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 $4$ .

#NumberTheory #CosinesGroup #Goldbach'sConjurersGroup #TorqueGroup #IntroduceYourself

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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.

Xuming Liang - 7 years, 2 months ago

Nice!

Eddie The Head - 7 years, 2 months ago

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.

Bogdan Simeonov - 7 years, 2 months ago

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 .......

Eddie The Head - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$