If n is sufficiently large depending on <math>\delta</math>, then any subset A of <math>{}(({\Bbb Z}/3{\Bbb Z})^n)^2</math> must contain a "corner" (x,y), (x+r,y), (x,y+r) with <math>r \neq 0</math>.


This result was first proven by [[Ajtai-Szemerédi's proof of the corners theorem|Ajtai and Szemerédi]].  A simpler proof, based on the [[triangle removal lemma]], was obtained by Solymosi.  The corners theorem implies Roth's theorem and is in turn implied by DHJ(3).
This result was first proven by [[Ajtai-Szemerédi's proof of the corners theorem|Ajtai and Szemerédi]].  A simpler proof, based on the [[triangle removal lemma]], was obtained by Solymosi.  The corners theorem implies [[Roth's theorem]] and is in turn implied by [[DHJ(3)]].

Revision as of 09:12, 14 February 2009

Corners theorem: ([math]\displaystyle{ {\Bbb Z}/N{\Bbb Z} }[/math] version) If N is sufficiently large depending on [math]\displaystyle{ \delta }[/math], then any subset A of [math]\displaystyle{ {}[N]^2 }[/math] must contain a "corner" (x,y), (x+r,y), (x,y+r) with [math]\displaystyle{ r \gt 0 }[/math].

Corners theorem: ([math]\displaystyle{ ({\Bbb Z}/3{\Bbb Z})^n }[/math] version) If n is sufficiently large depending on [math]\displaystyle{ \delta }[/math], then any subset A of [math]\displaystyle{ {}(({\Bbb Z}/3{\Bbb Z})^n)^2 }[/math] must contain a "corner" (x,y), (x+r,y), (x,y+r) with [math]\displaystyle{ r \neq 0 }[/math].

This result was first proven by Ajtai and Szemerédi. A simpler proof, based on the triangle removal lemma, was obtained by Solymosi. The corners theorem implies Roth's theorem and is in turn implied by DHJ(3).