Let ${\cal C}=\{0,1\}^n$ be the vertices of the unit cube. The polynomial $$ (1-x_1) x_1 x_2 \cdots x_n$$ has degree $n+1$ and vanishes at every point $(x_1, \ldots, x_n) \in {\cal C}$. Does there exist a polynomial of degree $n$ with this property?
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1Why not just $(1-x_1)x_1$ ? – Ted Apr 21 '13 at 21:11
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Oops, yeah, that works – jean Apr 21 '13 at 21:11
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1For $n \ge 4$, there is actually a degree 4 polynomial $\sum_{k=1}^{n} x_k^2 (1-x_k)^2$ which vanishes at and only at $\cal C$... – achille hui Apr 21 '13 at 21:43
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Hint: If $(x_1,\dots,x_n)\in\mathcal{C}$, then $x_1=0$ or $x_1=1$. What factors are sufficient to make the polynomial vanish on $\mathcal{C}$, then? Any polynomial of which those are a factor will do the trick.
Cameron Buie
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