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May 8, 2015 at 20:19 history edited Fedor Petrov CC BY-SA 3.0
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Oct 24, 2010 at 22:18 comment added Fedor Petrov In a sense, they are very similar. We assume a contrary and reformulate it as a system of algebraic equalities holding for free (independent) variables chosen from appropriate set. In case of Olson we know that all vectots $(\sum x_ia_i,\sum x_i b_i)$ belong to $\mathbb{F}_p^2\setminus (0,0)$ for $x_i\in \{0,1\}$. In case of Cauchy-Davenport we know that $x+y\in C$ for some $C$ of cardinality $|A|+|B|-2$ and $(x,y)\in A\times B$. Then we find one polynomial which respects all these conditions. It is a product of linear forms. Then we apply CN.
Oct 24, 2010 at 16:21 comment added gowers As an exercise, I tried doing the Cauchy-Davenport theorem without looking up how it is done. At first I couldn't see what to do at all, but eventually, after some false starts, I realized that the polynomial $\prod_{c\in A+B}(x+y-c)$ and the sets A and B would contradict the combinatorial nullstellensatz if Cauchy-Davenport was false. Somehow this application feels very different from the Olson theorem application.
Oct 24, 2010 at 10:23 history answered Fedor Petrov CC BY-SA 2.5