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Sep 15, 2016 at 9:52 comment added js21 Minor remark : the Croot-Lev-Pach-Ellenberg-Gijswijt method, as in dl.dropboxusercontent.com/u/15433464/f3_eng.pdf, gives a worse bound, namely $c = 4 \times 3^{-\frac{3}{4}} \simeq 1,75.$
Sep 15, 2016 at 9:25 vote accept Seva
Sep 15, 2016 at 9:22 comment added Ilya Bogdanov I've added an update with an example.
Sep 15, 2016 at 9:21 history edited Ilya Bogdanov CC BY-SA 3.0
Example added.
Sep 15, 2016 at 8:56 comment added Seva Good! Still, I wonder what are the best bounds known. The immediate probabilistic bound: consider a random set $A$ to which the elements of $\mathbb F_2^n$ are chosen independently with probability $p$, then the expected number of $2$-flats in $A$ is about $2^{3n}p^4$, while the expected size of $A$ is $2^np$; hence, for $p\approx 0.1\cdot 2^{-2n/3}$ we can destroy all $2$-flats removing not too many elements of $A$, and we are still left with, roughly, $2^np$ elements. That is, we get $c=2^{1/3}$ this way.
Sep 15, 2016 at 8:43 history answered Ilya Bogdanov CC BY-SA 3.0