Let $a_1, a_2, a_3,....a_n$be $n$ numbers such that $a_i$ is either $+1$ or $-1$. If $a_1a_2a_3a_4 + a_2a_3a_4a_5 +...+a_na_1a_2a_3=0$, then prove that $4$ divides $n$.
Well $2$ definitely divides $n$ because for every $+1$ there should be a $-1$. I also thought it would be helpful to the $a_4$ common from the first four terms and $a_8$ from the next four terms and so on and so forth. as I have already proved that $n$ is even I must try and prove that it is not possible to have exactly two terms left after performing the above operation. But, doesn't help...
Let $$\mathbb S=\left\{a_1a_2a_3a_4, a_2a_3a_4a_5 ,...,a_na_1a_2a_3\right\}$$
Note that the sum of elements of $\mathbb S$ is $0$.
Assume that $n \equiv 2 \pmod 4$. Since half of the $n$ elements in $s$ must be $-1$ this implies that a odd number of $-1$ in $S$.
But note $$-1=(-1)^{\frac{n}{2}}=\prod _{ x\in \mathbb S }^{ }{ x }=\prod _{ i=1 }^{ n }{ a_1^4 } $$
A contradiction!