Problem 698

Let $y_k = x_k x_{k+1} x_{k+2} x_{k+3}$ (clearly $y_{n-2} = x_{n-2}x_{n-1}x_nx_1$, $y_{n-1} = x_{n-1}x_nx_1x_2$, $y_n = x_nx_1x_2x_3$). All $y_k$ are $1$ or $-1$. Then, by the conditions of the problem, we get $y_1 + ... + y_n = 0$. Therefore $n = 2k$ and moreover exactly $k$ of the numbers $y_1, ..., y_n$ are equal to $1$ and the remaining $k$ are equal to $-1$. But then $y_1 \cdot ... \cdot y_n = (-1)^k$. On the other hand, notice that $y_1 \cdot ... \cdot y_n = x_1^4 \cdot x_2^4 \cdot ... \cdot x_n^4$, from where we get that $y_1 \cdot ... \cdot y_n = 1$, hence $k = 2t$, i.e. $n = 4t$.