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 $$(x_1 x_2+x_2 x_3+\cdots + x_n x_1 = 0)\wedge (\forall x_k \in \left\{ 1,-1 \right\} ) \Rightarrow (4 \mid n)$$
05-07-2010, 05:46 PM (This post was last modified: 05-11-2010 10:03 PM by elim.)
Post: #1
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
$$(x_1 x_2+x_2 x_3+\cdots + x_n x_1 = 0)\wedge (\forall x_k \in \left\{ 1,-1 \right\} ) \Rightarrow (4 \mid n)$$
Proof Let $$k$$ be the count of $$-1$$ terms in the sum $$x_1 x_2+x_2 x_3+\cdots + x_n x_1$$. Since the sum equals zero, $$n = 2k$$
Moreover, we then have $$(-1)^k = (x_1 x_2)(x_2 x_3)\cdots(x_n x_1) = (x_1x_2\cdots x_n)^2 =1$$ ans so $$k$$ is even. $$\square$$
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