$x_1 x_2 x_3 x_4 + x_2 x_3 x_4 x_5 +......+ x_n x_1 x_2 x_3 = 0$ 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.
 A: If we change $x_i$ to $-x_i$ the sum $S=x_1x_2x_3x_4+x_2x_3x_4x_5+\cdots+x_nx_1x_2x_3$ is invarient $\pmod{4}$, as four cyclically adjacent terms changes their sign. And if two of the terms are positive and two negatives, nothing changes. If one or three have the same sign, $S$ changes by $\pm 4$, finally if all four are of the same sign, $S$ changes by $\pm 8$. 
Now, initially we have $S=0$, means $S\equiv 0\pmod{4}$. Step by step, if we change the negative sign to positive, this does not change $S\pmod{4}$. And in the end, we will have $S\equiv 0\pmod{4}$. 
Also note that when we have the case where all $x_i=1$, we have $S=n$. Hence, $n\equiv 0\pmod{4}$. So, option $1$ is correct. 
A: There are restrictions on $x_i$ as stated in the question - $x_i = \pm 1$ are the only values they can take. So, any term $x_ix_jx_kx_l$ can only take the values $\pm 1$.
Now, there are n terms in the series which equals $0$. So, there must be an equal number of terms equalling $1$ and $-1$. Hence, half the terms = $1$ and half $-1$ i.e. $n$ has to be even.
A: 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.
A: The first part of the problem is:

"Each of the numbers $x_1,\ \cdots, x_n$ is equal to 1 or -1".

Thus, the restriction on the values of $x_i$ is that each and every one must be one of either $1$ or $-1$, and nothing else. They don't all have to be the same number, though they could be, but the only numbers allowed for each one individually are either $1$ or $-1$. So, if we had, say, $n = 5$, then
$$(1)(-1)(1)(1)(1)$$
would be an okay product, but
$$(1)(2)(1)(1)(1)$$
would not be, since now one of the $x_i$s (which one depends on where in the "cycle" we are imagining this product as representing) is $2$, and $2$ is not a value we are allowed to use.
Thus, you can then ask:


*

*Given that, what are the only possible values for any products formed from such $x_i$?

*After you find that, what must be true about the values of the specific products given, so that their sum is 0?


This should imply something about $n$, as for only certain $n$ values will you be able to make (2) work out with the constraint of allowed values by (1).
