$x_1 x_2 x_3 x_4 + x_2 x_3 x_4 x_5 +......+ x_n x_1 x_2 x_3 = 0$ then what is $n$? Can anyone please help me to understand what is the following problem saying?[!

Each of the numbers $x_1,x_2,\cdots,x_n,n>4$, is equal to $1$ or $-1$. Suppose 
  $$x_1x_2x_3x_4+x_2x_3x_4x_5+\cdots +x_nx_1x_2x_3=0$$
  then, 
  $(1)$ $n$ is even, $(2)$ $n$ is odd, $(3)$ $n$ is odd multiple of $3$, $(4)$ $n$ is prime.

I am really having a hard time to understand this problem. What I understood is I have to find a natural number $n$ for which the equation holds whatever the values of $x_i$ s are. That is impossible. That's why it seems that some conditions on $x_i$ s are necessary.

Let  $$I_1 = x_{1}x_{2}x_{3}x_4$$ $$I_2 = x_{2}x_{3}x_4x_5$$ $$\vdots $$ $$I_n = x_nx_{1}x_{2}x_{3}$$ Since each $I_k\in\{-1,1\}$ and $I_1+I_2+...+I_n=0$ we must have equaly $-1$ and $1$ so $n$ must be even.
