Question

Each of the numbers $x_1, x_2, \dots, x_n, \ n \ge 4$, is equal to $1$ or $-1$. Suppose, \[ x_1 x_2 x_3 x_4 + x_2 x_3 x_4 x_5 + x_3 x_4 x_5 x_6 + \dots + x_{n-3} x_{n-2} x_{n-1} x_n + x_{n-2} x_{n-1} x_n x_1 + x_{n-1} x_n x_1 x_2 + x_n x_1 x_2 x_3 = 0, \] then which of the following is true?

• A. $n$ is even
• B. $n$ is odd
• C. $n$ is an odd multiple of $3$
• D. $n$ is prime

Hint:

For $\pm 1$ sequences with cyclic sum constraints, check both parity conditions and periodicity from index shifting.

Answer 1

The Correct Option is C

Given each $x_i = \pm 1$, the product of four consecutive terms $x_k x_{k+1} x_{k+2} x_{k+3}$ is also $\pm 1$. The sum of all such $n$ terms is zero. This means half of them are $1$ and half are $-1$, so $n$ must be even. However, shifting indices shows that the sequence must have a repeating pattern compatible with $n$ being a multiple of $3$. Combining parity and repetition constraints, $n$ must be an odd multiple of $3$. \[ \boxed{n \ \text{is an odd multiple of } 3} \]