1. ## multiple of 4

Each of the numbers

$a_1,a_2,...,a_n$

is either 1 or -1.

If the sum

$S =a_1a_2a_3a_4+ a_2a_3a_4a_5+ a_3a_4a_5a_6 +.... + a_na_1a_2a_3= 0$,

prove that $n$

must be a multiple of $4$.

2. Originally Posted by perash
Each of the numbers
$a_1,a_2,...,a_n$
is either 1 or -1.
If the sum
$S =a_1a_2a_3a_4+ a_2a_3a_4a_5+ a_3a_4a_5a_6 +.... + a_na_1a_2a_3= 0$,
prove that $n$
must be a multiple of $4$.
If each of the numbers $a_1,a_2,...,a_n$ is either a 1 or a -1, then each of the products in the sum above must be 1 or -1. In order for the entire sum to be 0,
(1) the number of these products added must be even, meaning that n must be even, and
(2) half of these products (n/2) equal 1 and half of these products (n/2) equal -1.

Now, instead of taking the sum of these products, let's take the product of these products. This product would be
$P = (a_1a_2a_3a_4)(a_2a_3a_4a_5)(a_3a_4a_5a_6)....(a_n a_1a_2a_3) = (1)^{n/2}(-1)^{n/2} = (-1)^{n/2}$.

On the other hand, each of $a_1,a_2,...,a_n$ appears 4 times, so $(-1)^{n/2} = 1$. This means that n/2 must be even, which means that n must be a multiple of 4.

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