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Feb 27, 2020 at 7:12 comment added Fedor Petrov ... and the same trick works for any $k$
Feb 27, 2020 at 7:00 comment added Fedor Petrov moreover, $n+1$ do not suffice for $k=2$ and any set $A$ of size $n+1$. Consider the polynomial $x_1^2+\ldots+x_n^2+(x_1+\ldots+x_n-s)^2-t$, where $s=\sum_{a\in A} s$, $t=\sum_{a\in A} a^2$.
Feb 22, 2020 at 11:56 comment added Louis Deaett Oh, nice. Now I'm not sure whether I suspect the bound can be met in general.
Feb 22, 2020 at 6:30 history answered Aaron Meyerowitz CC BY-SA 4.0