$x_i=\prod_{j \neq i}x_j$ for $i=1,2,..,6$

Question

Find all real $x_i^s$ satisfying the system of equations $x_i=\prod_{j \neq i}x_j$ for all $i=1,2,..,6$ It is obvious that $(x_1,x_2,..x_6)=(0,0,..0),(1,1,...,1),(-1,-1,-1,...-1)$ are obvious solutions and so are $(1,1,-1,-1,-1,-1),(-1,-1,1,1,1,1)$

Actually we get the criteria $x_1^2=x_2^2=...x_6^2$ So, it seems it has finitely many solutions but the answer key claims it has infinitely many solutions. Where am I going wrong?

Answer 1

Define $P:=\prod_j x_j=x_i^2$ so $P^6=\prod_i x_i^2=P^2$ and $P^2(P^4-1)=0$, which allows finitely many values of $P$. Each gives at most two values for $x_i$, so I agree there are only finitely many choices for the $x_i$ with $1\le i\le 6$. The only sense in which there are infinitely many $x_i$ is if we have an infinite sequence, with no constraints on $x_i$ if $i\ge 7$. (However, I do think an answer key error is more likely here.)