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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
Post: #1
\( (x_1 x_2+x_2 x_3+\cdots + x_n x_1 = 0)\wedge (\forall x_k \in \left\{ 1,-1 \right\} ) \Rightarrow (4 \mid n)\)
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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