Chain of Integers

Certainly it's easy to see that $x=0$ is possible, by setting $x_1=1$ and $x_2=-1$ , so that the overall product is zero.

To see why this is the only possibility, assume that the product $x\cdot x_1 \cdots x_9$ is non-zero. Then none of the bracketed terms can be zero, and so none of the $x_i$ can be $\pm 1$ .

So $|x_i| \geq 2$ for all $i$ .

Write $P=(1+x_1)\cdots (1+x_9)$ and $M=(1-x_1)\cdots (1-x_9)$ .

Now, $\frac{1+x_i}{1-x_i}<0$ for all $i$ , so $\frac{P}{M}<0$ (we have the product of an odd number of negative numbers). But this directly contradicts the given information that $P=M$ ; hence the only possible answer is indeed $\boxed0$ .