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Oct 6, 2016 at 17:24 answer added js21 timeline score: 3
Oct 5, 2016 at 21:04 comment added Peter Mueller For $n=2$ and $n=3$ the optimal values are $12$ and $42$. For $n=4$ the naive program gets stuck, so far it only shows that the optimal value is $\ge144$. (And of course, a trivial upper bound is $4\cdot42=168$.)
Oct 5, 2016 at 17:25 comment added Seva @js21: it might be even possible to improve this slightly by finding a particular example for some small $n$ and then blowing it up. What would be of interest in this problem is to understand whether there is an upper bound of the form $c^n$ with $c<4$.
Oct 5, 2016 at 10:35 comment added js21 The probabilistic method yields squarefree sets of size $\asymp c^n$ with $c = 2^{\frac{5}{3}} \simeq 3,1748.$
Oct 3, 2016 at 19:57 comment added Victor Protsak You are right! Objection withdrawn.
Oct 3, 2016 at 19:54 comment added Seva @Victor Protsak: $|V|=2^{2n}$. Besides, using a different argument, I can related square-free sets to line-free sets in $(\mathbb Z/4\mathbb Z)^n$, which also gives the lower bound of $3^n$.
Oct 3, 2016 at 19:48 comment added Victor Protsak There are some issues with this argument. If $n$ is even, $|V|=2^{n+1}$ is not a power of 4. Thus $V$ cannot have a structure of vector space over $\Bbb{F}_4$.
Oct 3, 2016 at 8:26 comment added Ilya Bogdanov Set $V=\mathbb F_2^n\oplus \mathbb F_2^n$. Define $\varphi\colon V\to V$ by $\varphi(x,y)=(y,x+y)$. Then $\varphi^3=1$, and $\mathbb F_2[\varphi]\cong \mathbb F_4$.Thus $V$ gets a structure of $\mathbb F_4$-linear space, and a square becomes an affine line of a special form. In particular, every set containing no lines satisfies your requirements. Thus, at least $3^n$ is achievable...
Oct 3, 2016 at 7:51 history asked Seva CC BY-SA 3.0