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# Cool relationship regarding divisiblity

Each of the numbers $$x_1,x_2,....,x_n$$ equal $$1$$ or $$-1$$ and $x_1x_2x_3x_4 + x_2x_3x_4x_5 + x_3x_4x_5x_6+...+x_{n-1}x_nx_1x_2 + x_nx_1x_2x_3 = 0$

Prove that $$n$$ is divisible by $$4$$.

Note by Eddie The Head
2 years, 9 months ago

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In this proof the number of $$-1$$ in an expression means the number of variables $$x_i$$ that are equal to -1. Since each variable $$x_i$$ appears four times in the expression, the number of $$-1$$ on the LHS is a multiple of 4 or even($). Since each term is either 1 or -1 and they sum to 0, half of them must be -1(this implies n is even) and the other half 1. A term is equal to -1(1) $$\iff$$ it has an odd(even) number of $$-1$$, therefore there must be an even number of terms with an odd number of $$-1$$ to make statement ($) true, this means $$\frac {n}{2}=$$even so n is a multiple of 4. · 2 years, 9 months ago

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Nice! · 2 years, 9 months ago

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If we multiply them all together, we will get 1.So the number if $$-1$$'s of the form $$x_i.x_j.x_q.x_r$$ is even.But since the sum of all the terms is 0, then the $$-1$$'s will be half of all of them, which are n in total.Thus n is divisible by 4. · 2 years, 9 months ago

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Your idea is correct but you should explain it a bit more in my opinion because anyone who is not used to this type of problem may find it hard to decipher the solution ....... · 2 years, 9 months ago

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